{"id":5356,"date":"2022-08-19T22:21:50","date_gmt":"2022-08-19T22:21:50","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/?post_type=chapter&#038;p=5356"},"modified":"2022-08-19T22:26:45","modified_gmt":"2022-08-19T22:26:45","slug":"11c-preview","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/chapter\/11c-preview\/","title":{"raw":"11C Preview","rendered":"11C Preview"},"content":{"raw":"Preparing for the next class\r\n\r\nIn the next in-class activity, you will need to be able to describe what a P-value measures and identify how a P-value is represented in a statistical distribution.\r\n\r\nYou recently learned that a test statistic is used in hypothesis testing to help measure evidence against a null hypothesis. A test statistic allows us to take standard error into account when considering the evidence provided by a sample statistic. However, recall that the test statistic comes from one observed sample of data. Based on your knowledge of sampling distributions, you know that there are lots of different samples that could have been obtained. Each sample would have its own test statistic.\r\n\r\nIn In-Class Activity 11.A, we learned that the evidence used in hypothesis testing is probability. The statistical evidence that we gather is always evidence in support of the alternative hypothesis and against the null hypothesis. We ask ourselves the question, \u201cDo we have enough evidence to reject the null hypothesis?\u201d\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Question 1<\/h3>\r\nAccording to a 2021 report published by the Federal Trade Commission (FTC),\u00a0 29.4% of claims to the FTC in 2020 were due to identity theft cases.[footnote]Federal Trade Commission. (2021, February). <em>Consumer sentinel network<\/em>, data book 2020. https:\/\/www.ftc.gov\/system\/files\/documents\/reports\/consumer-sentinel-network-data-book-2020\/csn_annual_data_book_2020.pdf[\/footnote] Suppose that\u00a0 in the state of Florida, a random sample of 500 claims to the FTC are observed.\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>Based on the published national percentage of 29.4%, about how many of\u00a0 the 500 Florida claims would you expect to be due to identity theft?\r\nHint: What is 29.4% of 500?<\/li>\r\n \t<li>What is the null hypothesis value of [latex] p [\/latex]?<\/li>\r\n \t<li>In this example, we meet the following conditions for a one-sample z-test for\u00a0 proportions.\r\n<div class=\"textbox\">\r\n\r\nConditions for One-Sample Z-Test for Proportions *MISSING LATEX*\r\n<ol>\r\n \t<li>Large Counts: Check that 10 and\u00a0 10.<\/li>\r\n \t<li>Random Samples\/Assignment: Check that the samples are random samples.<\/li>\r\n \t<li>10% Population Size: Check that the sample size, , is less than 10% of the population size, : 0.10.<\/li>\r\n<\/ol>\r\n<\/div>\r\nThus, we can use the normal distribution to model the values of the sample proportion that would occur if the null hypothesis is true.\r\n\r\nWhat are the mean and standard deviation of the null distribution? Round to the nearest ten thousandth.\r\nHint: Recall that the mean of all sample proportions is calculated by\u00a0[latex] \\mu= p [\/latex] and the standard deviation of the sample proportions is [latex] \\sigma =\\sqrt{\\frac{1-p}{n}} [\/latex'<\/li>\r\n \t<li>Consider the sampling distribution of the sample proportion. In In-Class Activity 11.B, we used the Empirical Rule to determine if a test statistic was \u201cunusual.\u201dUsing the Empirical Rule, what value(s) of the test statistic would be \u201cunusual?\u201dHint: When is a z-score considered \u201cunusual?