{"id":317,"date":"2021-07-14T15:59:13","date_gmt":"2021-07-14T15:59:13","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/chapter\/facts-about-the-f-distribution\/"},"modified":"2023-12-05T09:52:04","modified_gmt":"2023-12-05T09:52:04","slug":"facts-about-the-f-distribution","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/chapter\/facts-about-the-f-distribution\/","title":{"raw":"Hypothesis Test: One-Way ANOVA","rendered":"Hypothesis Test: One-Way ANOVA"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<section>\r\n<ul id=\"fs-idp124304720\">\r\n \t<li>Conduct a one-way ANOVA and interpret the conclusion in context<\/li>\r\n<\/ul>\r\n<\/section><\/div>\r\n<strong>Here are some facts about the <em>F<\/em> distribution.<\/strong>\r\n<ol>\r\n \t<li>The curve is not symmetrical but skewed to the right.<\/li>\r\n \t<li>There is a different curve for each set of <em>df<\/em>s.<\/li>\r\n \t<li>The <em>F<\/em> statistic is greater than or equal to zero.<\/li>\r\n \t<li>As the degrees of freedom for the numerator and for the denominator get larger, the curve approximates the normal.<\/li>\r\n \t<li>Other uses for the <em>F<\/em> distribution include comparing two variances and two-way Analysis of Variance. Two-Way Analysis is beyond the scope of this chapter.<\/li>\r\n<\/ol>\r\n<img class=\"aligncenter wp-image-2328 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5668\/2021\/07\/11175124\/e70fc8d6ba90cb5d770e45ff88dc71543e211803.jpeg\" alt=\"This graph has an unmarked Y axis and then an X axis that ranges from 0.00 to 4.00. It has three plot lines. The plot line labelled F subscript 1, 5 starts near the top of the Y axis at the extreme left of the graph and drops quickly to near the bottom at 0.50, at which point is slowly decreases in a curved fashion to the 4.00 mark on the X axis. The plot line labelled F subscript 100, 100 remains at Y = 0 for much of its length, except for a distinct peak between 0.50 and 1.50. The peak is a smooth curve that reaches about half way up the Y axis at its peak. The plot line labeled F subscript 5, 10 increases slightly as it progresses from 0.00 to 0.50, after which it peaks and slowly decreases down the remainder of the X axis. The peak only reaches about one fifth up the height of the Y axis.\" width=\"605\" height=\"348\" \/>\r\n<div class=\"textbox exercises\">\r\n<h3>Example 1<\/h3>\r\nLet\u2019s return to the slicing tomato exercise in the <a href=\"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/chapter\/the-f-distribution-and-the-f-ratio\/\" target=\"_blank\" rel=\"noopener\">last section (try it 1)<\/a>. The means of the tomato yields under the five mulching conditions are represented by\u00a0<em data-effect=\"italics\">\u03bc<\/em><sub>1<\/sub>,\u00a0<em data-effect=\"italics\">\u03bc<\/em><sub>2<\/sub>,\u00a0<em data-effect=\"italics\">\u03bc<\/em><sub>3<\/sub>,\u00a0<em data-effect=\"italics\">\u03bc<\/em><sub>4<\/sub>,\u00a0<em data-effect=\"italics\">\u03bc<\/em><sub>5<\/sub>. We will conduct a hypothesis test to determine if all means are the same or at least one is different. Using a significance level of 5%, test the null hypothesis that there is no difference in mean yields among the five groups against the alternative hypothesis that at least one mean is different from the rest.\r\n[reveal-answer q=\"130941\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"130941\"]\r\n<p id=\"eip-573\" class=\" \">The null and alternative hypotheses are:<\/p>\r\n<p id=\"eip-idp93603520\" class=\" \"><em data-effect=\"italics\">H<sub>0<\/sub><\/em>:\u00a0<em data-effect=\"italics\">\u03bc<sub>1<\/sub><\/em>\u00a0=\u00a0<em data-effect=\"italics\">\u03bc<sub>2<\/sub><\/em>\u00a0=\u00a0<em data-effect=\"italics\">\u03bc<sub>3<\/sub><\/em>\u00a0=\u00a0<em data-effect=\"italics\">\u03bc<sub>4<\/sub><\/em>\u00a0=\u00a0<em data-effect=\"italics\">\u03bc<sub>5<\/sub><\/em><\/p>\r\n<p id=\"eip-idp134625200\" class=\" \"><em data-effect=\"italics\">H<sub>a<\/sub>: \u03bc<sub>i<\/sub>\u00a0\u2260 \u03bc<sub>j<\/sub><\/em>\u00a0some\u00a0<em data-effect=\"italics\">i \u2260 j<\/em><\/p>\r\n<p id=\"eip-idm22822112\" class=\" \">The one-way ANOVA results are shown below.<\/p>\r\n\r\n<table style=\"border-collapse: collapse; width: 100%; height: 48px;\" border=\"1\">\r\n<tbody>\r\n<tr style=\"height: 12px;\">\r\n<td style=\"width: 20%; height: 12px;\"><strong>Source of Variation<\/strong><\/td>\r\n<td style=\"width: 20%; height: 12px;\"><strong>Sum of Squares (<em>SS<\/em>)<\/strong><\/td>\r\n<td style=\"width: 20%; height: 12px;\"><strong>Degrees of Freedom (<em>df<\/em>)<\/strong><\/td>\r\n<td style=\"width: 20%; height: 12px;\"><strong>Mean Square (<em>MS<\/em>)<\/strong><\/td>\r\n<td style=\"width: 20%; height: 12px;\"><strong><em>F<\/em><\/strong><\/td>\r\n<\/tr>\r\n<tr style=\"height: 12px;\">\r\n<td style=\"width: 20%; height: 12px;\">Factor (Between)<\/td>\r\n<td style=\"width: 20%;\">36,648,561<\/td>\r\n<td style=\"width: 20%;\">5 \u2013 1 = 4<\/td>\r\n<td style=\"width: 20%;\">[latex]\\displaystyle\\frac{{{36},{648},{561}}}{{4}}={9},{162},{140}[\/latex]<\/td>\r\n<td style=\"width: 20%;\">[latex]\\displaystyle\\frac{{{9},{162},{140}}}{{{2},{044},{672.6}}}={4.4810}[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 12px;\">\r\n<td style=\"width: 20%; height: 12px;\">Error (Within)<\/td>\r\n<td style=\"width: 20%;\">20,446,726<\/td>\r\n<td style=\"width: 20%;\">15 \u2013 5 = 10<\/td>\r\n<td style=\"width: 20%;\">[latex]\\displaystyle\\frac{{{20},{446},{726}}}{{10}}={2},{044},{672.6}[\/latex]<\/td>\r\n<td style=\"width: 20%; height: 12px;\"><\/td>\r\n<\/tr>\r\n<tr style=\"height: 12px;\">\r\n<td style=\"width: 20%; height: 12px;\">Total<\/td>\r\n<td style=\"width: 20%;\">57,095,287<\/td>\r\n<td style=\"width: 20%;\">15 \u2013 1 = 14<\/td>\r\n<td style=\"width: 20%;\"><\/td>\r\n<td style=\"width: 20%; height: 12px;\"><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p id=\"eip-idm38415936\" class=\" \"><strong>Distribution for the test:\u00a0<em data-effect=\"italics\">F<\/em><sub>4,10<\/sub><\/strong><\/p>\r\n<p id=\"eip-idm45807632\" class=\" \"><em data-effect=\"italics\">df<\/em>(<em data-effect=\"italics\">num<\/em>) = 5 \u2013 1 = 4<\/p>\r\n<p id=\"eip-idm2595504\" class=\" \"><em data-effect=\"italics\">df<\/em>(<em data-effect=\"italics\">denom<\/em>) = 15 \u2013 5 = 10<\/p>\r\n<p id=\"eip-idm4421632\" class=\" \"><strong>Test statistic:<\/strong>\u00a0<em data-effect=\"italics\">F<\/em>\u00a0= 4.4810<\/p>\r\n<strong>Graph:<\/strong>\r\n\r\n<img class=\"aligncenter wp-image-2330 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5668\/2021\/07\/11175212\/47944008d54b3438f561c515dc26b48c1878de18.jpeg\" alt=\"This graph shows a nonsymmetrical F distribution curve. The horizontal axis extends from 0 - 5, and the vertical axis ranges from 0 - 0.7. The curve is strongly skewed to the right.\" width=\"488\" height=\"325\" \/>\r\n<p id=\"eip-idm4421376\" class=\" \"><strong>Probability Statement:<\/strong>\u00a0<em data-effect=\"italics\">p<\/em>-value =\u00a0<em data-effect=\"italics\">P<\/em>(<em data-effect=\"italics\">F<\/em>\u00a0&gt; 4.481) = 0.0248.<\/p>\r\n<p id=\"eip-idm4421120\" class=\" \"><strong>Compare\u00a0<em data-effect=\"italics\">\u03b1<\/em>\u00a0and the\u00a0<em data-effect=\"italics\">p<\/em>-value:<\/strong>\u00a0<em data-effect=\"italics\">\u03b1<\/em>\u00a0= 0.05,\u00a0<em data-effect=\"italics\">p<\/em>-value = 0.0248<\/p>\r\n<p id=\"eip-idm141674128\" class=\" \"><strong>Make a decision:<\/strong>\u00a0Since\u00a0<em data-effect=\"italics\">\u03b1<\/em>\u00a0&gt;\u00a0<em data-effect=\"italics\">p<\/em>-value, we reject\u00a0<em data-effect=\"italics\">H<sub>0<\/sub><\/em>.<\/p>\r\n<p id=\"eip-idm156731472\" class=\" \"><strong>Conclusion:<\/strong>\u00a0At the 5% significance level, we have reasonably strong evidence that differences in mean yields for slicing tomato plants grown under different mulching conditions are unlikely to be due to chance alone. We may conclude that at least some of the mulches led to different mean yields.