{"id":272,"date":"2021-07-14T15:59:03","date_gmt":"2021-07-14T15:59:03","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/chapter\/a-single-population-mean-using-the-student-t-distribution\/"},"modified":"2023-12-05T09:28:39","modified_gmt":"2023-12-05T09:28:39","slug":"a-single-population-mean-using-the-student-t-distribution","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/chapter\/a-single-population-mean-using-the-student-t-distribution\/","title":{"raw":"The Student's t-Distribution","rendered":"The Student&#8217;s t-Distribution"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<section>\r\n<ul id=\"list12315\">\r\n \t<li>Using the formula for creating a confidence interval or technology, construct a confidence interval for a population mean based on a t-distribution<\/li>\r\n<\/ul>\r\n<\/section><\/div>\r\nIn practice, we rarely know the population <strong>standard deviation<\/strong>. In the past, when the sample size was large, this did not present a problem to statisticians. They used the sample standard deviation <em>s<\/em> as an estimate for <em>\u03c3\u00a0<\/em>and proceeded as before to calculate a confidence interval with close enough results. However, statisticians ran into problems when the sample size was small. A small sample size caused inaccuracies in the confidence interval.\r\n\r\nWilliam S. Goset (1876\u20131937) of the Guinness brewery in Dublin, Ireland ran into this problem. His experiments with hops and barley produced very few samples. Just replacing <em>\u03c3<\/em> with <em>s<\/em> did not produce accurate results when he tried to calculate a confidence interval. He realized that he could not use a normal distribution for the calculation; he found that the actual distribution depends on the sample size. This problem led him to \"discover\" what is called the <strong>Student's t-distribution<\/strong>. The name comes from the fact that Gosset wrote under the pen name \"Student.\"\r\n\r\nUp until the mid-1970s, some statisticians used the <strong>normal distribution <\/strong>approximation for large sample sizes and only used the Student's t-distribution only for sample sizes of at most 30. With graphing calculators and computers, the practice now is to use the Student's t-distribution whenever <em>s\u00a0<\/em>is used as an estimate for <em>\u03c3<\/em>.\r\n\r\nIf you draw a simple random sample of size <em>n<\/em> from a population that has an approximately a normal distribution with mean <em>\u03bc<\/em> and unknown population standard deviation <em>\u03c3<\/em> and calculate the <em>t<\/em>-score: [latex]\\displaystyle{t}=\\dfrac{{\\overline{{x}}-\\mu}}{{\\frac{{s}}{\\sqrt{{n}}}}}[\/latex] is from its mean <em>\u03bc<\/em>. For each sample size <em>n<\/em>, there is a different Student's t-distribution.\r\n\r\nThe <strong>degrees of freedom, <em data-redactor-tag=\"em\">n<\/em> \u2013 1<\/strong>, come from the calculation of the sample standard deviation <em>s<\/em>. Because the sum of the deviations is zero, we can find the last deviation once we know the other <em>n<\/em> \u2013 1 deviations. The other <strong><em>n<\/em> \u2013 1<\/strong> deviations can change or vary freely. <strong>We call the number <em data-redactor-tag=\"em\">n<\/em> \u2013 1 the degrees of freedom (df)<\/strong>.\r\n<h3>Properties of the Student's t-Distribution<\/h3>\r\n<ul>\r\n \t<li>The graph for the Student's t-distribution is similar to the standard normal curve.<\/li>\r\n \t<li>The mean for the Student's t-distribution is zero and the distribution is symmetric about zero.<\/li>\r\n \t<li>The Student's t-distribution has more probability in its tails than the standard normal distribution because the spread of the t-distribution is greater than the spread of the standard normal. So the graph of the Student's t-distribution will be thicker in the tails and shorter in the center than the graph of the standard normal distribution.<\/li>\r\n \t<li>The exact shape of the Student's t-distribution depends on the degrees of freedom. As the degrees of freedom increases, the graph of Student's t-distribution becomes more like the graph of the standard normal distribution.<\/li>\r\n \t<li>The underlying population of individual observations is assumed to be normally distributed with unknown population mean <em>\u03bc<\/em> and unknown population standard deviation <em>\u03c3<\/em>. The size of the underlying population is generally not relevant unless it is very small. If it is bell shaped (normal) then the assumption is met and doesn't need discussion. Random sampling is assumed, but that is a completely separate assumption from normality.<\/li>\r\n<\/ul>\r\nCalculators and computers can easily calculate any Student's t-probabilities. The TI-83,83+, and 84+ have a tcdf function to find the probability for given values of <em>t<\/em>. The grammar for the tcdf command is tcdf(lower bound, upper bound, degrees of freedom). However for confidence intervals, we need to use <strong>inverse<\/strong> probability to find the value of <em>t<\/em> when we know the probability.\r\n\r\nFor the TI-84+ you can use the invT command on the DISTRibution menu. The invT command works similarly to the invnorm. The invT command requires two inputs: <strong>invT(area to the left, degrees of freedom)<\/strong> The output is the t-score that corresponds to the area we specified.