\u201d<\/li>\r\n<\/ol>\r\n<\/div>\r\nBased on the answers to Parts A and B, we would expect about 29.5% (147) of the\u00a0 claims to the FTC in Florida to be because of identity theft. What if 29.6% (148) were\u00a0 due to identity theft? Or 29.8% (149)?\r\n\r\nAt what point does the number seem unusual? That is, at what point does it appear that\u00a0 Florida identity theft complaints surpass that of the national percentage? How do we\u00a0 measure that?\r\n\r\nIn statistical testing, we define the P-value as the probability of obtaining a test statistic\u00a0 at least as extreme (in the direction of the alternative hypothesis) as the one that is actually seen if the null hypothesis is true. That is, a P-value answers the question:\r\n\r\n\u201cHow unlikely is the sample data given the null hypothesis is true?\u201d\r\n\r\nIt is important to remember that a P-value is a probability, which means that it is a number between 0 and 1.\r\n<ul>\r\n \t<li aria-level=\"1\">The smaller the P-value is, the more unlikely it is to observe the sample data given the null hypothesis is true. Thus, the evidence against the null hypothesis is stronger and is in favor of the alternative hypothesis.<\/li>\r\n \t<li aria-level=\"1\">The larger the P-value is, the more likely it is to observe the sample data. Thus, the evidence against the null hypothesis is weaker.<\/li>\r\n<\/ul>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Question 2<\/h3>\r\nWhich of the following values could not be a P-value? There may be more than one correct answer.\r\n<ol>\r\n \t<li>a) 0.02<\/li>\r\n \t<li>b) -05<\/li>\r\n \t<li>c) 2\u00d710-3<\/li>\r\n \t<li>d) 1.2<\/li>\r\n \t<li>e) 0.88\r\nHint: Remember that the \u201cP\u201d in P-value stands for probability!<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Question 3<\/h3>\r\nOf the following options, which P-value would provide the strongest evidence against\u00a0 a null hypothesis?\r\n<ol>\r\n \t<li>a) 0.9<\/li>\r\n \t<li>b) 0.5<\/li>\r\n \t<li>c) 0.05<\/li>\r\n \t<li>d) 0.01<\/li>\r\n<\/ol>\r\nHint: Remember that the smaller a P-value is, the stronger the evidence is against\u00a0 the null hypothesis.\r\n\r\n<\/div>\r\nOnce a test statistic has been calculated, we calculate the P-value by using what we\u00a0 know about the distribution of the test statistic. For the test of proportions that meets the\u00a0 sample conditions (like in Question 1), we use the standard normal curve to calculate\u00a0 the P-value as an area under the standard normal curve. Since the P-value provides\u00a0 evidence used in support of the alternative hypothesis, the area we measure depends\u00a0 on the alternative hypothesis.\r\n\r\nAsk yourself: \u201cIf the null is in fact true, how likely are the data that you\u2019ve gathered?\u201d[footnote]Lesson 7.4 - <em>Introduction to hypothesis tests.<\/em> (2021). Skew The Script. Retrieved from https:\/\/skewthescript.org\/7-4[\/footnote]\r\n\r\nQuestions 4\u20136: Go to https:\/\/dcmathpathways.shinyapps.io\/NormalDist\/ and select \u201cFind\u00a0 Probability.\u201d\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Question 4, 5, 6<\/h3>\r\n<div align=\"left\">\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td><\/td>\r\n<td>When the alternative is:<\/td>\r\n<td>The P-value is:<\/td>\r\n<td>The P-value equals:<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>4)<\/td>\r\n<td>Lower-tailed\r\n\r\n([latex]p &lt; [\/latex] null value)<\/td>\r\n<td>How \u201cunlikely\u201d it is that we observed the sample data that resulted in a test statistic of [latex] z=-1.2[\/latex] or lower.\r\n\r\nThe area on the left of the test statistic under the standard normal curve.