<\/p>\r\n\r\n<header>\r\n<h2 class=\"os-title\" data-type=\"title\"><span class=\"os-title-label\">USING THE TI-83, 83+, 84, 84+ CALCULATOR<\/span><\/h2>\r\n<\/header>\r\n<ul>\r\n \t<li id=\"eip-idm156731216\" class=\" \">Press STAT. Press 1:EDIT. Put the data into the lists\u00a0<em data-effect=\"italics\">L<sub>1<\/sub><\/em>,\u00a0<em data-effect=\"italics\">L<sub>2<\/sub><\/em>,\u00a0<em data-effect=\"italics\">L<sub>3<\/sub><\/em>,\u00a0<em data-effect=\"italics\">L<sub>4<\/sub><\/em>,\u00a0<em data-effect=\"italics\">L<sub>5<\/sub><\/em>.<\/li>\r\n \t<li class=\" \">Press STAT, arrow over to TESTS, and arrow down to ANOVA. Press ENTER, and then enter\u00a0<em style=\"font-size: 1rem; orphans: 1; text-align: initial;\" data-effect=\"italics\">L<sub>1<\/sub><\/em><span style=\"font-size: 1rem; orphans: 1; text-align: initial;\">,\u00a0<\/span><em style=\"font-size: 1rem; orphans: 1; text-align: initial;\" data-effect=\"italics\">L<sub>2<\/sub><\/em><span style=\"font-size: 1rem; orphans: 1; text-align: initial;\">,\u00a0<\/span><em style=\"font-size: 1rem; orphans: 1; text-align: initial;\" data-effect=\"italics\">L<sub>3<\/sub><\/em><span style=\"font-size: 1rem; orphans: 1; text-align: initial;\">,\u00a0<\/span><em style=\"font-size: 1rem; orphans: 1; text-align: initial;\" data-effect=\"italics\">L<sub>4<\/sub><\/em><span style=\"font-size: 1rem; orphans: 1; text-align: initial;\">,\u00a0<\/span><em style=\"font-size: 1rem; orphans: 1; text-align: initial;\" data-effect=\"italics\">L<sub>5<\/sub><\/em><span style=\"font-size: 1rem; orphans: 1; text-align: initial;\">).<\/span><\/li>\r\n \t<li class=\" \">Press ENTER. You will see that the values in the foregoing ANOVA table are easily produced by the calculator, including the test statistic and the\u00a0<em>p<\/em>-value of the test.<\/li>\r\n \t<li class=\" \">The calculator displays:\r\n<ul>\r\n \t<li class=\" \"><em style=\"font-size: 1rem; orphans: 1; text-align: initial;\" data-effect=\"italics\">F<\/em><span style=\"font-size: 1rem; orphans: 1; text-align: initial;\">\u00a0= 4.4810<\/span><\/li>\r\n \t<li class=\" \"><em style=\"font-size: 1rem; orphans: 1; text-align: initial;\" data-effect=\"italics\">p<\/em><span style=\"font-size: 1rem; orphans: 1; text-align: initial;\">\u00a0= 0.0248 (<\/span><em style=\"font-size: 1rem; orphans: 1; text-align: initial;\" data-effect=\"italics\">p<\/em><span style=\"font-size: 1rem; orphans: 1; text-align: initial;\">-value)<\/span><\/li>\r\n \t<li class=\" \">Factor\u00a0<em style=\"font-size: 1rem; orphans: 1; text-align: initial;\" data-effect=\"italics\">df<\/em><span style=\"font-size: 1rem; orphans: 1; text-align: initial;\">\u00a0= 4<\/span><\/li>\r\n \t<li class=\" \"><em style=\"font-size: 1rem; orphans: 1; text-align: initial;\" data-effect=\"italics\">SS<\/em><span style=\"font-size: 1rem; orphans: 1; text-align: initial;\">\u00a0= 36648560.9<\/span><\/li>\r\n \t<li class=\" \"><em style=\"font-size: 1rem; orphans: 1; text-align: initial;\" data-effect=\"italics\">MS<\/em><span style=\"font-size: 1rem; orphans: 1; text-align: initial;\">\u00a0= 9162140.23<\/span><\/li>\r\n \t<li class=\" \">Error\u00a0<em style=\"font-size: 1rem; orphans: 1; text-align: initial;\" data-effect=\"italics\">df<\/em><span style=\"font-size: 1rem; orphans: 1; text-align: initial;\">\u00a0= 10<\/span><\/li>\r\n \t<li class=\" \"><em style=\"font-size: 1rem; orphans: 1; text-align: initial;\" data-effect=\"italics\">SS<\/em><span style=\"font-size: 1rem; orphans: 1; text-align: initial;\">\u00a0= 20446726<\/span><\/li>\r\n \t<li class=\" \"><em style=\"font-size: 1rem; orphans: 1; text-align: initial;\" data-effect=\"italics\">MS<\/em><span style=\"font-size: 1rem; orphans: 1; text-align: initial;\">\u00a0= 2044672.6<\/span><\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ul>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it 1<\/h3>\r\nMRSA, or <em>Staphylococcus aureus<\/em>, can cause serious bacterial infections in hospital patients. This table shows various colony counts from different patients who may or may not have MRSA.\r\n<table>\r\n<thead>\r\n<tr>\r\n<th>Conc = 0.6<\/th>\r\n<th>Conc = 0.8<\/th>\r\n<th>Conc = 1.0<\/th>\r\n<th>Conc = 1.2<\/th>\r\n<th>Conc = 1.4<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>9<\/td>\r\n<td>16<\/td>\r\n<td>22<\/td>\r\n<td>30<\/td>\r\n<td>27<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>66<\/td>\r\n<td>93<\/td>\r\n<td>147<\/td>\r\n<td>199<\/td>\r\n<td>168<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>98<\/td>\r\n<td>82<\/td>\r\n<td>120<\/td>\r\n<td>148<\/td>\r\n<td>132<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nPlot of the data for the different concentrations:\r\n\r\n<img class=\"aligncenter wp-image-2332 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5668\/2021\/07\/11175328\/9d1cdcd57e7f58e7af6badb0855358d780c243bd.png\" alt=\"This graph is a scatterplot for the data provided. The horizontal axis is labeled 'Colony counts' and extends from 0 - 200. The vertical axis is labeled 'Tryptone concentrations' and extends from 0.6 - 1.4.\" width=\"477\" height=\"284\" \/>\r\n\r\nTest whether the mean number of colonies is the same or different. Construct the ANOVA table (by hand or by using a TI-83, 83+, or 84+ calculator), find the <em>p<\/em>-value, and state your conclusion. Use a 5% significance level.\r\n[reveal-answer q=\"66762\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"66762\"]\r\n\r\nWhile there are differences in the spreads between the groups, the differences do not appear to be big enough to cause concern.\r\n\r\nWe test for the equality of the mean number of colonies:\r\n\r\n<em>H<sub data-redactor-tag=\"sub\">0<\/sub><\/em> : <em>\u03bc<\/em><sub>1<\/sub> = <em>\u03bc<\/em><sub>2<\/sub> = <em>\u03bc<\/em><sub>3<\/sub> = <em>\u03bc<\/em><sub>4<\/sub> = <em>\u03bc<\/em><sub>5<\/sub><em>H<sub data-redactor-tag=\"sub\">a<\/sub><\/em>: <em>\u03bc<sup data-redactor-tag=\"sup\">i<\/sup><\/em> \u2260 <em>\u03bc<sup data-redactor-tag=\"sup\">j<\/sup><\/em> some <em>i<\/em> \u2260 <em>j<\/em>\r\n\r\nThe one-way ANOVA table results are shown below.\r\n<table style=\"border-collapse: collapse; width: 99.8837%; height: 84px;\" border=\"1\">\r\n<thead>\r\n<tr style=\"height: 12px;\">\r\n<th style=\"width: 19.9069%; height: 12px;\">Source of Variation<\/th>\r\n<th style=\"width: 20.0233%; height: 12px;\">Sum of Squares (<em>SS<\/em>)<\/th>\r\n<th style=\"width: 1.1655%; height: 12px;\">Degrees of Freedom (<em>df<\/em>)<\/th>\r\n<th style=\"width: 38.7647%; height: 12px;\">Mean Square (<em>MS<\/em>)<\/th>\r\n<th style=\"width: 20.0233%; height: 12px;\"><em>F<\/em><\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr style=\"height: 36px;\">\r\n<td style=\"width: 19.9069%; height: 36px;\">Factor (Between)<\/td>\r\n<td style=\"width: 20.0233%; height: 36px;\">10,233<\/td>\r\n<td style=\"width: 1.1655%; height: 36px;\">5 \u2013 1 = 4<\/td>\r\n<td style=\"width: 38.7647%; height: 36px;\">[latex]\\displaystyle\\frac{{{10},{233}}}{{4}}={2},{558.25}[\/latex]<\/td>\r\n<td style=\"width: 20.0233%; height: 36px;\">[latex]\\displaystyle\\frac{{{2},{558.25}}}{{{4},{194.9}}}={0.6099}[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 12px;\">\r\n<td style=\"width: 19.9069%; height: 12px;\">Error (Within)<\/td>\r\n<td style=\"width: 20.0233%; height: 12px;\">41,949<\/td>\r\n<td style=\"width: 1.1655%; height: 12px;\">15 \u2013 5 = 10<\/td>\r\n<td style=\"width: 38.7647%; height: 12px;\"><\/td>\r\n<td style=\"width: 20.0233%; height: 12px;\"><\/td>\r\n<\/tr>\r\n<tr style=\"height: 24px;\">\r\n<td style=\"width: 19.9069%; height: 24px;\">Total<\/td>\r\n<td style=\"width: 20.0233%; height: 24px;\">52,182<\/td>\r\n<td style=\"width: 1.1655%; height: 24px;\">15 \u2013 1 = 14<\/td>\r\n<td style=\"width: 38.7647%; height: 24px;\">[latex]\\displaystyle\\frac{{{41},{949}}}{{10}}={4},{194.9}[\/latex]<\/td>\r\n<td style=\"width: 20.0233%; height: 24px;\"><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<strong>Graph:<\/strong>\r\n\r\n<img class=\"aligncenter\" style=\"width: 442px; height: 299.055319148936px;\" src=\"https:\/\/textimgs.s3.amazonaws.com\/DE\/stats\/2qg4-x87q657i#fixme#fixme#fixme\" alt=\"This graph shows a nonsymmetrical F distribution curve. The curve is skewed to the right. A vertical upward line extends from 0.6649 to the curve. This line is just to the right of the graph's peak and the region to the right of the line is shaded to represent the p-value.