\r\n\r\nThe TI-83 and 83+ do not have the invT command. (The TI-89 has an inverse T command.)\r\n\r\nA probability table for the Student's t-distribution can also be used. The table gives t-scores that correspond to the confidence level (column) and degrees of freedom (row). (The TI-86 does not have an invT program or command, so if you are using that calculator, you need to use a probability table for the Student's t-distribution.) When using a <em>t<\/em>-table, note that some tables are formatted to show the confidence level in the column headings, while the column headings in some tables may show only corresponding area in one or both tails.\r\n\r\nA <a href=\"http:\/\/www.itl.nist.gov\/div898\/handbook\/eda\/section3\/eda3672.htm\" target=\"_blank\" rel=\"noopener\">Student's t table<\/a> gives <em>t<\/em>-scores given the degrees of freedom and the right-tailed probability. The table is very limited. <strong>Calculators and computers can easily calculate any Student's t-probabilities.<\/strong>\r\n\r\n<strong>The notation for the Student's t-distribution (using <em>T<\/em> as the random variable) is:<\/strong>\r\n<ul>\r\n \t<li><em>T ~ t<sub data-redactor-tag=\"sub\">df<\/sub><\/em> where <em>df<\/em> = <em>n<\/em> \u2013 1.<\/li>\r\n \t<li>For example, if we have a sample of size <em>n<\/em> = 20 items, then we calculate the degrees of freedom as <em>df<\/em> = <em>n<\/em> - 1 = 20 - 1 = 19 and we write the distribution as <em>T ~ t<sub data-redactor-tag=\"sub\">19<\/sub><\/em>.<\/li>\r\n<\/ul>\r\n<strong>If the population standard deviation is not known, the error bound for a population mean is:<\/strong>\r\n<ul>\r\n \t<li>EBM = [latex]\\displaystyle({t}_{\\frac{{\\alpha}}{{2}}})(\\frac{{s}}{{\\sqrt{n}}})[\/latex],<\/li>\r\n \t<li>[latex]\\displaystyle{t}_{\\frac{{\\alpha}}{{2}}}[\/latex] is the t-score with area to the right equal to[latex]\\displaystyle\\frac{{\\alpha}}{{2}}[\/latex],<\/li>\r\n \t<li>use <em data-redactor-tag=\"em\">df<\/em> = <em data-redactor-tag=\"em\">n<\/em> \u2013 1 degrees of freedom, and<\/li>\r\n \t<li><em data-redactor-tag=\"em\">s<\/em> = sample standard deviation.<\/li>\r\n<\/ul>\r\n<strong>The format for the confidence interval is:<\/strong>\r\n<p class=\"p1\"><span class=\"s1\">([latex]\\displaystyle\\overline{x}[\/latex] - EBM, [latex]\\displaystyle\\overline{x}[\/latex] + EBM).<\/span><\/p>\r\n\r\n<header>\r\n<h3 class=\"os-title\" data-type=\"title\"><span class=\"os-title-label\">USING THE TI-83, 83+, 84, 84+ CALCULATOR<\/span><\/h3>\r\n<\/header>To calculate the confidence interval directly:\r\n<ul>\r\n \t<li>Press STAT.<\/li>\r\n \t<li>Arrow over to TESTS.<\/li>\r\n \t<li>Arrow down to 8:TInterval and press ENTER (or just press 8).<\/li>\r\n<\/ul>\r\nhttps:\/\/youtu.be\/bFefxSE5bmo\r\n<div class=\"textbox exercises\">\r\n<h3>Example 1<\/h3>\r\nSuppose you do a study of acupuncture to determine how effective it is in relieving pain. You measure sensory rates for 15 subjects with the results given. Use the sample data to construct a 95% confidence interval for the mean sensory rate for the population (assumed normal) from which you took the data.\r\n\r\nThe solution is shown step-by-step and by using the TI-83, 83+, or 84+ calculators.\r\n\r\n8.6 9.4 7.9 6.8 8.3 7.3 9.2 9.6 8.7 11.4 10.3 5.4 8.1 5.5 6.9\r\n\r\nThe first solution is step-by-step.\r\n[reveal-answer q=\"776542\"]Show Solution A[\/reveal-answer]\r\n[hidden-answer a=\"776542\"]To find the confidence interval, you need the sample mean, [latex]{\\overline{x}}[\/latex], and the <em data-redactor-tag=\"em\">EBM<\/em>.\r\n<p class=\"p1\"><span class=\"s1\">[latex]\\displaystyle\\overline{x}[\/latex] = 8.2267<\/span><\/p>\r\n<p class=\"p1\">s = 1.6722<\/p>\r\n<p class=\"p1\">n = 15<\/p>\r\n<p class=\"p1\"><em data-redactor-tag=\"em\">df<\/em> = 15 \u2013 1 = 14 <em data-redactor-tag=\"em\">CL<\/em> so <em data-redactor-tag=\"em\">\u03b1<\/em> = 1 \u2013 <em data-redactor-tag=\"em\">CL<\/em> = 1 \u2013 0.95 = 0.05<\/p>\r\n<p class=\"p1\"><span class=\"s1\">[latex]\\displaystyle\\frac{{\\alpha}}{{2}}[\/latex] = 0.025<\/span><\/p>\r\n<p class=\"p1\"><span class=\"s1\">[latex]\\displaystyle{t}_{\\frac{{\\alpha}}{{2}}}[\/latex] = t<sub>0.025<\/sub><\/span><\/p>\r\n<p class=\"p1\">The area to the right of t<sub>0.025<\/sub> is 0.025, and the area to the left of\u00a0t<sub>0.025<\/sub> is\u00a01 \u2013 0.025 = 0.975<\/p>\r\n<p class=\"p1\">[latex]\\displaystyle{t}_{\\frac{{\\alpha}}{{2}}}[\/latex] = t<sub>0.025<\/sub> = 2.14 using invT(.975,14) on the TI-84+ calculator<\/p>\r\n<p class=\"p1\"><span class=\"s1\"><em>EBM<\/em> =\u00a0[latex]\\displaystyle({t}_{\\frac{{\\alpha}}{{2}}})(\\frac{{s}}{{\\sqrt{n}}})[\/latex]<\/span><\/p>\r\n<p class=\"p1\"><em>EBM<\/em> = (2.14)[latex](\\frac{{1.6722}}{{\\sqrt{15}}})[\/latex] = 0.924<\/p>\r\n<p class=\"p1\">[latex]\\displaystyle\\overline{x}[\/latex] - EBM = 8.2267 - 0.9240 = 7.3<\/p>\r\n<p class=\"p1\">[latex]\\displaystyle\\overline{x}[\/latex] + EBM = 8.2267 + 0.9240 = 9.15<\/p>\r\nThe 95% confidence interval is (7.30, 9.15).\r\n\r\nWe estimate with 95% confidence that the true population mean sensory rate is between 7.30 and 9.15.