<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>5)<\/td>\r\n<td>Upper-tailed\r\n\r\n([latex] p &gt;[\/latex] null value)<\/td>\r\n<td>How \u201cunlikely\u201d it is that we observed the sample data that resulted in a test statistic of[latex] z=-1.2 [\/latex] or higher.\r\n\r\nThe area on the right side of the test statistic under the standard normal curve.<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>6)<\/td>\r\n<td>Two-tailed\r\n\r\n([latex] p \\neq [\/latex] null value)<\/td>\r\n<td>How \u201cunlikely\u201d it is that we observed the sample data that resulted in a test statistic of [latex] z=-1.2 [\/latex] or lower OR [latex]z=1.2 [\/latex] or higher.\r\n\r\nThe area on the right of the absolute value of the test statistic and the area on the left of the negative absolute value of the test statistic (i.e., more extreme).<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nHint: Remember that a z-test statistic is from a standard normal distribution, which\u00a0 has a mean of 0 and a standard deviation of 1.\r\n\r\n<\/div>\r\n<\/div>\r\nLooking ahead\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Question 7<\/h3>\r\nIn the scenario discussed in Question 1, suppose that a commission in Florida is\u00a0 asked to investigate the types of claims made to the FTC.\r\n\r\nIf Florida follows the national trend, we determined that in a sample of 500 claims, observing 147 (or 29.4%) due to identity theft would not be unusual. Would 148\u00a0 (29.6%) be unusual? How about 150 (30%)? Or 170 (34%)?\r\n\r\nIf Florida follows the national trend, we determined that in a sample of 500 claims, observing 147 (or 29.4%) due to identity theft would not be unusual. Would 148 (29.6%) be unusual? How about 150 (30%)? Or 170 (34%)?\r\n\r\nAt what point would you start to think that there is convincing evidence that Florida is exceeding the national trend?\r\n\r\nUse the data analysis tool at <a href=\"https:\/\/dcmathpathways.shinyapps.io\/NormalDist\/\">https:\/\/dcmathpathways.shinyapps.io\/NormalDist\/<\/a> to complete the following table and identify a sample proportion which you think is unusual and provides convincing evidence that Florida is exceeding the national trend. Try at least three additional sample proportions.\r\n\r\nNote that in this exercise, if Florida is exceeding the national trend, then we have an upper-tailed test.\r\n<div align=\"left\">\r\n<table style=\"height: 245px;\">\r\n<tbody>\r\n<tr style=\"height: 33px;\">\r\n<td style=\"height: 33px; width: 412.383px;\">Number of complaints due to identity theft (out of 500)<\/td>\r\n<td style=\"height: 33px; width: 252.117px;\">Value of [latex] \\hat{p} [\/latex], the sample proportion<\/td>\r\n<td style=\"height: 33px; width: 153.617px;\">[latex] z = \\frac{\\hat{p}-0.294}{0.0204} [\/latex]<\/td>\r\n<td style=\"height: 33px; width: 54.9833px;\">P-value<\/td>\r\n<td style=\"height: 33px; width: 778.35px;\">Do you think we have convincing evidence to suggest that Florida is exceeding the national trend? Why?<\/td>\r\n<\/tr>\r\n<tr style=\"height: 67px;\">\r\n<td style=\"height: 67px; width: 412.383px;\">148<\/td>\r\n<td style=\"height: 67px; width: 252.117px;\">0.296<\/td>\r\n<td style=\"height: 67px; width: 153.617px;\">[latex] z = \\frac{0.296-0.294}{0.0204} [\/latex]\r\n\r\nz=0.098<\/td>\r\n<td style=\"height: 67px; width: 54.9833px;\">0.461<\/td>\r\n<td style=\"height: 67px; width: 778.35px;\">No, because a sample proportion of 0.296 is not that unlikely given the national trend of 0.294.<\/td>\r\n<\/tr>\r\n<tr style=\"height: 33px;\">\r\n<td style=\"height: 33px; width: 412.383px;\">150<\/td>\r\n<td style=\"height: 33px; width: 252.117px;\">0.3<\/td>\r\n<td style=\"height: 33px; width: 153.617px;\">z= 0.29<\/td>\r\n<td style=\"height: 33px; width: 54.9833px;\">0.3859<\/td>\r\n<td style=\"height: 33px; width: 778.35px;\">No, because a sample proportion of 0.30 is not that unlikely given the national trend of 0.294.