\" \/>\r\n\r\n<strong>Distribution for the test:<\/strong> <em>F<\/em><sub>4,10<\/sub>\r\n\r\n<strong>Probability Statement: <\/strong><em>p<\/em>-value = <em>P<\/em>(<em>F<\/em> &gt; 0.6099) = 0.6649.\r\n\r\n<strong>Compare <em data-redactor-tag=\"em\">\u03b1<\/em> and the <em>p<\/em>-value:<\/strong> <em>\u03b1<\/em> = 0.05, <em>p<\/em>-value = 0.669, <em>\u03b1<\/em>\u00a0&lt;\u00a0<em>p<\/em>-value\r\n\r\n<strong>Make a decision:<\/strong> Since <em>\u03b1<\/em>\u00a0&lt;\u00a0<em>p<\/em>-value, we do not reject <em>H<\/em>0.\r\n\r\n<strong>Conclusion:<\/strong> At the 5% significance level, there is insufficient evidence from these data that different levels of tryptone will cause a significant difference in the mean number of bacterial colonies formed.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example 2<\/h3>\r\nFour sororities took a random sample of sisters regarding their grade means for the past term. The results are shown in the table.\r\n\r\nMean Grades for Four Sororities\r\n<table>\r\n<thead>\r\n<tr>\r\n<th>Sorority 1<\/th>\r\n<th>Sorority 2<\/th>\r\n<th>Sorority 3<\/th>\r\n<th>Sorority 4<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>2.17<\/td>\r\n<td>2.63<\/td>\r\n<td>2.63<\/td>\r\n<td>3.79<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>1.85<\/td>\r\n<td>1.77<\/td>\r\n<td>3.78<\/td>\r\n<td>3.45<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>2.83<\/td>\r\n<td>3.25<\/td>\r\n<td>4.00<\/td>\r\n<td>3.08<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>1.69<\/td>\r\n<td>1.86<\/td>\r\n<td>2.55<\/td>\r\n<td>2.26<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>3.33<\/td>\r\n<td>2.21<\/td>\r\n<td>2.45<\/td>\r\n<td>3.18<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nUsing a significance level of 1%, is there a difference in mean grades among the sororities?\r\n\r\n[reveal-answer q=\"901953\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"901953\"]\r\n\r\nLet <em>\u03bc<sub data-redactor-tag=\"sub\">1<\/sub><\/em>, <em>\u03bc<sub data-redactor-tag=\"sub\">2<\/sub><\/em>, <em>\u03bc<sub data-redactor-tag=\"sub\">3<\/sub><\/em>, <em>\u03bc<sub data-redactor-tag=\"sub\">4<\/sub><\/em> be the population means of the sororities. Remember that the null hypothesis claims that the sorority groups are from the same normal distribution. The alternate hypothesis says that at least two of the sorority groups come from populations with different normal distributions. Notice that the four sample sizes are each five.\r\n<h4>Note<\/h4>\r\nThis is an example of a <strong>balanced design<\/strong>\u00a0because each factor (i.e., sorority) has the same number of observations.\r\n\r\n<em>H<sub data-redactor-tag=\"sub\">0<\/sub><\/em>: <em>\u03bc<sub data-redactor-tag=\"sub\">1<\/sub><\/em> = <em>\u03bc<sub data-redactor-tag=\"sub\">2<\/sub><\/em> = <em>\u03bc<sub data-redactor-tag=\"sub\">3<\/sub><\/em> = <em>\u03bc<sub data-redactor-tag=\"sub\">4<\/sub><\/em>\r\n\r\n<em>H<sub data-redactor-tag=\"sub\">a<\/sub><\/em>: Not all of the means <em>\u03bc<sub data-redactor-tag=\"sub\">1<\/sub><\/em>, <em>\u03bc<sub data-redactor-tag=\"sub\">2<\/sub><\/em>, <em>\u03bc<sub data-redactor-tag=\"sub\">3<\/sub><\/em>, <em>\u03bc<sub data-redactor-tag=\"sub\">4<\/sub><\/em> are equal.\r\n\r\n<strong>Distribution for the test:<\/strong> <em>F<\/em><sub>3,16<\/sub>\r\n\r\nwhere <em>k<\/em> = 4 groups and <em>n<\/em> = 20 samples in total\r\n\r\n<em>df<\/em>(<em>num<\/em>)= <em>k<\/em> \u2013 1 = 4 \u2013 1 = 3\r\n\r\n<em>df<\/em>(<em>denom<\/em>) = <em>n<\/em> \u2013 <em>k<\/em> = 20 \u2013 4 = 16\r\n\r\n<strong>Calculate the test statistic:<\/strong> <em>F<\/em> = 2.23\r\n\r\n<strong>Graph:<\/strong>\r\n\r\n<img class=\"aligncenter wp-image-2334 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5668\/2021\/07\/11175436\/400f26530a962102a5c7068ab6a3f2a2adfb2d17.jpeg\" alt=\"This graph shows a nonsymmetrical F distribution curve with values of 0 and 2.23 on the x-axis representing the test statistic of sorority grade averages. The curve is slightly skewed to the right, but is approximately normal. A vertical upward line extends from 2.23 to the curve and the area to the right of this is shaded to represent the p-value.\" width=\"487\" height=\"275\" \/>\r\n\r\n<strong>Probability statement:<\/strong> <em>p<\/em>-value = <em>P<\/em>(<em>F<\/em> &gt; 2.23) = 0.1241\r\n\r\n<strong>Compare <em data-redactor-tag=\"em\">\u03b1<\/em> and the <em>p<\/em>-value:<\/strong> <em>\u03b1<\/em> = 0.01\r\n\r\n<em>p<\/em>-value = 0.1241\r\n\r\n<em>\u03b1<\/em> &lt; <em>p<\/em>-value\r\n\r\n<strong>Make a decision:<\/strong> Since <em>\u03b1<\/em> &lt; <em>p<\/em>-value, you cannot reject <em>H0<\/em>.\r\n\r\n<strong>Conclusion: <\/strong>There is not sufficient evidence to conclude that there is a difference among the mean grades for the sororities.\r\n<h2 class=\"os-title\" data-type=\"title\"><span class=\"os-title-label\">USING THE TI-83, 83+, 84, 84+ CALCULATOR<\/span><\/h2>\r\n<ul>\r\n \t<li>Put the data into lists L1, L2, L3, and L4. Press <code>STAT<\/code> and arrow over to <code>TESTS<\/code>. Arrow down to <code>F:ANOVA<\/code>. Press <code>ENTER<\/code>and Enter (<code>L1,L2,L3,L4<\/code>).<\/li>\r\n \t<li>The calculator displays the F statistic, the <em style=\"font-size: 1rem; orphans: 1; text-align: initial;\">p<\/em><span style=\"font-size: 1rem; orphans: 1; text-align: initial;\">-value and the values for the one-way ANOVA table:<\/span>\r\n<ul>\r\n \t<li><em style=\"font-size: 1rem; orphans: 1; text-align: initial;\">F<\/em><span style=\"font-size: 1rem; orphans: 1; text-align: initial;\"> = 2.2303<\/span><\/li>\r\n \t<li><em style=\"font-size: 1rem; orphans: 1; text-align: initial;\">p<\/em><span style=\"font-size: 1rem; orphans: 1; text-align: initial;\"> = 0.1241 (<\/span><em style=\"font-size: 1rem; orphans: 1; text-align: initial;\">p<\/em><span style=\"font-size: 1rem; orphans: 1; text-align: initial;\">-value)<\/span><\/li>\r\n \t<li>Factor <em style=\"font-size: 1rem; orphans: 1; text-align: initial;\">df<\/em><span style=\"font-size: 1rem; orphans: 1; text-align: initial;\"> = 3<\/span><\/li>\r\n \t<li><em style=\"font-size: 1rem; orphans: 1; text-align: initial;\">SS<\/em><span style=\"font-size: 1rem; orphans: 1; text-align: initial;\"> = 2.88732<\/span><\/li>\r\n \t<li><em style=\"font-size: 1rem; orphans: 1; text-align: initial;\">MS<\/em><span style=\"font-size: 1rem; orphans: 1; text-align: initial;\"> = 0.96244<\/span><\/li>\r\n \t<li>Error <em style=\"font-size: 1rem; orphans: 1; text-align: initial;\">df<\/em><span style=\"font-size: 1rem; orphans: 1; text-align: initial;\"> = 16<\/span><\/li>\r\n \t<li><em style=\"font-size: 1rem; orphans: 1; text-align: initial;\">SS<\/em><span style=\"font-size: 1rem; orphans: 1; text-align: initial;\"> = 6.9044<\/span><\/li>\r\n \t<li><em style=\"font-size: 1rem; orphans: 1; text-align: initial;\">MS<\/em><span style=\"font-size: 1rem; orphans: 1; text-align: initial;\"> = 0.431525<\/span><\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ul>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it 2<\/h3>\r\nFour sports teams took a random sample of players regarding their GPAs for the last year. The results are shown below:\r\n\r\nGPAs for Four Sports Teams\r\n<table>\r\n<thead>\r\n<tr>\r\n<th>Basketball<\/th>\r\n<th>Baseball<\/th>\r\n<th>Hockey<\/th>\r\n<th>Lacrosse<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>3.6<\/td>\r\n<td>2.1<\/td>\r\n<td>4.0<\/td>\r\n<td>2.0<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>2.9<\/td>\r\n<td>2.6<\/td>\r\n<td>2.0<\/td>\r\n<td>3.6<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>2.5<\/td>\r\n<td>3.9<\/td>\r\n<td>2.6<\/td>\r\n<td>3.9<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>3.3<\/td>\r\n<td>3.1<\/td>\r\n<td>3.2<\/td>\r\n<td>2.7<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>3.8<\/td>\r\n<td>3.4<\/td>\r\n<td>3.2<\/td>\r\n<td>2.5<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nUse a significance level of 5%, and determine if there is a difference in GPA among the teams.