\r\n\r\n[\/hidden-answer]\r\n\r\n&nbsp;\r\n\r\nThe second solution uses the TI-83+ and TI-84 calculators.\r\n[reveal-answer q=\"496358\"]Show Solution B[\/reveal-answer]\r\n[hidden-answer a=\"496358\"]\r\n\r\nPress <code data-redactor-tag=\"code\">STAT<\/code> and arrow over to\u00a0<code data-redactor-tag=\"code\">TESTS<\/code>.\r\nArrow down to <code data-redactor-tag=\"code\">8:TInterval<\/code> and press <code data-redactor-tag=\"code\">ENTER<\/code> (or you can just press\u00a0<code data-redactor-tag=\"code\">8<\/code>).\r\nArrow to <code data-redactor-tag=\"code\">Data<\/code> and press <code data-redactor-tag=\"code\">ENTER<\/code>.\r\nArrow down to <code data-redactor-tag=\"code\">List<\/code> and enter the list name where you put the data.\r\nThere should be a 1 after <code data-redactor-tag=\"code\">Freq<\/code>.\r\nArrow down to <code data-redactor-tag=\"code\">C-level<\/code> and enter 0.95\r\nArrow down to <code data-redactor-tag=\"code\">Calculate<\/code> and press <code data-redactor-tag=\"code\">ENTER<\/code>.\r\nThe 95% confidence interval is (7.3006, 9.1527)[\/hidden-answer]\r\n\r\n<strong data-redactor-tag=\"strong\">Note: <\/strong>When calculating the error bound, a probability table for the Student's t-distribution can also be used to find the value of <em data-redactor-tag=\"em\">t<\/em>. The table gives\u00a0<em data-redactor-tag=\"em\">t<\/em>-scores that correspond to the confidence level (column) and degrees of freedom (row); the <em data-redactor-tag=\"em\">t<\/em>-score is found where the row and column intersect in the table.\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it 1<\/h3>\r\nYou do a study of hypnotherapy to determine how effective it is in increasing the number of hourse of sleep subjects get each night. You measure hours of sleep for 12 subjects with the following results. Construct a 95% confidence interval for the mean number of hours slept for the population (assumed normal) from which you took the data.\r\n\r\n8.2; 9.1; 7.7; 8.6; 6.9; 11.2; 10.1; 9.9; 8.9; 9.2; 7.5; 10.5\r\n[reveal-answer q=\"742079\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"742079\"]\r\n\r\n(8.1634, 9.8032)\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h3><\/h3>\r\n<div class=\"textbox exercises\">\r\n<h3>Example 2<\/h3>\r\nThe Human Toxome Project (HTP) is working to understand the scope of industrial pollution in the human body. Industrial chemicals may enter the body through pollution or as ingredients in consumer products. In October 2008, the scientists at HTP tested cord blood samples for 20 newborn infants in the United States. The cord blood of the \"In utero\/newborn\" group was tested for 430 industrial compounds, pollutants, and other chemicals, including chemicals linked to brain and nervous system toxicity, immune system toxicity, reproductive toxicity, and fertility problems. There are health concerns about the effects of some chemicals on the brain and nervous system. This table shows how many of the targeted chemicals were found in each infant's cord blood.\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td>79<\/td>\r\n<td>145<\/td>\r\n<td>147<\/td>\r\n<td>160<\/td>\r\n<td>116<\/td>\r\n<td>100<\/td>\r\n<td>159<\/td>\r\n<td>151<\/td>\r\n<td>156<\/td>\r\n<td>126<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>137<\/td>\r\n<td>83<\/td>\r\n<td>156<\/td>\r\n<td>94<\/td>\r\n<td>121<\/td>\r\n<td>144<\/td>\r\n<td>123<\/td>\r\n<td>114<\/td>\r\n<td>139<\/td>\r\n<td>99<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nUse this sample data to construct a 90% confidence interval for the mean number of targeted industrial chemicals to be found in an in infant's blood.\r\n\r\nThe first solution is step-by-step.\r\n[reveal-answer q=\"583137\"]Show Solution A[\/reveal-answer]\r\n[hidden-answer a=\"583137\"]\r\n<p class=\"p1\">From the sample, you can calculate [latex]\\displaystyle\\overline{x}[\/latex] = 127.45 and <em data-redactor-tag=\"em\">s<\/em> = 25.965. There are 20 infants in the sample, so <em data-redactor-tag=\"em\">n<\/em> = 20, and <em data-redactor-tag=\"em\">df<\/em> = 20 \u2013 1 = 19.<\/p>\r\nYou are asked to calculate a 90% confidence interval: CL = 0.90, so [latex]\\displaystyle\\alpha[\/latex] = 1 - CL = 1 - 0.90 = 0.10\r\n<p class=\"p1\"><span class=\"s1\">[latex]\\displaystyle\\frac{{\\alpha}}{{2}}[\/latex] = 0.05<\/span><\/p>\r\n<p class=\"p1\">[latex]\\displaystyle{t}_{\\frac{{\\alpha}}{{2}}}={t}_{0.05}[\/latex]<\/p>\r\nBy definition, the area to the right of <em data-redactor-tag=\"em\">t<\/em><sub data-redactor-tag=\"sub\">0.05<\/sub> is 0.05 and so the area to the left of <em data-redactor-tag=\"em\">t<\/em><sub data-redactor-tag=\"sub\">0.05<\/sub> is 1 \u2013 0.05 = 0.95.\r\n\r\nUse a table, calculator, or computer to find that <em data-redactor-tag=\"em\">t<\/em><sub data-redactor-tag=\"sub\">0.05<\/sub> = 1.729.\r\n\r\n<em>EBM<\/em> =\u00a0[latex]\\displaystyle{t}_{\\frac{{\\alpha}}{{2}}}(\\frac{{s}}{{\\sqrt{n}}})[\/latex] = 1.729 ([latex]\\displaystyle\\frac{{25.965}}{{\\sqrt{20}}}) [\/latex] = 10.038\r\n\r\n[latex]\\displaystyle\\overline{x}[\/latex] - EBM = 127.45 - 10.038 = 117.412\r\n\r\n[latex]\\displaystyle\\overline{x}[\/latex] + EBM =\u00a0127.45 + 10.038 = 137.488\r\n\r\nWe estimate with 90% confidence that the mean number of all targeted industrial chemicals found in cord blood in the United States is between 117.412 and 137.488.