<\/td>\r\n<\/tr>\r\n<tr style=\"height: 33px;\">\r\n<td style=\"height: 33px; width: 412.383px;\"><\/td>\r\n<td style=\"height: 33px; width: 252.117px;\"><\/td>\r\n<td style=\"height: 33px; width: 153.617px;\"><\/td>\r\n<td style=\"height: 33px; width: 54.9833px;\"><\/td>\r\n<td style=\"height: 33px; width: 778.35px;\"><\/td>\r\n<\/tr>\r\n<tr style=\"height: 33px;\">\r\n<td style=\"height: 33px; width: 412.383px;\"><\/td>\r\n<td style=\"height: 33px; width: 252.117px;\"><\/td>\r\n<td style=\"height: 33px; width: 153.617px;\"><\/td>\r\n<td style=\"height: 33px; width: 54.9833px;\"><\/td>\r\n<td style=\"height: 33px; width: 778.35px;\"><\/td>\r\n<\/tr>\r\n<tr style=\"height: 46px;\">\r\n<td style=\"height: 46px; width: 412.383px;\"><\/td>\r\n<td style=\"height: 46px; width: 252.117px;\"><\/td>\r\n<td style=\"height: 46px; width: 153.617px;\"><\/td>\r\n<td style=\"height: 46px; width: 54.9833px;\"><\/td>\r\n<td style=\"height: 46px; width: 778.35px;\"><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\n<\/div>\r\n&nbsp;\r\n\r\n&nbsp;","rendered":"<p>Preparing for the next class<\/p>\n<p>In the next in-class activity, you will need to be able to describe what a P-value measures and identify how a P-value is represented in a statistical distribution.<\/p>\n<p>You recently learned that a test statistic is used in hypothesis testing to help measure evidence against a null hypothesis. A test statistic allows us to take standard error into account when considering the evidence provided by a sample statistic. However, recall that the test statistic comes from one observed sample of data. Based on your knowledge of sampling distributions, you know that there are lots of different samples that could have been obtained. Each sample would have its own test statistic.<\/p>\n<p>In In-Class Activity 11.A, we learned that the evidence used in hypothesis testing is probability. The statistical evidence that we gather is always evidence in support of the alternative hypothesis and against the null hypothesis. We ask ourselves the question, \u201cDo we have enough evidence to reject the null hypothesis?\u201d<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Question 1<\/h3>\n<p>According to a 2021 report published by the Federal Trade Commission (FTC),\u00a0 29.4% of claims to the FTC in 2020 were due to identity theft cases.<a class=\"footnote\" title=\"Federal Trade Commission. (2021, February). Consumer sentinel network, data book 2020. https:\/\/www.ftc.gov\/system\/files\/documents\/reports\/consumer-sentinel-network-data-book-2020\/csn_annual_data_book_2020.pdf\" id=\"return-footnote-5356-1\" href=\"#footnote-5356-1\" aria-label=\"Footnote 1\"><sup class=\"footnote\">[1]<\/sup><\/a> Suppose that\u00a0 in the state of Florida, a random sample of 500 claims to the FTC are observed.<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>Based on the published national percentage of 29.4%, about how many of\u00a0 the 500 Florida claims would you expect to be due to identity theft?<br \/>\nHint: What is 29.4% of 500?<\/li>\n<li>What is the null hypothesis value of [latex]p[\/latex]?<\/li>\n<li>In this example, we meet the following conditions for a one-sample z-test for\u00a0 proportions.\n<div class=\"textbox\">\n<p>Conditions for One-Sample Z-Test for Proportions *MISSING LATEX*<\/p>\n<ol>\n<li>Large Counts: Check that 10 and\u00a0 10.<\/li>\n<li>Random Samples\/Assignment: Check that the samples are random samples.<\/li>\n<li>10% Population Size: Check that the sample size, , is less than 10% of the population size, : 0.10.