\r\n[reveal-answer q=\"964208\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"964208\"]\r\n\r\nWith a <em>p<\/em>-value of 0.9271, we decline to reject the null hypothesis. There is not sufficient evidence to conclude that there is a difference among the GPAs for the sports teams.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example 3<\/h3>\r\nA fourth-grade class is studying the environment. One of the assignments is to grow bean plants in different soils. Tommy chose to grow his bean plants in soil found outside his classroom mixed with dryer lint. Tara chose to grow her bean plants in potting soil bought at the local nursery. Nick chose to grow his bean plants in soil from his mother's garden. No chemicals were used on the plants, only water. They were grown inside the classroom next to a large window. Each child grew five plants. At the end of the growing period, each plant was measured, producing the data (in inches) in this table.\r\n<table>\r\n<thead>\r\n<tr>\r\n<th>Tommy's Plants<\/th>\r\n<th>Tara's Plants<\/th>\r\n<th>Nick's Plants<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>24<\/td>\r\n<td>25<\/td>\r\n<td>23<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>21<\/td>\r\n<td>31<\/td>\r\n<td>27<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>23<\/td>\r\n<td>23<\/td>\r\n<td>22<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>30<\/td>\r\n<td>20<\/td>\r\n<td>30<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>23<\/td>\r\n<td>28<\/td>\r\n<td>20<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nDoes it appear that the three media in which the bean plants were grown produce the same mean height? Test at a 3% level of significance.\r\n\r\n[reveal-answer q=\"127409\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"127409\"]\r\n\r\nThis time, we will perform the calculations that lead to the <em>F'<\/em>statistic. Notice that each group has the same number of plants, so we will use the formula [latex]\\displaystyle{F}'={\\dfrac{n \\cdot {s_{\\overline x}}^{2}}{{s^2}_{pooled}}}[\/latex].\r\n\r\nFirst, calculate the sample mean and sample variance of each group.\r\n<table style=\"width: 428px;\">\r\n<thead>\r\n<tr>\r\n<th style=\"width: 116px;\"><\/th>\r\n<th style=\"width: 117px;\">Tommy's Plants<\/th>\r\n<th style=\"width: 97px;\">Tara's Plants<\/th>\r\n<th style=\"width: 98px;\">Nick's Plants<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td style=\"width: 116px;\">Sample Mean<\/td>\r\n<td style=\"width: 117px;\">24.2<\/td>\r\n<td style=\"width: 97px;\">25.4<\/td>\r\n<td style=\"width: 98px;\">24.4<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 116px;\">Sample Variance<\/td>\r\n<td style=\"width: 117px;\">11.7<\/td>\r\n<td style=\"width: 97px;\">18.3<\/td>\r\n<td style=\"width: 98px;\">16.3<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nNext, calculate the variance of the three group means (Calculate the variance of 24.2, 25.4, and 24.4). <strong>Variance of the group means = 0.413<\/strong> = [latex]{s_{\\overline x}}^{2}[\/latex]\r\n\r\nThen [latex]\\displaystyle{M}{S}_{{\\text{between}}}={n}{s_{\\overline x}}^{2}={({5})}{({0.413})}[\/latex] where [latex]n = {5}[\/latex] is the sample size (number of plants each child grew).\r\n\r\nCalculate the mean of the three sample variances (Calculate the mean of 11.7, 18.3, and 16.3). <strong>Mean of the sample variances = 15.433 = <em>s<sup>2<\/sup>pooled<\/em><\/strong>\r\n\r\nThen [latex]\\displaystyle{M}{S}_{{\\text{within}}}[\/latex] =\u00a0<em>s<sup>2<\/sup><sub>pooled\u00a0<\/sub><\/em>= 15.433.\r\n\r\nThe <em>F<\/em> statistic (or <em>F<\/em> ratio) is [latex]{\\text{F}}[\/latex] = [latex]{\\dfrac{MS_{between}}{MS_{within}}}[\/latex] = [latex]\\dfrac{n{s_{\\overline x}}^2}{{s^2}_{pooled}}[\/latex] = [latex]{\\dfrac{(5)(0.413)}{15.433}}[\/latex] = 0.134\r\n\r\nThe <em>dfs<\/em> for the numerator = the number of groups \u2013 1 = 3 \u2013 1 = 2.\r\n\r\nThe <em>dfs<\/em> for the denominator = the total number of samples \u2013 the number of groups = 15 \u2013 3 = 12\r\n\r\nThe distribution for the test is <em>F<\/em><sub>2,12<\/sub> and the <em>F<\/em> statistic is <em>F<\/em> = 0.134\r\n\r\nThe <em>p<\/em>-value is <em>P<\/em>(<em>F<\/em> &gt; 0.134) = 0.8759.\r\n\r\n<strong>Decision:<\/strong> Since <em>\u03b1<\/em> = 0.03 and the <em>p<\/em>-value = 0.8759, do not reject <em>H<sub>0<\/sub><\/em>. (Why?)\r\n\r\n<strong>Conclusion:<\/strong> With a 3% level of significance, from the sample data, the evidence is not sufficient to conclude that the mean heights of the bean plants are different.\r\n<h2 class=\"os-title\" data-type=\"title\"><span class=\"os-title-label\">USING THE TI-83, 83+, 84, 84+ CALCULATOR<\/span><\/h2>\r\nTo calculate the <em>p<\/em>-value:\r\n<ul>\r\n \t<li>Press <code style=\"line-height: 1.6em;\">2nd DISTR<\/code><\/li>\r\n \t<li>Arrow down to <code style=\"line-height: 1.6em;\">Fcdf<\/code>(and press<code style=\"line-height: 1.6em;\">ENTER<\/code>.<\/li>\r\n \t<li>Enter 0.134, <code style=\"line-height: 1.6em;\">E99<\/code>, 2, 12)<\/li>\r\n \t<li>Press <code style=\"line-height: 1.6em;\">ENTER<\/code><\/li>\r\n \t<li>The <em style=\"font-size: 1rem; orphans: 1; text-align: initial;\">p<\/em><span style=\"font-size: 1rem; orphans: 1; text-align: initial;\">-value is 0.8759.<\/span><\/li>\r\n<\/ul>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it 3<\/h3>\r\nAnother fourth-grader also grew bean plants, but this time in a jelly-like mass. The heights were (in inches) 24, 28, 25, 30, and 32. Do a one-way ANOVA test on the four groups. Are the heights of the bean plants different? Use the same method as shown in Example 3.\r\n[reveal-answer q=\"11368\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"11368\"]\r\n<ul>\r\n \t<li><em>F<\/em> = 0.9496<\/li>\r\n \t<li><em>p<\/em>-value = 0.4402<\/li>\r\n<\/ul>\r\nFrom the sample data, the evidence is not sufficient to conclude that the mean heights of the bean plants are different.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\"><header>\r\n<h3 class=\"title\" data-type=\"title\">Activity<\/h3>\r\n<\/header>From the class, create four groups of the same size as follows: identified as men under 22, identified as men at least 22, identified as women under 22, identified as women at least 22. Have each member of each group record the number of states in the United States he or she has visited. Run an ANOVA test to determine if the average number of states visited in the four groups is the same. Test at a 1% level of significance.\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<section>\n<ul id=\"fs-idp124304720\">\n<li>Conduct a one-way ANOVA and interpret the conclusion in context<\/li>\n<\/ul>\n<\/section>\n<\/div>\n<p><strong>Here are some facts about the <em>F<\/em> distribution.<\/strong><\/p>\n<ol>\n<li>The curve is not symmetrical but skewed to the right.<\/li>\n<li>There is a different curve for each set of <em>df<\/em>s.<\/li>\n<li>The <em>F<\/em> statistic is greater than or equal to zero.<\/li>\n<li>As the degrees of freedom for the numerator and for the denominator get larger, the curve approximates the normal.<\/li>\n<li>Other uses for the <em>F<\/em> distribution include comparing two variances and two-way Analysis of Variance. Two-Way Analysis is beyond the scope of this chapter.