\r\n\r\n[\/hidden-answer]\r\n\r\n&nbsp;\r\n\r\nThe second solution uses the TI-83+ and TI-84 calculators.\r\n[reveal-answer q=\"738187\"]Show Solution B[\/reveal-answer]\r\n[hidden-answer a=\"738187\"]\r\n\r\n<header>\r\n<h4 class=\"os-title\" data-type=\"title\"><span class=\"os-title-label\">USING THE TI-83, 83+, 84, 84+ CALCULATOR<\/span><\/h4>\r\n<\/header><section>\r\n<div class=\"os-note-body\"><\/div>\r\n<\/section>Enter the data as a list. Press <code data-redactor-tag=\"code\">STAT<\/code> and arrow over to <code data-redactor-tag=\"code\">TESTS<\/code>.Arrow down to <code data-redactor-tag=\"code\">8:TInterval<\/code> and press <code data-redactor-tag=\"code\">ENTER<\/code> (or you can just press\u00a0<code data-redactor-tag=\"code\">8<\/code>). Arrow to Data and press\u00a0<code data-redactor-tag=\"code\">ENTER<\/code>.\r\n\r\nArrow down to <code data-redactor-tag=\"code\">List<\/code> and enter the list name where you put the data.\r\n\r\nArrow down to <code data-redactor-tag=\"code\">Freq<\/code> and enter 1.\r\n\r\nArrow down to <code data-redactor-tag=\"code\">C-level<\/code> and enter 0.90\r\n\r\nArrow down to <code data-redactor-tag=\"code\">Calculate<\/code> and press <code data-redactor-tag=\"code\">ENTER<\/code>.\r\n\r\nThe 90% confidence interval is (117.41, 137.49).\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h3><\/h3>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It 2<\/h3>\r\nA random sample of statistics students were asked to estimate the total number of hours they spend watching television in an average week. The responses are recorded in this table. Use this sample data to construct a 98% confidence interval for the mean number of hours statistics students will spend watching television in one week.\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td>0<\/td>\r\n<td>3<\/td>\r\n<td>1<\/td>\r\n<td>20<\/td>\r\n<td>9<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>5<\/td>\r\n<td>10<\/td>\r\n<td>1<\/td>\r\n<td>10<\/td>\r\n<td>4<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>14<\/td>\r\n<td>2<\/td>\r\n<td>4<\/td>\r\n<td>4<\/td>\r\n<td>5<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nThe first solution is step-by-step.\r\n[reveal-answer q=\"363487\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"363487\"]\r\n<p class=\"p1\"><span class=\"s1\">[latex]\\displaystyle\\overline{x}[\/latex] = 6.133, s = 5.514, n = 15, and df = 15-1=14<\/span><\/p>\r\n<p class=\"p1\">CL = 0.98, so [latex]\\displaystyle\\alpha[\/latex] = 1 - CL = 1.0.98 = 0.02<\/p>\r\n<p class=\"p1\">[latex]\\displaystyle{t}_{\\frac{{\\alpha}}{{2}}}={t}_{0.01}[\/latex]<\/p>\r\n<p class=\"p1\">[latex]\\displaystyle{t}_{\\frac{{\\alpha}}{{2}}}={t}_{0.01}={2.624}[\/latex]<\/p>\r\n<em>EBM<\/em> =\u00a0[latex]\\displaystyle{t}_{\\frac{{\\alpha}}{{2}}}(\\frac{{s}}{{\\sqrt{n}}})={2.624}(\\frac{{5.514}}{{\\sqrt{15}}})={3.736}[\/latex]\r\n\r\n[latex]\\displaystyle\\overline{x}[\/latex] - EBM = 6.133\u00a0- 3.736 = 2.397\r\n\r\n[latex]\\displaystyle\\overline{x}[\/latex] + EBM =\u00a016.133\u00a0-+3.736 =\u00a09.869\r\n\r\nWe estimate with 98% confidence that the mean number of all hours that statistics students spend watching television in one week is between 2.397 and 9.869.\r\n\r\n[\/hidden-answer]\r\n\r\n&nbsp;\r\n\r\nThe second solution uses the TI-83+ and TI-84 calculators.\r\n[reveal-answer q=\"621578\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"621578\"]\r\n\r\n<header>\r\n<h4 class=\"os-title\" data-type=\"title\"><span class=\"os-title-label\">USING THE TI-83, 83+, 84, 84+ CALCULATOR<\/span><\/h4>\r\n<\/header>Enter the data as a list.\r\n\r\nPress\u00a0<code data-redactor-tag=\"code\">STAT<\/code> and arrow over to <code data-redactor-tag=\"code\">TESTS<\/code>. Arrow down to\u00a0<code data-redactor-tag=\"code\">8:TInterval<\/code>.\r\n\r\nPress\u00a0<code data-redactor-tag=\"code\">ENTER<\/code>.\r\n\r\nArrow to\u00a0<code data-redactor-tag=\"code\">Data<\/code> and press\u00a0<code data-redactor-tag=\"code\">ENTER<\/code>.\r\n\r\nArrow down and enter the name of the list where the data is stored.\r\n\r\nEnter\u00a0<code data-redactor-tag=\"code\">Freq<\/code>: 1Enter <code data-redactor-tag=\"code\">C-Level<\/code>: 0.98\r\n\r\nArrow down to\u00a0<code data-redactor-tag=\"code\">Calculate<\/code>and press\u00a0<code data-redactor-tag=\"code\">Enter<\/code>.\r\n\r\nThe 98% confidence interval is (2.3965, 9,8702).\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<section>\n<ul id=\"list12315\">\n<li>Using the formula for creating a confidence interval or technology, construct a confidence interval for a population mean based on a t-distribution<\/li>\n<\/ul>\n<\/section>\n<\/div>\n<p>In practice, we rarely know the population <strong>standard deviation<\/strong>. In the past, when the sample size was large, this did not present a problem to statisticians. They used the sample standard deviation <em>s<\/em> as an estimate for <em>\u03c3\u00a0<\/em>and proceeded as before to calculate a confidence interval with close enough results. However, statisticians ran into problems when the sample size was small. A small sample size caused inaccuracies in the confidence interval.