<\/li>\n<\/ol>\n<\/div>\n<p>Thus, we can use the normal distribution to model the values of the sample proportion that would occur if the null hypothesis is true.<\/p>\n<p>What are the mean and standard deviation of the null distribution? Round to the nearest ten thousandth.<br \/>\nHint: Recall that the mean of all sample proportions is calculated by\u00a0[latex]\\mu= p[\/latex] and the standard deviation of the sample proportions is [latex]\\sigma =\\sqrt{\\frac{1-p}{n}} [\/latex'<\/li>\n<li>Consider the sampling distribution of the sample proportion. In In-Class Activity 11.B, we used the Empirical Rule to determine if a test statistic was \u201cunusual.\u201dUsing the Empirical Rule, what value(s) of the test statistic would be \u201cunusual?\u201dHint: When is a z-score considered \u201cunusual?\u201d<\/li>\n<\/ol><\/div>\n<p>  Based on the answers to Parts A and B, we would expect about 29.5% (147) of the\u00a0 claims to the FTC in Florida to be because of identity theft. What if 29.6% (148) were\u00a0 due to identity theft? Or 29.8% (149)?    At what point does the number seem unusual? That is, at what point does it appear that\u00a0 Florida identity theft complaints surpass that of the national percentage? How do we\u00a0 measure that?    In statistical testing, we define the P-value as the probability of obtaining a test statistic\u00a0 at least as extreme (in the direction of the alternative hypothesis) as the one that is actually seen if the null hypothesis is true. That is, a P-value answers the question:    \u201cHow unlikely is the sample data given the null hypothesis is true?\u201d    It is important to remember that a P-value is a probability, which means that it is a number between 0 and 1.  <\/p>\n<ul>\n<li aria-level=\"1\">The smaller the P-value is, the more unlikely it is to observe the sample data given the null hypothesis is true. Thus, the evidence against the null hypothesis is stronger and is in favor of the alternative hypothesis.<\/li>\n<li aria-level=\"1\">The larger the P-value is, the more likely it is to observe the sample data. Thus, the evidence against the null hypothesis is weaker.<\/li>\n<\/ul>\n<div class=\"textbox key-takeaways\">\n<h3>Question 2<\/h3>\n<p>  Which of the following values could not be a P-value? There may be more than one correct answer.  <\/p>\n<ol>\n<li>a) 0.02<\/li>\n<li>b) -05<\/li>\n<li>c) 2\u00d710-3<\/li>\n<li>d) 1.2<\/li>\n<li>e) 0.88  Hint: Remember that the \u201cP\u201d in P-value stands for probability!<\/li>\n<\/ol><\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Question 3<\/h3>\n<p>  Of the following options, which P-value would provide the strongest evidence against\u00a0 a null hypothesis?  <\/p>\n<ol>\n<li>a) 0.9<\/li>\n<li>b) 0.5<\/li>\n<li>c) 0.05<\/li>\n<li>d) 0.01<\/li>\n<\/ol>\n<p>  Hint: Remember that the smaller a P-value is, the stronger the evidence is against\u00a0 the null hypothesis.    <\/p><\/div>\n<p>  Once a test statistic has been calculated, we calculate the P-value by using what we\u00a0 know about the distribution of the test statistic. For the test of proportions that meets the\u00a0 sample conditions (like in Question 1), we use the standard normal curve to calculate\u00a0 the P-value as an area under the standard normal curve. Since the P-value provides\u00a0 evidence used in support of the alternative hypothesis, the area we measure depends\u00a0 on the alternative hypothesis.    Ask yourself: \u201cIf the null is in fact true, how likely are the data that you\u2019ve gathered?\u201d[footnote]Lesson 7.4 - <em>Introduction to hypothesis tests.