<\/li>\n<\/ol>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-2328 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5668\/2021\/07\/11175124\/e70fc8d6ba90cb5d770e45ff88dc71543e211803.jpeg\" alt=\"This graph has an unmarked Y axis and then an X axis that ranges from 0.00 to 4.00. It has three plot lines. The plot line labelled F subscript 1, 5 starts near the top of the Y axis at the extreme left of the graph and drops quickly to near the bottom at 0.50, at which point is slowly decreases in a curved fashion to the 4.00 mark on the X axis. The plot line labelled F subscript 100, 100 remains at Y = 0 for much of its length, except for a distinct peak between 0.50 and 1.50. The peak is a smooth curve that reaches about half way up the Y axis at its peak. The plot line labeled F subscript 5, 10 increases slightly as it progresses from 0.00 to 0.50, after which it peaks and slowly decreases down the remainder of the X axis. The peak only reaches about one fifth up the height of the Y axis.\" width=\"605\" height=\"348\" \/><\/p>\n<div class=\"textbox exercises\">\n<h3>Example 1<\/h3>\n<p>Let\u2019s return to the slicing tomato exercise in the <a href=\"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/chapter\/the-f-distribution-and-the-f-ratio\/\" target=\"_blank\" rel=\"noopener\">last section (try it 1)<\/a>. The means of the tomato yields under the five mulching conditions are represented by\u00a0<em data-effect=\"italics\">\u03bc<\/em><sub>1<\/sub>,\u00a0<em data-effect=\"italics\">\u03bc<\/em><sub>2<\/sub>,\u00a0<em data-effect=\"italics\">\u03bc<\/em><sub>3<\/sub>,\u00a0<em data-effect=\"italics\">\u03bc<\/em><sub>4<\/sub>,\u00a0<em data-effect=\"italics\">\u03bc<\/em><sub>5<\/sub>. We will conduct a hypothesis test to determine if all means are the same or at least one is different. Using a significance level of 5%, test the null hypothesis that there is no difference in mean yields among the five groups against the alternative hypothesis that at least one mean is different from the rest.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q130941\">Show Answer<\/span><\/p>\n<div id=\"q130941\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"eip-573\" class=\"\">The null and alternative hypotheses are:<\/p>\n<p id=\"eip-idp93603520\" class=\"\"><em data-effect=\"italics\">H<sub>0<\/sub><\/em>:\u00a0<em data-effect=\"italics\">\u03bc<sub>1<\/sub><\/em>\u00a0=\u00a0<em data-effect=\"italics\">\u03bc<sub>2<\/sub><\/em>\u00a0=\u00a0<em data-effect=\"italics\">\u03bc<sub>3<\/sub><\/em>\u00a0=\u00a0<em data-effect=\"italics\">\u03bc<sub>4<\/sub><\/em>\u00a0=\u00a0<em data-effect=\"italics\">\u03bc<sub>5<\/sub><\/em><\/p>\n<p id=\"eip-idp134625200\" class=\"\"><em data-effect=\"italics\">H<sub>a<\/sub>: \u03bc<sub>i<\/sub>\u00a0\u2260 \u03bc<sub>j<\/sub><\/em>\u00a0some\u00a0<em data-effect=\"italics\">i \u2260 j<\/em><\/p>\n<p id=\"eip-idm22822112\" class=\"\">The one-way ANOVA results are shown below.<\/p>\n<table style=\"border-collapse: collapse; width: 100%; height: 48px;\">\n<tbody>\n<tr style=\"height: 12px;\">\n<td style=\"width: 20%; height: 12px;\"><strong>Source of Variation<\/strong><\/td>\n<td style=\"width: 20%; height: 12px;\"><strong>Sum of Squares (<em>SS<\/em>)<\/strong><\/td>\n<td style=\"width: 20%; height: 12px;\"><strong>Degrees of Freedom (<em>df<\/em>)<\/strong><\/td>\n<td style=\"width: 20%; height: 12px;\"><strong>Mean Square (<em>MS<\/em>)<\/strong><\/td>\n<td style=\"width: 20%; height: 12px;\"><strong><em>F<\/em><\/strong><\/td>\n<\/tr>\n<tr style=\"height: 12px;\">\n<td style=\"width: 20%; height: 12px;\">Factor (Between)<\/td>\n<td style=\"width: 20%;\">36,648,561<\/td>\n<td style=\"width: 20%;\">5 \u2013 1 = 4<\/td>\n<td style=\"width: 20%;\">[latex]\\displaystyle\\frac{{{36},{648},{561}}}{{4}}={9},{162},{140}[\/latex]<\/td>\n<td style=\"width: 20%;\">[latex]\\displaystyle\\frac{{{9},{162},{140}}}{{{2},{044},{672.6}}}={4.4810}[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 12px;\">\n<td style=\"width: 20%; height: 12px;\">Error (Within)<\/td>\n<td style=\"width: 20%;\">20,446,726<\/td>\n<td style=\"width: 20%;\">15 \u2013 5 = 10<\/td>\n<td style=\"width: 20%;\">[latex]\\displaystyle\\frac{{{20},{446},{726}}}{{10}}={2},{044},{672.6}[\/latex]<\/td>\n<td style=\"width: 20%; height: 12px;\"><\/td>\n<\/tr>\n<tr style=\"height: 12px;\">\n<td style=\"width: 20%; height: 12px;\">Total<\/td>\n<td style=\"width: 20%;\">57,095,287<\/td>\n<td style=\"width: 20%;\">15 \u2013 1 = 14<\/td>\n<td style=\"width: 20%;\"><\/td>\n<td style=\"width: 20%; height: 12px;\"><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"eip-idm38415936\" class=\"\"><strong>Distribution for the test:\u00a0<em data-effect=\"italics\">F<\/em><sub>4,10<\/sub><\/strong><\/p>\n<p id=\"eip-idm45807632\" class=\"\"><em data-effect=\"italics\">df<\/em>(<em data-effect=\"italics\">num<\/em>) = 5 \u2013 1 = 4<\/p>\n<p id=\"eip-idm2595504\" class=\"\"><em data-effect=\"italics\">df<\/em>(<em data-effect=\"italics\">denom<\/em>) = 15 \u2013 5 = 10<\/p>\n<p id=\"eip-idm4421632\" class=\"\"><strong>Test statistic:<\/strong>\u00a0<em data-effect=\"italics\">F<\/em>\u00a0= 4.4810<\/p>\n<p><strong>Graph:<\/strong><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-2330 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5668\/2021\/07\/11175212\/47944008d54b3438f561c515dc26b48c1878de18.jpeg\" alt=\"This graph shows a nonsymmetrical F distribution curve. The horizontal axis extends from 0 - 5, and the vertical axis ranges from 0 - 0.7. The curve is strongly skewed to the right.\" width=\"488\" height=\"325\" \/><\/p>\n<p id=\"eip-idm4421376\" class=\"\"><strong>Probability Statement:<\/strong>\u00a0<em data-effect=\"italics\">p<\/em>-value =\u00a0<em data-effect=\"italics\">P<\/em>(<em data-effect=\"italics\">F<\/em>\u00a0&gt; 4.481) = 0.0248.<\/p>\n<p id=\"eip-idm4421120\" class=\"\"><strong>Compare\u00a0<em data-effect=\"italics\">\u03b1<\/em>\u00a0and the\u00a0<em data-effect=\"italics\">p<\/em>-value:<\/strong>\u00a0<em data-effect=\"italics\">\u03b1<\/em>\u00a0= 0.05,\u00a0<em data-effect=\"italics\">p<\/em>-value = 0.0248<\/p>\n<p id=\"eip-idm141674128\" class=\"\"><strong>Make a decision:<\/strong>\u00a0Since\u00a0<em data-effect=\"italics\">\u03b1<\/em>\u00a0&gt;\u00a0<em data-effect=\"italics\">p<\/em>-value, we reject\u00a0<em data-effect=\"italics\">H<sub>0<\/sub><\/em>.<\/p>\n<p id=\"eip-idm156731472\" class=\"\"><strong>Conclusion:<\/strong>\u00a0At the 5% significance level, we have reasonably strong evidence that differences in mean yields for slicing tomato plants grown under different mulching conditions are unlikely to be due to chance alone. We may conclude that at least some of the mulches led to different mean yields.<\/p>\n<header>\n<h2 class=\"os-title\" data-type=\"title\"><span class=\"os-title-label\">USING THE TI-83, 83+, 84, 84+ CALCULATOR<\/span><\/h2>\n<\/header>\n<ul>\n<li id=\"eip-idm156731216\" class=\"\">Press STAT. Press 1:EDIT. Put the data into the lists\u00a0<em data-effect=\"italics\">L<sub>1<\/sub><\/em>,\u00a0<em data-effect=\"italics\">L<sub>2<\/sub><\/em>,\u00a0<em data-effect=\"italics\">L<sub>3<\/sub><\/em>,\u00a0<em data-effect=\"italics\">L<sub>4<\/sub><\/em>,\u00a0<em data-effect=\"italics\">L<sub>5<\/sub><\/em>.<\/li>\n<li class=\"\">Press STAT, arrow over to TESTS, and arrow down to ANOVA. Press ENTER, and then enter\u00a0<em style=\"font-size: 1rem; orphans: 1; text-align: initial;\" data-effect=\"italics\">L<sub>1<\/sub><\/em><span style=\"font-size: 1rem; orphans: 1; text-align: initial;\">,\u00a0<\/span><em style=\"font-size: 1rem; orphans: 1; text-align: initial;\" data-effect=\"italics\">L<sub>2<\/sub><\/em><span style=\"font-size: 1rem; orphans: 1; text-align: initial;\">,\u00a0<\/span><em style=\"font-size: 1rem; orphans: 1; text-align: initial;\" data-effect=\"italics\">L<sub>3<\/sub><\/em><span style=\"font-size: 1rem; orphans: 1; text-align: initial;\">,\u00a0<\/span><em style=\"font-size: 1rem; orphans: 1; text-align: initial;\" data-effect=\"italics\">L<sub>4<\/sub><\/em><span style=\"font-size: 1rem; orphans: 1; text-align: initial;\">,\u00a0<\/span><em style=\"font-size: 1rem; orphans: 1; text-align: initial;\" data-effect=\"italics\">L<sub>5<\/sub><\/em><span style=\"font-size: 1rem; orphans: 1; text-align: initial;\">).