<\/p>\n<p>William S. Goset (1876\u20131937) of the Guinness brewery in Dublin, Ireland ran into this problem. His experiments with hops and barley produced very few samples. Just replacing <em>\u03c3<\/em> with <em>s<\/em> did not produce accurate results when he tried to calculate a confidence interval. He realized that he could not use a normal distribution for the calculation; he found that the actual distribution depends on the sample size. This problem led him to &#8220;discover&#8221; what is called the <strong>Student&#8217;s t-distribution<\/strong>. The name comes from the fact that Gosset wrote under the pen name &#8220;Student.&#8221;<\/p>\n<p>Up until the mid-1970s, some statisticians used the <strong>normal distribution <\/strong>approximation for large sample sizes and only used the Student&#8217;s t-distribution only for sample sizes of at most 30. With graphing calculators and computers, the practice now is to use the Student&#8217;s t-distribution whenever <em>s\u00a0<\/em>is used as an estimate for <em>\u03c3<\/em>.<\/p>\n<p>If you draw a simple random sample of size <em>n<\/em> from a population that has an approximately a normal distribution with mean <em>\u03bc<\/em> and unknown population standard deviation <em>\u03c3<\/em> and calculate the <em>t<\/em>-score: [latex]\\displaystyle{t}=\\dfrac{{\\overline{{x}}-\\mu}}{{\\frac{{s}}{\\sqrt{{n}}}}}[\/latex] is from its mean <em>\u03bc<\/em>. For each sample size <em>n<\/em>, there is a different Student&#8217;s t-distribution.<\/p>\n<p>The <strong>degrees of freedom, <em data-redactor-tag=\"em\">n<\/em> \u2013 1<\/strong>, come from the calculation of the sample standard deviation <em>s<\/em>. Because the sum of the deviations is zero, we can find the last deviation once we know the other <em>n<\/em> \u2013 1 deviations. The other <strong><em>n<\/em> \u2013 1<\/strong> deviations can change or vary freely. <strong>We call the number <em data-redactor-tag=\"em\">n<\/em> \u2013 1 the degrees of freedom (df)<\/strong>.<\/p>\n<h3>Properties of the Student&#8217;s t-Distribution<\/h3>\n<ul>\n<li>The graph for the Student&#8217;s t-distribution is similar to the standard normal curve.<\/li>\n<li>The mean for the Student&#8217;s t-distribution is zero and the distribution is symmetric about zero.<\/li>\n<li>The Student&#8217;s t-distribution has more probability in its tails than the standard normal distribution because the spread of the t-distribution is greater than the spread of the standard normal. So the graph of the Student&#8217;s t-distribution will be thicker in the tails and shorter in the center than the graph of the standard normal distribution.<\/li>\n<li>The exact shape of the Student&#8217;s t-distribution depends on the degrees of freedom. As the degrees of freedom increases, the graph of Student&#8217;s t-distribution becomes more like the graph of the standard normal distribution.<\/li>\n<li>The underlying population of individual observations is assumed to be normally distributed with unknown population mean <em>\u03bc<\/em> and unknown population standard deviation <em>\u03c3<\/em>. The size of the underlying population is generally not relevant unless it is very small. If it is bell shaped (normal) then the assumption is met and doesn&#8217;t need discussion. Random sampling is assumed, but that is a completely separate assumption from normality.<\/li>\n<\/ul>\n<p>Calculators and computers can easily calculate any Student&#8217;s t-probabilities. The TI-83,83+, and 84+ have a tcdf function to find the probability for given values of <em>t<\/em>. The grammar for the tcdf command is tcdf(lower bound, upper bound, degrees of freedom). However for confidence intervals, we need to use <strong>inverse<\/strong> probability to find the value of <em>t<\/em> when we know the probability.<\/p>\n<p>For the TI-84+ you can use the invT command on the DISTRibution menu. The invT command works similarly to the invnorm. The invT command requires two inputs: <strong>invT(area to the left, degrees of freedom)<\/strong> The output is the t-score that corresponds to the area we specified.<\/p>\n<p>The TI-83 and 83+ do not have the invT command. (The TI-89 has an inverse T command.)<\/p>\n<p>A probability table for the Student&#8217;s t-distribution can also be used. The table gives t-scores that correspond to the confidence level (column) and degrees of freedom (row). (The TI-86 does not have an invT program or command, so if you are using that calculator, you need to use a probability table for the Student&#8217;s t-distribution.) When using a <em>t<\/em>-table, note that some tables are formatted to show the confidence level in the column headings, while the column headings in some tables may show only corresponding area in one or both tails.<\/p>\n<p>A <a href=\"http:\/\/www.itl.nist.gov\/div898\/handbook\/eda\/section3\/eda3672.htm\" target=\"_blank\" rel=\"noopener\">Student&#8217;s t table<\/a> gives <em>t<\/em>-scores given the degrees of freedom and the right-tailed probability. The table is very limited. <strong>Calculators and computers can easily calculate any Student&#8217;s t-probabilities.