<\/em> (2021). Skew The Script. Retrieved from https:\/\/skewthescript.org\/7-4[\/footnote]    Questions 4\u20136: Go to https:\/\/dcmathpathways.shinyapps.io\/NormalDist\/ and select \u201cFind\u00a0 Probability.\u201d  <\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Question 4, 5, 6<\/h3>\n<div style=\"text-align: left;\">\n<table>\n<tbody>\n<tr>\n<td><\/td>\n<td>When the alternative is:<\/td>\n<td>The P-value is:<\/td>\n<td>The P-value equals:<\/td>\n<\/tr>\n<tr>\n<td>4)<\/td>\n<td>Lower-tailed    ([latex]p <[\/latex] null value)<\/td>\n<td>How \u201cunlikely\u201d it is that we observed the sample data that resulted in a test statistic of [latex]z=-1.2[\/latex] or lower.<\/p>\n<p>The area on the left of the test statistic under the standard normal curve.<\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td>5)<\/td>\n<td>Upper-tailed<\/p>\n<p>([latex]p >[\/latex] null value)<\/td>\n<td>How \u201cunlikely\u201d it is that we observed the sample data that resulted in a test statistic of[latex]z=-1.2[\/latex] or higher.<\/p>\n<p>The area on the right side of the test statistic under the standard normal curve.<\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td>6)<\/td>\n<td>Two-tailed<\/p>\n<p>([latex]p \\neq[\/latex] null value)<\/td>\n<td>How \u201cunlikely\u201d it is that we observed the sample data that resulted in a test statistic of [latex]z=-1.2[\/latex] or lower OR [latex]z=1.2[\/latex] or higher.<\/p>\n<p>The area on the right of the absolute value of the test statistic and the area on the left of the negative absolute value of the test statistic (i.e., more extreme).<\/td>\n<td><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Hint: Remember that a z-test statistic is from a standard normal distribution, which\u00a0 has a mean of 0 and a standard deviation of 1.<\/p>\n<\/div>\n<\/div>\n<p>Looking ahead<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Question 7<\/h3>\n<p>In the scenario discussed in Question 1, suppose that a commission in Florida is\u00a0 asked to investigate the types of claims made to the FTC.<\/p>\n<p>If Florida follows the national trend, we determined that in a sample of 500 claims, observing 147 (or 29.4%) due to identity theft would not be unusual. Would 148\u00a0 (29.6%) be unusual? How about 150 (30%)? Or 170 (34%)?<\/p>\n<p>If Florida follows the national trend, we determined that in a sample of 500 claims, observing 147 (or 29.4%) due to identity theft would not be unusual. Would 148 (29.6%) be unusual? How about 150 (30%)? Or 170 (34%)?<\/p>\n<p>At what point would you start to think that there is convincing evidence that Florida is exceeding the national trend?<\/p>\n<p>Use the data analysis tool at <a href=\"https:\/\/dcmathpathways.shinyapps.io\/NormalDist\/\">https:\/\/dcmathpathways.shinyapps.io\/NormalDist\/<\/a> to complete the following table and identify a sample proportion which you think is unusual and provides convincing evidence that Florida is exceeding the national trend. Try at least three additional sample proportions.<\/p>\n<p>Note that in this exercise, if Florida is exceeding the national trend, then we have an upper-tailed test.<\/p>\n<div style=\"text-align: left;\">\n<table style=\"height: 245px;\">\n<tbody>\n<tr style=\"height: 33px;\">\n<td style=\"height: 33px; width: 412.383px;\">Number of complaints due to identity theft (out of 500)<\/td>\n<td style=\"height: 33px; width: 252.117px;\">Value of [latex]\\hat{p}[\/latex], the sample proportion<\/td>\n<td style=\"height: 33px; width: 153.617px;\">[latex]z = \\frac{\\hat{p}-0.294}{0.0204}[\/latex]<\/td>\n<td style=\"height: 33px; width: 54.9833px;\">P-value<\/td>\n<td style=\"height: 33px; width: 778.35px;\">Do you think we have convincing evidence to suggest that Florida is exceeding the national trend? Why?