<\/span><\/li>\n<li class=\"\">Press ENTER. You will see that the values in the foregoing ANOVA table are easily produced by the calculator, including the test statistic and the\u00a0<em>p<\/em>-value of the test.<\/li>\n<li class=\"\">The calculator displays:\n<ul>\n<li class=\"\"><em style=\"font-size: 1rem; orphans: 1; text-align: initial;\" data-effect=\"italics\">F<\/em><span style=\"font-size: 1rem; orphans: 1; text-align: initial;\">\u00a0= 4.4810<\/span><\/li>\n<li class=\"\"><em style=\"font-size: 1rem; orphans: 1; text-align: initial;\" data-effect=\"italics\">p<\/em><span style=\"font-size: 1rem; orphans: 1; text-align: initial;\">\u00a0= 0.0248 (<\/span><em style=\"font-size: 1rem; orphans: 1; text-align: initial;\" data-effect=\"italics\">p<\/em><span style=\"font-size: 1rem; orphans: 1; text-align: initial;\">-value)<\/span><\/li>\n<li class=\"\">Factor\u00a0<em style=\"font-size: 1rem; orphans: 1; text-align: initial;\" data-effect=\"italics\">df<\/em><span style=\"font-size: 1rem; orphans: 1; text-align: initial;\">\u00a0= 4<\/span><\/li>\n<li class=\"\"><em style=\"font-size: 1rem; orphans: 1; text-align: initial;\" data-effect=\"italics\">SS<\/em><span style=\"font-size: 1rem; orphans: 1; text-align: initial;\">\u00a0= 36648560.9<\/span><\/li>\n<li class=\"\"><em style=\"font-size: 1rem; orphans: 1; text-align: initial;\" data-effect=\"italics\">MS<\/em><span style=\"font-size: 1rem; orphans: 1; text-align: initial;\">\u00a0= 9162140.23<\/span><\/li>\n<li class=\"\">Error\u00a0<em style=\"font-size: 1rem; orphans: 1; text-align: initial;\" data-effect=\"italics\">df<\/em><span style=\"font-size: 1rem; orphans: 1; text-align: initial;\">\u00a0= 10<\/span><\/li>\n<li class=\"\"><em style=\"font-size: 1rem; orphans: 1; text-align: initial;\" data-effect=\"italics\">SS<\/em><span style=\"font-size: 1rem; orphans: 1; text-align: initial;\">\u00a0= 20446726<\/span><\/li>\n<li class=\"\"><em style=\"font-size: 1rem; orphans: 1; text-align: initial;\" data-effect=\"italics\">MS<\/em><span style=\"font-size: 1rem; orphans: 1; text-align: initial;\">\u00a0= 2044672.6<\/span><\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it 1<\/h3>\n<p>MRSA, or <em>Staphylococcus aureus<\/em>, can cause serious bacterial infections in hospital patients. This table shows various colony counts from different patients who may or may not have MRSA.<\/p>\n<table>\n<thead>\n<tr>\n<th>Conc = 0.6<\/th>\n<th>Conc = 0.8<\/th>\n<th>Conc = 1.0<\/th>\n<th>Conc = 1.2<\/th>\n<th>Conc = 1.4<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>9<\/td>\n<td>16<\/td>\n<td>22<\/td>\n<td>30<\/td>\n<td>27<\/td>\n<\/tr>\n<tr>\n<td>66<\/td>\n<td>93<\/td>\n<td>147<\/td>\n<td>199<\/td>\n<td>168<\/td>\n<\/tr>\n<tr>\n<td>98<\/td>\n<td>82<\/td>\n<td>120<\/td>\n<td>148<\/td>\n<td>132<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Plot of the data for the different concentrations:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-2332 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5668\/2021\/07\/11175328\/9d1cdcd57e7f58e7af6badb0855358d780c243bd.png\" alt=\"This graph is a scatterplot for the data provided. The horizontal axis is labeled 'Colony counts' and extends from 0 - 200. The vertical axis is labeled 'Tryptone concentrations' and extends from 0.6 - 1.4.\" width=\"477\" height=\"284\" \/><\/p>\n<p>Test whether the mean number of colonies is the same or different. Construct the ANOVA table (by hand or by using a TI-83, 83+, or 84+ calculator), find the <em>p<\/em>-value, and state your conclusion. Use a 5% significance level.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q66762\">Show Answer<\/span><\/p>\n<div id=\"q66762\" class=\"hidden-answer\" style=\"display: none\">\n<p>While there are differences in the spreads between the groups, the differences do not appear to be big enough to cause concern.<\/p>\n<p>We test for the equality of the mean number of colonies:<\/p>\n<p><em>H<sub data-redactor-tag=\"sub\">0<\/sub><\/em> : <em>\u03bc<\/em><sub>1<\/sub> = <em>\u03bc<\/em><sub>2<\/sub> = <em>\u03bc<\/em><sub>3<\/sub> = <em>\u03bc<\/em><sub>4<\/sub> = <em>\u03bc<\/em><sub>5<\/sub><em>H<sub data-redactor-tag=\"sub\">a<\/sub><\/em>: <em>\u03bc<sup data-redactor-tag=\"sup\">i<\/sup><\/em> \u2260 <em>\u03bc<sup data-redactor-tag=\"sup\">j<\/sup><\/em> some <em>i<\/em> \u2260 <em>j<\/em><\/p>\n<p>The one-way ANOVA table results are shown below.<\/p>\n<table style=\"border-collapse: collapse; width: 99.8837%; height: 84px;\">\n<thead>\n<tr style=\"height: 12px;\">\n<th style=\"width: 19.9069%; height: 12px;\">Source of Variation<\/th>\n<th style=\"width: 20.0233%; height: 12px;\">Sum of Squares (<em>SS<\/em>)<\/th>\n<th style=\"width: 1.1655%; height: 12px;\">Degrees of Freedom (<em>df<\/em>)<\/th>\n<th style=\"width: 38.7647%; height: 12px;\">Mean Square (<em>MS<\/em>)<\/th>\n<th style=\"width: 20.0233%; height: 12px;\"><em>F<\/em><\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr style=\"height: 36px;\">\n<td style=\"width: 19.9069%; height: 36px;\">Factor (Between)<\/td>\n<td style=\"width: 20.0233%; height: 36px;\">10,233<\/td>\n<td style=\"width: 1.1655%; height: 36px;\">5 \u2013 1 = 4<\/td>\n<td style=\"width: 38.7647%; height: 36px;\">[latex]\\displaystyle\\frac{{{10},{233}}}{{4}}={2},{558.25}[\/latex]<\/td>\n<td style=\"width: 20.0233%; height: 36px;\">[latex]\\displaystyle\\frac{{{2},{558.25}}}{{{4},{194.9}}}={0.6099}[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 12px;\">\n<td style=\"width: 19.9069%; height: 12px;\">Error (Within)<\/td>\n<td style=\"width: 20.0233%; height: 12px;\">41,949<\/td>\n<td style=\"width: 1.1655%; height: 12px;\">15 \u2013 5 = 10<\/td>\n<td style=\"width: 38.7647%; height: 12px;\"><\/td>\n<td style=\"width: 20.0233%; height: 12px;\"><\/td>\n<\/tr>\n<tr style=\"height: 24px;\">\n<td style=\"width: 19.9069%; height: 24px;\">Total<\/td>\n<td style=\"width: 20.0233%; height: 24px;\">52,182<\/td>\n<td style=\"width: 1.1655%; height: 24px;\">15 \u2013 1 = 14<\/td>\n<td style=\"width: 38.7647%; height: 24px;\">[latex]\\displaystyle\\frac{{{41},{949}}}{{10}}={4},{194.9}[\/latex]<\/td>\n<td style=\"width: 20.0233%; height: 24px;\"><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><strong>Graph:<\/strong><\/p>\n<p><img decoding=\"async\" class=\"aligncenter\" style=\"width: 442px; height: 299.055319148936px;\" src=\"https:\/\/textimgs.s3.amazonaws.com\/DE\/stats\/2qg4-x87q657i#fixme#fixme#fixme\" alt=\"This graph shows a nonsymmetrical F distribution curve. The curve is skewed to the right. A vertical upward line extends from 0.6649 to the curve. This line is just to the right of the graph's peak and the region to the right of the line is shaded to represent the p-value.\" \/><\/p>\n<p><strong>Distribution for the test:<\/strong> <em>F<\/em><sub>4,10<\/sub><\/p>\n<p><strong>Probability Statement: <\/strong><em>p<\/em>-value = <em>P<\/em>(<em>F<\/em> &gt; 0.6099) = 0.6649.<\/p>\n<p><strong>Compare <em data-redactor-tag=\"em\">\u03b1<\/em> and the <em>p<\/em>-value:<\/strong> <em>\u03b1<\/em> = 0.05, <em>p<\/em>-value = 0.669, <em>\u03b1<\/em>\u00a0&lt;\u00a0<em>p<\/em>-value<\/p>\n<p><strong>Make a decision:<\/strong> Since <em>\u03b1<\/em>\u00a0&lt;\u00a0<em>p<\/em>-value, we do not reject <em>H<\/em>0.<\/p>\n<p><strong>Conclusion:<\/strong> At the 5% significance level, there is insufficient evidence from these data that different levels of tryptone will cause a significant difference in the mean number of bacterial colonies formed.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example 2<\/h3>\n<p>Four sororities took a random sample of sisters regarding their grade means for the past term. The results are shown in the table.