<\/strong><\/p>\n<p><strong>The notation for the Student&#8217;s t-distribution (using <em>T<\/em> as the random variable) is:<\/strong><\/p>\n<ul>\n<li><em>T ~ t<sub data-redactor-tag=\"sub\">df<\/sub><\/em> where <em>df<\/em> = <em>n<\/em> \u2013 1.<\/li>\n<li>For example, if we have a sample of size <em>n<\/em> = 20 items, then we calculate the degrees of freedom as <em>df<\/em> = <em>n<\/em> &#8211; 1 = 20 &#8211; 1 = 19 and we write the distribution as <em>T ~ t<sub data-redactor-tag=\"sub\">19<\/sub><\/em>.<\/li>\n<\/ul>\n<p><strong>If the population standard deviation is not known, the error bound for a population mean is:<\/strong><\/p>\n<ul>\n<li>EBM = [latex]\\displaystyle({t}_{\\frac{{\\alpha}}{{2}}})(\\frac{{s}}{{\\sqrt{n}}})[\/latex],<\/li>\n<li>[latex]\\displaystyle{t}_{\\frac{{\\alpha}}{{2}}}[\/latex] is the t-score with area to the right equal to[latex]\\displaystyle\\frac{{\\alpha}}{{2}}[\/latex],<\/li>\n<li>use <em data-redactor-tag=\"em\">df<\/em> = <em data-redactor-tag=\"em\">n<\/em> \u2013 1 degrees of freedom, and<\/li>\n<li><em data-redactor-tag=\"em\">s<\/em> = sample standard deviation.<\/li>\n<\/ul>\n<p><strong>The format for the confidence interval is:<\/strong><\/p>\n<p class=\"p1\"><span class=\"s1\">([latex]\\displaystyle\\overline{x}[\/latex] &#8211; EBM, [latex]\\displaystyle\\overline{x}[\/latex] + EBM).<\/span><\/p>\n<header>\n<h3 class=\"os-title\" data-type=\"title\"><span class=\"os-title-label\">USING THE TI-83, 83+, 84, 84+ CALCULATOR<\/span><\/h3>\n<\/header>\n<p>To calculate the confidence interval directly:<\/p>\n<ul>\n<li>Press STAT.<\/li>\n<li>Arrow over to TESTS.<\/li>\n<li>Arrow down to 8:TInterval and press ENTER (or just press 8).<\/li>\n<\/ul>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Confidence Intervals for One Mean:  Sigma Not Known (t Method)\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/bFefxSE5bmo?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<div class=\"textbox exercises\">\n<h3>Example 1<\/h3>\n<p>Suppose you do a study of acupuncture to determine how effective it is in relieving pain. You measure sensory rates for 15 subjects with the results given. Use the sample data to construct a 95% confidence interval for the mean sensory rate for the population (assumed normal) from which you took the data.<\/p>\n<p>The solution is shown step-by-step and by using the TI-83, 83+, or 84+ calculators.<\/p>\n<p>8.6 9.4 7.9 6.8 8.3 7.3 9.2 9.6 8.7 11.4 10.3 5.4 8.1 5.5 6.9<\/p>\n<p>The first solution is step-by-step.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q776542\">Show Solution A<\/span><\/p>\n<div id=\"q776542\" class=\"hidden-answer\" style=\"display: none\">To find the confidence interval, you need the sample mean, [latex]{\\overline{x}}[\/latex], and the <em data-redactor-tag=\"em\">EBM<\/em>.<\/p>\n<p class=\"p1\"><span class=\"s1\">[latex]\\displaystyle\\overline{x}[\/latex] = 8.2267<\/span><\/p>\n<p class=\"p1\">s = 1.6722<\/p>\n<p class=\"p1\">n = 15<\/p>\n<p class=\"p1\"><em data-redactor-tag=\"em\">df<\/em> = 15 \u2013 1 = 14 <em data-redactor-tag=\"em\">CL<\/em> so <em data-redactor-tag=\"em\">\u03b1<\/em> = 1 \u2013 <em data-redactor-tag=\"em\">CL<\/em> = 1 \u2013 0.95 = 0.05<\/p>\n<p class=\"p1\"><span class=\"s1\">[latex]\\displaystyle\\frac{{\\alpha}}{{2}}[\/latex] = 0.025<\/span><\/p>\n<p class=\"p1\"><span class=\"s1\">[latex]\\displaystyle{t}_{\\frac{{\\alpha}}{{2}}}[\/latex] = t<sub>0.025<\/sub><\/span><\/p>\n<p class=\"p1\">The area to the right of t<sub>0.025<\/sub> is 0.025, and the area to the left of\u00a0t<sub>0.025<\/sub> is\u00a01 \u2013 0.025 = 0.975<\/p>\n<p class=\"p1\">[latex]\\displaystyle{t}_{\\frac{{\\alpha}}{{2}}}[\/latex] = t<sub>0.025<\/sub> = 2.14 using invT(.975,14) on the TI-84+ calculator<\/p>\n<p class=\"p1\"><span class=\"s1\"><em>EBM<\/em> =\u00a0[latex]\\displaystyle({t}_{\\frac{{\\alpha}}{{2}}})(\\frac{{s}}{{\\sqrt{n}}})[\/latex]<\/span><\/p>\n<p class=\"p1\"><em>EBM<\/em> = (2.14)[latex](\\frac{{1.6722}}{{\\sqrt{15}}})[\/latex] = 0.924<\/p>\n<p class=\"p1\">[latex]\\displaystyle\\overline{x}[\/latex] &#8211; EBM = 8.2267 &#8211; 0.9240 = 7.3<\/p>\n<p class=\"p1\">[latex]\\displaystyle\\overline{x}[\/latex] + EBM = 8.2267 + 0.9240 = 9.15<\/p>\n<p>The 95% confidence interval is (7.30, 9.15).<\/p>\n<p>We estimate with 95% confidence that the true population mean sensory rate is between 7.30 and 9.15.<\/p>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<p>The second solution uses the TI-83+ and TI-84 calculators.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q496358\">Show Solution B<\/span><\/p>\n<div id=\"q496358\" class=\"hidden-answer\" style=\"display: none\">\n<p>Press <code data-redactor-tag=\"code\">STAT<\/code> and arrow over to\u00a0<code data-redactor-tag=\"code\">TESTS<\/code>.<br \/>\nArrow down to <code data-redactor-tag=\"code\">8:TInterval<\/code> and press <code data-redactor-tag=\"code\">ENTER<\/code> (or you can just press\u00a0<code data-redactor-tag=\"code\">8<\/code>).<br \/>\nArrow to <code data-redactor-tag=\"code\">Data<\/code> and press <code data-redactor-tag=\"code\">ENTER<\/code>.<br \/>\nArrow down to <code data-redactor-tag=\"code\">List<\/code> and enter the list name where you put the data.