<\/td>\n<\/tr>\n<tr style=\"height: 67px;\">\n<td style=\"height: 67px; width: 412.383px;\">148<\/td>\n<td style=\"height: 67px; width: 252.117px;\">0.296<\/td>\n<td style=\"height: 67px; width: 153.617px;\">[latex]z = \\frac{0.296-0.294}{0.0204}[\/latex]<\/p>\n<p>z=0.098<\/td>\n<td style=\"height: 67px; width: 54.9833px;\">0.461<\/td>\n<td style=\"height: 67px; width: 778.35px;\">No, because a sample proportion of 0.296 is not that unlikely given the national trend of 0.294.<\/td>\n<\/tr>\n<tr style=\"height: 33px;\">\n<td style=\"height: 33px; width: 412.383px;\">150<\/td>\n<td style=\"height: 33px; width: 252.117px;\">0.3<\/td>\n<td style=\"height: 33px; width: 153.617px;\">z= 0.29<\/td>\n<td style=\"height: 33px; width: 54.9833px;\">0.3859<\/td>\n<td style=\"height: 33px; width: 778.35px;\">No, because a sample proportion of 0.30 is not that unlikely given the national trend of 0.294.<\/td>\n<\/tr>\n<tr style=\"height: 33px;\">\n<td style=\"height: 33px; width: 412.383px;\"><\/td>\n<td style=\"height: 33px; width: 252.117px;\"><\/td>\n<td style=\"height: 33px; width: 153.617px;\"><\/td>\n<td style=\"height: 33px; width: 54.9833px;\"><\/td>\n<td style=\"height: 33px; width: 778.35px;\"><\/td>\n<\/tr>\n<tr style=\"height: 33px;\">\n<td style=\"height: 33px; width: 412.383px;\"><\/td>\n<td style=\"height: 33px; width: 252.117px;\"><\/td>\n<td style=\"height: 33px; width: 153.617px;\"><\/td>\n<td style=\"height: 33px; width: 54.9833px;\"><\/td>\n<td style=\"height: 33px; width: 778.35px;\"><\/td>\n<\/tr>\n<tr style=\"height: 46px;\">\n<td style=\"height: 46px; width: 412.383px;\"><\/td>\n<td style=\"height: 46px; width: 252.117px;\"><\/td>\n<td style=\"height: 46px; width: 153.617px;\"><\/td>\n<td style=\"height: 46px; width: 54.9833px;\"><\/td>\n<td style=\"height: 46px; width: 778.35px;\"><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<hr class=\"before-footnotes clear\" \/><div class=\"footnotes\"><ol><li id=\"footnote-5356-1\">Federal Trade Commission. (2021, February). <em>Consumer sentinel network<\/em>, data book 2020. https:\/\/www.ftc.gov\/system\/files\/documents\/reports\/consumer-sentinel-network-data-book-2020\/csn_annual_data_book_2020.pdf <a href=\"#return-footnote-5356-1\" class=\"return-footnote\" aria-label=\"Return to footnote 1\">&crarr;<\/a><\/li><\/ol><\/div>","protected":false},"author":574340,"menu_order":9,"template":"","meta":{"_candela_citation":"[]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-5356","chapter","type-chapter","status-publish","hentry"],"part":5305,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/wp-json\/pressbooks\/v2\/chapters\/5356","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/wp-json\/wp\/v2\/users\/574340"}],"version-history":[{"count":4,"href":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/wp-json\/pressbooks\/v2\/chapters\/5356\/revisions"}],"predecessor-version":[{"id":5360,"href":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/wp-json\/pressbooks\/v2\/chapters\/5356\/revisions\/5360"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/wp-json\/pressbooks\/v2\/parts\/5305"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/wp-json\/pressbooks\/v2\/chapters\/5356\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/wp-json\/wp\/v2\/media?parent=5356"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/wp-json\/pressbooks\/v2\/chapter-type?post=5356"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/wp-json\/wp\/v2\/contributor?post=5356"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/wp-json\/wp\/v2\/license?post=5356"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}