<\/p>\n<p>Mean Grades for Four Sororities<\/p>\n<table>\n<thead>\n<tr>\n<th>Sorority 1<\/th>\n<th>Sorority 2<\/th>\n<th>Sorority 3<\/th>\n<th>Sorority 4<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>2.17<\/td>\n<td>2.63<\/td>\n<td>2.63<\/td>\n<td>3.79<\/td>\n<\/tr>\n<tr>\n<td>1.85<\/td>\n<td>1.77<\/td>\n<td>3.78<\/td>\n<td>3.45<\/td>\n<\/tr>\n<tr>\n<td>2.83<\/td>\n<td>3.25<\/td>\n<td>4.00<\/td>\n<td>3.08<\/td>\n<\/tr>\n<tr>\n<td>1.69<\/td>\n<td>1.86<\/td>\n<td>2.55<\/td>\n<td>2.26<\/td>\n<\/tr>\n<tr>\n<td>3.33<\/td>\n<td>2.21<\/td>\n<td>2.45<\/td>\n<td>3.18<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Using a significance level of 1%, is there a difference in mean grades among the sororities?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q901953\">Show Answer<\/span><\/p>\n<div id=\"q901953\" class=\"hidden-answer\" style=\"display: none\">\n<p>Let <em>\u03bc<sub data-redactor-tag=\"sub\">1<\/sub><\/em>, <em>\u03bc<sub data-redactor-tag=\"sub\">2<\/sub><\/em>, <em>\u03bc<sub data-redactor-tag=\"sub\">3<\/sub><\/em>, <em>\u03bc<sub data-redactor-tag=\"sub\">4<\/sub><\/em> be the population means of the sororities. Remember that the null hypothesis claims that the sorority groups are from the same normal distribution. The alternate hypothesis says that at least two of the sorority groups come from populations with different normal distributions. Notice that the four sample sizes are each five.<\/p>\n<h4>Note<\/h4>\n<p>This is an example of a <strong>balanced design<\/strong>\u00a0because each factor (i.e., sorority) has the same number of observations.<\/p>\n<p><em>H<sub data-redactor-tag=\"sub\">0<\/sub><\/em>: <em>\u03bc<sub data-redactor-tag=\"sub\">1<\/sub><\/em> = <em>\u03bc<sub data-redactor-tag=\"sub\">2<\/sub><\/em> = <em>\u03bc<sub data-redactor-tag=\"sub\">3<\/sub><\/em> = <em>\u03bc<sub data-redactor-tag=\"sub\">4<\/sub><\/em><\/p>\n<p><em>H<sub data-redactor-tag=\"sub\">a<\/sub><\/em>: Not all of the means <em>\u03bc<sub data-redactor-tag=\"sub\">1<\/sub><\/em>, <em>\u03bc<sub data-redactor-tag=\"sub\">2<\/sub><\/em>, <em>\u03bc<sub data-redactor-tag=\"sub\">3<\/sub><\/em>, <em>\u03bc<sub data-redactor-tag=\"sub\">4<\/sub><\/em> are equal.<\/p>\n<p><strong>Distribution for the test:<\/strong> <em>F<\/em><sub>3,16<\/sub><\/p>\n<p>where <em>k<\/em> = 4 groups and <em>n<\/em> = 20 samples in total<\/p>\n<p><em>df<\/em>(<em>num<\/em>)= <em>k<\/em> \u2013 1 = 4 \u2013 1 = 3<\/p>\n<p><em>df<\/em>(<em>denom<\/em>) = <em>n<\/em> \u2013 <em>k<\/em> = 20 \u2013 4 = 16<\/p>\n<p><strong>Calculate the test statistic:<\/strong> <em>F<\/em> = 2.23<\/p>\n<p><strong>Graph:<\/strong><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-2334 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5668\/2021\/07\/11175436\/400f26530a962102a5c7068ab6a3f2a2adfb2d17.jpeg\" alt=\"This graph shows a nonsymmetrical F distribution curve with values of 0 and 2.23 on the x-axis representing the test statistic of sorority grade averages. The curve is slightly skewed to the right, but is approximately normal. A vertical upward line extends from 2.23 to the curve and the area to the right of this is shaded to represent the p-value.\" width=\"487\" height=\"275\" \/><\/p>\n<p><strong>Probability statement:<\/strong> <em>p<\/em>-value = <em>P<\/em>(<em>F<\/em> &gt; 2.23) = 0.1241<\/p>\n<p><strong>Compare <em data-redactor-tag=\"em\">\u03b1<\/em> and the <em>p<\/em>-value:<\/strong> <em>\u03b1<\/em> = 0.01<\/p>\n<p><em>p<\/em>-value = 0.1241<\/p>\n<p><em>\u03b1<\/em> &lt; <em>p<\/em>-value<\/p>\n<p><strong>Make a decision:<\/strong> Since <em>\u03b1<\/em> &lt; <em>p<\/em>-value, you cannot reject <em>H0<\/em>.<\/p>\n<p><strong>Conclusion: <\/strong>There is not sufficient evidence to conclude that there is a difference among the mean grades for the sororities.<\/p>\n<h2 class=\"os-title\" data-type=\"title\"><span class=\"os-title-label\">USING THE TI-83, 83+, 84, 84+ CALCULATOR<\/span><\/h2>\n<ul>\n<li>Put the data into lists L1, L2, L3, and L4. Press <code>STAT<\/code> and arrow over to <code>TESTS<\/code>. Arrow down to <code>F:ANOVA<\/code>. Press <code>ENTER<\/code>and Enter (<code>L1,L2,L3,L4<\/code>).<\/li>\n<li>The calculator displays the F statistic, the <em style=\"font-size: 1rem; orphans: 1; text-align: initial;\">p<\/em><span style=\"font-size: 1rem; orphans: 1; text-align: initial;\">-value and the values for the one-way ANOVA table:<\/span>\n<ul>\n<li><em style=\"font-size: 1rem; orphans: 1; text-align: initial;\">F<\/em><span style=\"font-size: 1rem; orphans: 1; text-align: initial;\"> = 2.2303<\/span><\/li>\n<li><em style=\"font-size: 1rem; orphans: 1; text-align: initial;\">p<\/em><span style=\"font-size: 1rem; orphans: 1; text-align: initial;\"> = 0.1241 (<\/span><em style=\"font-size: 1rem; orphans: 1; text-align: initial;\">p<\/em><span style=\"font-size: 1rem; orphans: 1; text-align: initial;\">-value)<\/span><\/li>\n<li>Factor <em style=\"font-size: 1rem; orphans: 1; text-align: initial;\">df<\/em><span style=\"font-size: 1rem; orphans: 1; text-align: initial;\"> = 3<\/span><\/li>\n<li><em style=\"font-size: 1rem; orphans: 1; text-align: initial;\">SS<\/em><span style=\"font-size: 1rem; orphans: 1; text-align: initial;\"> = 2.88732<\/span><\/li>\n<li><em style=\"font-size: 1rem; orphans: 1; text-align: initial;\">MS<\/em><span style=\"font-size: 1rem; orphans: 1; text-align: initial;\"> = 0.96244<\/span><\/li>\n<li>Error <em style=\"font-size: 1rem; orphans: 1; text-align: initial;\">df<\/em><span style=\"font-size: 1rem; orphans: 1; text-align: initial;\"> = 16<\/span><\/li>\n<li><em style=\"font-size: 1rem; orphans: 1; text-align: initial;\">SS<\/em><span style=\"font-size: 1rem; orphans: 1; text-align: initial;\"> = 6.9044<\/span><\/li>\n<li><em style=\"font-size: 1rem; orphans: 1; text-align: initial;\">MS<\/em><span style=\"font-size: 1rem; orphans: 1; text-align: initial;\"> = 0.431525<\/span><\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it 2<\/h3>\n<p>Four sports teams took a random sample of players regarding their GPAs for the last year. The results are shown below:<\/p>\n<p>GPAs for Four Sports Teams<\/p>\n<table>\n<thead>\n<tr>\n<th>Basketball<\/th>\n<th>Baseball<\/th>\n<th>Hockey<\/th>\n<th>Lacrosse<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>3.6<\/td>\n<td>2.1<\/td>\n<td>4.0<\/td>\n<td>2.0<\/td>\n<\/tr>\n<tr>\n<td>2.9<\/td>\n<td>2.6<\/td>\n<td>2.0<\/td>\n<td>3.6<\/td>\n<\/tr>\n<tr>\n<td>2.5<\/td>\n<td>3.9<\/td>\n<td>2.6<\/td>\n<td>3.9<\/td>\n<\/tr>\n<tr>\n<td>3.3<\/td>\n<td>3.1<\/td>\n<td>3.2<\/td>\n<td>2.7<\/td>\n<\/tr>\n<tr>\n<td>3.8<\/td>\n<td>3.4<\/td>\n<td>3.2<\/td>\n<td>2.5<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Use a significance level of 5%, and determine if there is a difference in GPA among the teams.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q964208\">Show Answer<\/span><\/p>\n<div id=\"q964208\" class=\"hidden-answer\" style=\"display: none\">\n<p>With a <em>p<\/em>-value of 0.9271, we decline to reject the null hypothesis. There is not sufficient evidence to conclude that there is a difference among the GPAs for the sports teams.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example 3<\/h3>\n<p>A fourth-grade class is studying the environment. One of the assignments is to grow bean plants in different soils. Tommy chose to grow his bean plants in soil found outside his classroom mixed with dryer lint. Tara chose to grow her bean plants in potting soil bought at the local nursery. Nick chose to grow his bean plants in soil from his mother&#8217;s garden. No chemicals were used on the plants, only water. They were grown inside the classroom next to a large window. Each child grew five plants. At the end of the growing period, each plant was measured, producing the data (in inches) in this table.<\/p>\n<table>\n<thead>\n<tr>\n<th>Tommy&#8217;s Plants<\/th>\n<th>Tara&#8217;s Plants<\/th>\n<th>Nick&#8217;s Plants<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>24<\/td>\n<td>25<\/td>\n<td>23<\/td>\n<\/tr>\n<tr>\n<td>21<\/td>\n<td>31<\/td>\n<td>27<\/td>\n<\/tr>\n<tr>\n<td>23<\/td>\n<td>23<\/td>\n<td>22<\/td>\n<\/tr>\n<tr>\n<td>30<\/td>\n<td>20<\/td>\n<td>30<\/td>\n<\/tr>\n<tr>\n<td>23<\/td>\n<td>28<\/td>\n<td>20<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Does it appear that the three media in which the bean plants were grown produce the same mean height? Test at a 3% level of significance.