<br \/>\nThere should be a 1 after <code data-redactor-tag=\"code\">Freq<\/code>.<br \/>\nArrow down to <code data-redactor-tag=\"code\">C-level<\/code> and enter 0.95<br \/>\nArrow down to <code data-redactor-tag=\"code\">Calculate<\/code> and press <code data-redactor-tag=\"code\">ENTER<\/code>.<br \/>\nThe 95% confidence interval is (7.3006, 9.1527)<\/div>\n<\/div>\n<p><strong data-redactor-tag=\"strong\">Note: <\/strong>When calculating the error bound, a probability table for the Student&#8217;s t-distribution can also be used to find the value of <em data-redactor-tag=\"em\">t<\/em>. The table gives\u00a0<em data-redactor-tag=\"em\">t<\/em>-scores that correspond to the confidence level (column) and degrees of freedom (row); the <em data-redactor-tag=\"em\">t<\/em>-score is found where the row and column intersect in the table.<\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it 1<\/h3>\n<p>You do a study of hypnotherapy to determine how effective it is in increasing the number of hourse of sleep subjects get each night. You measure hours of sleep for 12 subjects with the following results. Construct a 95% confidence interval for the mean number of hours slept for the population (assumed normal) from which you took the data.<\/p>\n<p>8.2; 9.1; 7.7; 8.6; 6.9; 11.2; 10.1; 9.9; 8.9; 9.2; 7.5; 10.5<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q742079\">Show Answer<\/span><\/p>\n<div id=\"q742079\" class=\"hidden-answer\" style=\"display: none\">\n<p>(8.1634, 9.8032)<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h3><\/h3>\n<div class=\"textbox exercises\">\n<h3>Example 2<\/h3>\n<p>The Human Toxome Project (HTP) is working to understand the scope of industrial pollution in the human body. Industrial chemicals may enter the body through pollution or as ingredients in consumer products. In October 2008, the scientists at HTP tested cord blood samples for 20 newborn infants in the United States. The cord blood of the &#8220;In utero\/newborn&#8221; group was tested for 430 industrial compounds, pollutants, and other chemicals, including chemicals linked to brain and nervous system toxicity, immune system toxicity, reproductive toxicity, and fertility problems. There are health concerns about the effects of some chemicals on the brain and nervous system. This table shows how many of the targeted chemicals were found in each infant&#8217;s cord blood.<\/p>\n<table>\n<tbody>\n<tr>\n<td>79<\/td>\n<td>145<\/td>\n<td>147<\/td>\n<td>160<\/td>\n<td>116<\/td>\n<td>100<\/td>\n<td>159<\/td>\n<td>151<\/td>\n<td>156<\/td>\n<td>126<\/td>\n<\/tr>\n<tr>\n<td>137<\/td>\n<td>83<\/td>\n<td>156<\/td>\n<td>94<\/td>\n<td>121<\/td>\n<td>144<\/td>\n<td>123<\/td>\n<td>114<\/td>\n<td>139<\/td>\n<td>99<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Use this sample data to construct a 90% confidence interval for the mean number of targeted industrial chemicals to be found in an in infant&#8217;s blood.<\/p>\n<p>The first solution is step-by-step.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q583137\">Show Solution A<\/span><\/p>\n<div id=\"q583137\" class=\"hidden-answer\" style=\"display: none\">\n<p class=\"p1\">From the sample, you can calculate [latex]\\displaystyle\\overline{x}[\/latex] = 127.45 and <em data-redactor-tag=\"em\">s<\/em> = 25.965. There are 20 infants in the sample, so <em data-redactor-tag=\"em\">n<\/em> = 20, and <em data-redactor-tag=\"em\">df<\/em> = 20 \u2013 1 = 19.<\/p>\n<p>You are asked to calculate a 90% confidence interval: CL = 0.90, so [latex]\\displaystyle\\alpha[\/latex] = 1 &#8211; CL = 1 &#8211; 0.90 = 0.10<\/p>\n<p class=\"p1\"><span class=\"s1\">[latex]\\displaystyle\\frac{{\\alpha}}{{2}}[\/latex] = 0.05<\/span><\/p>\n<p class=\"p1\">[latex]\\displaystyle{t}_{\\frac{{\\alpha}}{{2}}}={t}_{0.05}[\/latex]<\/p>\n<p>By definition, the area to the right of <em data-redactor-tag=\"em\">t<\/em><sub data-redactor-tag=\"sub\">0.05<\/sub> is 0.05 and so the area to the left of <em data-redactor-tag=\"em\">t<\/em><sub data-redactor-tag=\"sub\">0.05<\/sub> is 1 \u2013 0.05 = 0.95.<\/p>\n<p>Use a table, calculator, or computer to find that <em data-redactor-tag=\"em\">t<\/em><sub data-redactor-tag=\"sub\">0.05<\/sub> = 1.729.<\/p>\n<p><em>EBM<\/em> =\u00a0[latex]\\displaystyle{t}_{\\frac{{\\alpha}}{{2}}}(\\frac{{s}}{{\\sqrt{n}}})[\/latex] = 1.729 ([latex]\\displaystyle\\frac{{25.965}}{{\\sqrt{20}}})[\/latex] = 10.038<\/p>\n<p>[latex]\\displaystyle\\overline{x}[\/latex] &#8211; EBM = 127.45 &#8211; 10.038 = 117.412<\/p>\n<p>[latex]\\displaystyle\\overline{x}[\/latex] + EBM =\u00a0127.45 + 10.038 = 137.488<\/p>\n<p>We estimate with 90% confidence that the mean number of all targeted industrial chemicals found in cord blood in the United States is between 117.412 and 137.488.<\/p>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<p>The second solution uses the TI-83+ and TI-84 calculators.