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q127409\">Show Answer<\/span><\/p>\n<div id=\"q127409\" class=\"hidden-answer\" style=\"display: none\">\n<p>This time, we will perform the calculations that lead to the <em>F&#8217;<\/em>statistic. Notice that each group has the same number of plants, so we will use the formula [latex]\\displaystyle{F}'={\\dfrac{n \\cdot {s_{\\overline x}}^{2}}{{s^2}_{pooled}}}[\/latex].<\/p>\n<p>First, calculate the sample mean and sample variance of each group.<\/p>\n<table style=\"width: 428px;\">\n<thead>\n<tr>\n<th style=\"width: 116px;\"><\/th>\n<th style=\"width: 117px;\">Tommy&#8217;s Plants<\/th>\n<th style=\"width: 97px;\">Tara&#8217;s Plants<\/th>\n<th style=\"width: 98px;\">Nick&#8217;s Plants<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td style=\"width: 116px;\">Sample Mean<\/td>\n<td style=\"width: 117px;\">24.2<\/td>\n<td style=\"width: 97px;\">25.4<\/td>\n<td style=\"width: 98px;\">24.4<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 116px;\">Sample Variance<\/td>\n<td style=\"width: 117px;\">11.7<\/td>\n<td style=\"width: 97px;\">18.3<\/td>\n<td style=\"width: 98px;\">16.3<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Next, calculate the variance of the three group means (Calculate the variance of 24.2, 25.4, and 24.4). <strong>Variance of the group means = 0.413<\/strong> = [latex]{s_{\\overline x}}^{2}[\/latex]<\/p>\n<p>Then [latex]\\displaystyle{M}{S}_{{\\text{between}}}={n}{s_{\\overline x}}^{2}={({5})}{({0.413})}[\/latex] where [latex]n = {5}[\/latex] is the sample size (number of plants each child grew).<\/p>\n<p>Calculate the mean of the three sample variances (Calculate the mean of 11.7, 18.3, and 16.3). <strong>Mean of the sample variances = 15.433 = <em>s<sup>2<\/sup>pooled<\/em><\/strong><\/p>\n<p>Then [latex]\\displaystyle{M}{S}_{{\\text{within}}}[\/latex] =\u00a0<em>s<sup>2<\/sup><sub>pooled\u00a0<\/sub><\/em>= 15.433.<\/p>\n<p>The <em>F<\/em> statistic (or <em>F<\/em> ratio) is [latex]{\\text{F}}[\/latex] = [latex]{\\dfrac{MS_{between}}{MS_{within}}}[\/latex] = [latex]\\dfrac{n{s_{\\overline x}}^2}{{s^2}_{pooled}}[\/latex] = [latex]{\\dfrac{(5)(0.413)}{15.433}}[\/latex] = 0.134<\/p>\n<p>The <em>dfs<\/em> for the numerator = the number of groups \u2013 1 = 3 \u2013 1 = 2.<\/p>\n<p>The <em>dfs<\/em> for the denominator = the total number of samples \u2013 the number of groups = 15 \u2013 3 = 12<\/p>\n<p>The distribution for the test is <em>F<\/em><sub>2,12<\/sub> and the <em>F<\/em> statistic is <em>F<\/em> = 0.134<\/p>\n<p>The <em>p<\/em>-value is <em>P<\/em>(<em>F<\/em> &gt; 0.134) = 0.8759.<\/p>\n<p><strong>Decision:<\/strong> Since <em>\u03b1<\/em> = 0.03 and the <em>p<\/em>-value = 0.8759, do not reject <em>H<sub>0<\/sub><\/em>. (Why?)<\/p>\n<p><strong>Conclusion:<\/strong> With a 3% level of significance, from the sample data, the evidence is not sufficient to conclude that the mean heights of the bean plants are different.<\/p>\n<h2 class=\"os-title\" data-type=\"title\"><span class=\"os-title-label\">USING THE TI-83, 83+, 84, 84+ CALCULATOR<\/span><\/h2>\n<p>To calculate the <em>p<\/em>-value:<\/p>\n<ul>\n<li>Press <code style=\"line-height: 1.6em;\">2nd DISTR<\/code><\/li>\n<li>Arrow down to <code style=\"line-height: 1.6em;\">Fcdf<\/code>(and press<code style=\"line-height: 1.6em;\">ENTER<\/code>.<\/li>\n<li>Enter 0.134, <code style=\"line-height: 1.6em;\">E99<\/code>, 2, 12)<\/li>\n<li>Press <code style=\"line-height: 1.6em;\">ENTER<\/code><\/li>\n<li>The <em style=\"font-size: 1rem; orphans: 1; text-align: initial;\">p<\/em><span style=\"font-size: 1rem; orphans: 1; text-align: initial;\">-value is 0.8759.<\/span><\/li>\n<\/ul>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it 3<\/h3>\n<p>Another fourth-grader also grew bean plants, but this time in a jelly-like mass. The heights were (in inches) 24, 28, 25, 30, and 32. Do a one-way ANOVA test on the four groups. Are the heights of the bean plants different? Use the same method as shown in Example 3.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q11368\">Show Answer<\/span><\/p>\n<div id=\"q11368\" class=\"hidden-answer\" style=\"display: none\">\n<ul>\n<li><em>F<\/em> = 0.9496<\/li>\n<li><em>p<\/em>-value = 0.4402<\/li>\n<\/ul>\n<p>From the sample data, the evidence is not sufficient to conclude that the mean heights of the bean plants are different.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox tryit\">\n<header>\n<h3 class=\"title\" data-type=\"title\">Activity<\/h3>\n<\/header>\n<p>From the class, create four groups of the same size as follows: identified as men under 22, identified as men at least 22, identified as women under 22, identified as women at least 22. Have each member of each group record the number of states in the United States he or she has visited. Run an ANOVA test to determine if the average number of states visited in the four groups is the same. Test at a 1% level of significance.<\/p>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-317\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Facts About the F Distribution. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/books\/introductory-statistics\/pages\/13-3-facts-about-the-f-distribution\">https:\/\/openstax.org\/books\/introductory-statistics\/pages\/13-3-facts-about-the-f-distribution<\/a>. <strong>Project<\/strong>: Introductory Statistics. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Access for free at https:\/\/openstax.org\/books\/introductory-statistics\/pages\/1-introduction<\/li><li>Introductory Statistics. <strong>Authored by<\/strong>: Barbara Illowsky, Susan Dean. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/books\/introductory-statistics\/pages\/1-introduction\">https:\/\/openstax.org\/books\/introductory-statistics\/pages\/1-introduction<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Access for free at https:\/\/openstax.org\/books\/introductory-statistics\/pages\/1-introduction<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":169134,"menu_order":13,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Facts About the F Distribution\",\"author\":\"\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/books\/introductory-statistics\/pages\/13-3-facts-about-the-f-distribution\",\"project\":\"Introductory Statistics\",\"license\":\"cc-by\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/introductory-statistics\/pages\/1-introduction\"},{\"type\":\"cc\",\"description\":\"Introductory Statistics\",\"author\":\"Barbara Illowsky, Susan Dean\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/books\/introductory-statistics\/pages\/1-introduction\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/introductory-statistics\/pages\/1-introduction\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-317","chapter","type-chapter","status-publish","hentry"],"part":313,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/wp-json\/pressbooks\/v2\/chapters\/317","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/wp-json\/wp\/v2\/users\/169134"}],"version-history":[{"count":43,"href":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/wp-json\/pressbooks\/v2\/chapters\/317\/revisions"}],"predecessor-version":[{"id":4024,"href":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/wp-json\/pressbooks\/v2\/chapters\/317\/revisions\/4024"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/wp-json\/pressbooks\/v2\/parts\/313"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/wp-json\/pressbooks\/v2\/chapters\/317\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/wp-json\/wp\/v2\/media?parent=317"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/wp-json\/pressbooks\/v2\/chapter-type?post=317"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/wp-json\/wp\/v2\/contributor?post=317"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/wp-json\/wp\/v2\/license?post=317"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}