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q738187\">Show Solution B<\/span><\/p>\n<div id=\"q738187\" class=\"hidden-answer\" style=\"display: none\">\n<header>\n<h4 class=\"os-title\" data-type=\"title\"><span class=\"os-title-label\">USING THE TI-83, 83+, 84, 84+ CALCULATOR<\/span><\/h4>\n<\/header>\n<section>\n<div class=\"os-note-body\"><\/div>\n<\/section>\n<p>Enter the data as a list. Press <code data-redactor-tag=\"code\">STAT<\/code> and arrow over to <code data-redactor-tag=\"code\">TESTS<\/code>.Arrow down to <code data-redactor-tag=\"code\">8:TInterval<\/code> and press <code data-redactor-tag=\"code\">ENTER<\/code> (or you can just press\u00a0<code data-redactor-tag=\"code\">8<\/code>). Arrow to Data and press\u00a0<code data-redactor-tag=\"code\">ENTER<\/code>.<\/p>\n<p>Arrow down to <code data-redactor-tag=\"code\">List<\/code> and enter the list name where you put the data.<\/p>\n<p>Arrow down to <code data-redactor-tag=\"code\">Freq<\/code> and enter 1.<\/p>\n<p>Arrow down to <code data-redactor-tag=\"code\">C-level<\/code> and enter 0.90<\/p>\n<p>Arrow down to <code data-redactor-tag=\"code\">Calculate<\/code> and press <code data-redactor-tag=\"code\">ENTER<\/code>.<\/p>\n<p>The 90% confidence interval is (117.41, 137.49).<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h3><\/h3>\n<div class=\"textbox key-takeaways\">\n<h3>Try It 2<\/h3>\n<p>A random sample of statistics students were asked to estimate the total number of hours they spend watching television in an average week. The responses are recorded in this table. Use this sample data to construct a 98% confidence interval for the mean number of hours statistics students will spend watching television in one week.<\/p>\n<table>\n<tbody>\n<tr>\n<td>0<\/td>\n<td>3<\/td>\n<td>1<\/td>\n<td>20<\/td>\n<td>9<\/td>\n<\/tr>\n<tr>\n<td>5<\/td>\n<td>10<\/td>\n<td>1<\/td>\n<td>10<\/td>\n<td>4<\/td>\n<\/tr>\n<tr>\n<td>14<\/td>\n<td>2<\/td>\n<td>4<\/td>\n<td>4<\/td>\n<td>5<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>The first solution is step-by-step.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q363487\">Show Answer<\/span><\/p>\n<div id=\"q363487\" class=\"hidden-answer\" style=\"display: none\">\n<p class=\"p1\"><span class=\"s1\">[latex]\\displaystyle\\overline{x}[\/latex] = 6.133, s = 5.514, n = 15, and df = 15-1=14<\/span><\/p>\n<p class=\"p1\">CL = 0.98, so [latex]\\displaystyle\\alpha[\/latex] = 1 &#8211; CL = 1.0.98 = 0.02<\/p>\n<p class=\"p1\">[latex]\\displaystyle{t}_{\\frac{{\\alpha}}{{2}}}={t}_{0.01}[\/latex]<\/p>\n<p class=\"p1\">[latex]\\displaystyle{t}_{\\frac{{\\alpha}}{{2}}}={t}_{0.01}={2.624}[\/latex]<\/p>\n<p><em>EBM<\/em> =\u00a0[latex]\\displaystyle{t}_{\\frac{{\\alpha}}{{2}}}(\\frac{{s}}{{\\sqrt{n}}})={2.624}(\\frac{{5.514}}{{\\sqrt{15}}})={3.736}[\/latex]<\/p>\n<p>[latex]\\displaystyle\\overline{x}[\/latex] &#8211; EBM = 6.133\u00a0&#8211; 3.736 = 2.397<\/p>\n<p>[latex]\\displaystyle\\overline{x}[\/latex] + EBM =\u00a016.133\u00a0-+3.736 =\u00a09.869<\/p>\n<p>We estimate with 98% confidence that the mean number of all hours that statistics students spend watching television in one week is between 2.397 and 9.869.<\/p>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<p>The second solution uses the TI-83+ and TI-84 calculators.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q621578\">Show Answer<\/span><\/p>\n<div id=\"q621578\" class=\"hidden-answer\" style=\"display: none\">\n<header>\n<h4 class=\"os-title\" data-type=\"title\"><span class=\"os-title-label\">USING THE TI-83, 83+, 84, 84+ CALCULATOR<\/span><\/h4>\n<\/header>\n<p>Enter the data as a list.<\/p>\n<p>Press\u00a0<code data-redactor-tag=\"code\">STAT<\/code> and arrow over to <code data-redactor-tag=\"code\">TESTS<\/code>. Arrow down to\u00a0<code data-redactor-tag=\"code\">8:TInterval<\/code>.<\/p>\n<p>Press\u00a0<code data-redactor-tag=\"code\">ENTER<\/code>.<\/p>\n<p>Arrow to\u00a0<code data-redactor-tag=\"code\">Data<\/code> and press\u00a0<code data-redactor-tag=\"code\">ENTER<\/code>.<\/p>\n<p>Arrow down and enter the name of the list where the data is stored.<\/p>\n<p>Enter\u00a0<code data-redactor-tag=\"code\">Freq<\/code>: 1Enter <code data-redactor-tag=\"code\">C-Level<\/code>: 0.98<\/p>\n<p>Arrow down to\u00a0<code data-redactor-tag=\"code\">Calculate<\/code>and press\u00a0<code data-redactor-tag=\"code\">Enter<\/code>.<\/p>\n<p>The 98% confidence interval is (2.3965, 9,8702).<\/p>\n<\/div>\n<\/div>\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-272\">\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>A Single Population Mean using the Student t Distribution. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/books\/introductory-statistics\/pages\/8-2-a-single-population-mean-using-the-student-t-distribution\">https:\/\/openstax.org\/books\/introductory-statistics\/pages\/8-2-a-single-population-mean-using-the-student-t-distribution<\/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><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 class=\"license-attribution-dropdown-subheading\">All rights reserved content<\/div><ul class=\"citation-list\"><li>Confidence Intervals for One Mean: Sigma Not Known (t Method). <strong>Authored by<\/strong>: jbstatistics. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/bFefxSE5bmo\">https:\/\/youtu.be\/bFefxSE5bmo<\/a>. <strong>License<\/strong>: <em>All Rights Reserved<\/em>. <strong>License Terms<\/strong>: Standard YouTube License<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t 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