{"id":214574,"date":"2025-05-15T04:30:04","date_gmt":"2025-05-15T04:30:04","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=214574"},"modified":"2025-05-15T04:30:07","modified_gmt":"2025-05-15T04:30:07","slug":"the-following-data-represent-the-age-in-weeks-at-which-babies-first-crawl-based-on-a-survey-of-12-mothers-2","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/15\/the-following-data-represent-the-age-in-weeks-at-which-babies-first-crawl-based-on-a-survey-of-12-mothers-2\/","title":{"rendered":"The following data represent the age (in weeks) at which babies first crawl based on a survey of 12 mothers"},"content":{"rendered":"\n<p class=\"wp-block-paragraph\">The following data represent the age (in weeks) at which babies first crawl based on a survey of 12 mothers. The data are normally distributed and s = 9.950 weeks. Construct and interpret a 90% confidence interval for the population standard deviation of the age (in weeks) at which babies first crawl. Click the icon to view the table of critical values of the chi-square distribution. Select the correct choice below and fill in the answer boxes to complete your choice. (Use ascending order. Round to three decimal places as needed.)<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">A. There is a 90% probability that the true population standard deviation is between and<br>B. There is 90% confidence that the population standard deviation is between and<br>C. If repeated samples are taken, 90% of them will have the sample standard deviation between and<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-1-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">To construct a <strong>90% confidence interval<\/strong> for the <strong>population standard deviation<\/strong> when the population is normally distributed, we use the <strong>chi-square distribution<\/strong>. The formula to construct a confidence interval for the population <strong>variance<\/strong> \u03c32\\sigma^2 is: ((n\u22121)s2\u03c7R2,(n\u22121)s2\u03c7L2)\\left( \\frac{(n &#8211; 1)s^2}{\\chi^2_{R}}, \\frac{(n &#8211; 1)s^2}{\\chi^2_{L}} \\right)<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>nn = sample size = 12<\/li>\n\n\n\n<li>ss = sample standard deviation = 9.950 weeks<\/li>\n\n\n\n<li>s2s^2 = sample variance = 9.9502=99.009.950^2 = 99.00<\/li>\n\n\n\n<li>\u03c7L2\\chi^2_{L} and \u03c7R2\\chi^2_{R} are the <strong>critical chi-square values<\/strong> for the left and right tails of the distribution with n\u22121=11n &#8211; 1 = 11 degrees of freedom<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">From the chi-square distribution table for 11 degrees of freedom:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>\u03c7L2=\u03c70.952=4.575\\chi^2_{L} = \\chi^2_{0.95} = 4.575 (right tail)<\/li>\n\n\n\n<li>\u03c7R2=\u03c70.052=19.675\\chi^2_{R} = \\chi^2_{0.05} = 19.675 (left tail)<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">Plug into the confidence interval for variance: ((11)(99.00)19.675,(11)(99.00)4.575)=(108919.675,10894.575)\\left( \\frac{(11)(99.00)}{19.675}, \\frac{(11)(99.00)}{4.575} \\right) = \\left( \\frac{1089}{19.675}, \\frac{1089}{4.575} \\right) \u21d2(55.337,237.967)\\Rightarrow \\left( 55.337, 237.967 \\right)<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Now take the square root of each to get the confidence interval for the <strong>standard deviation<\/strong>: (55.337,237.967)=(7.437,15.429)\\left( \\sqrt{55.337}, \\sqrt{237.967} \\right) = \\left( 7.437, 15.429 \\right)<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Final Answer:<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>B. There is 90% confidence that the population standard deviation is between 7.437 and 15.429.<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation (300+ words):<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">In statistics, when a variable is normally distributed and the standard deviation is unknown, we can estimate the population standard deviation using a confidence interval. A <strong>confidence interval<\/strong> provides a range of plausible values for a population parameter\u2014in this case, the <strong>population standard deviation<\/strong>\u2014based on the sample data.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">The data set given includes a sample of 12 mothers reporting the age at which their babies first crawled. The sample standard deviation is 9.950 weeks, and since the data are normally distributed, it is appropriate to use the <strong>chi-square distribution<\/strong> to estimate the population standard deviation.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Because we&#8217;re dealing with standard deviation and not the mean, we use a different approach involving the <strong>chi-square distribution<\/strong> with n\u22121n &#8211; 1 degrees of freedom. This is because variance (and by extension, standard deviation) is a function of squared deviations, which follow a chi-square distribution under the normality assumption.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">We calculated the <strong>confidence interval for the variance first<\/strong> and then took square roots to obtain the interval for the standard deviation. Using critical chi-square values for a 90% confidence level (which leaves 5% in each tail of the chi-square distribution), we found the range in which the true standard deviation likely falls.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">The interpretation of this result is best reflected in <strong>Choice B<\/strong>, which aligns with proper statistical language. Saying we are \u201c90% confident\u201d means that <strong>if we repeated the sampling process many times<\/strong>, about 90% of the calculated intervals from those samples would contain the true population standard deviation. It does <strong>not<\/strong> mean there is a 90% probability that the true value lies within this specific interval\u2014that would imply the population parameter is variable, which it is not.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Thus, our confidence interval for the population standard deviation is <strong>between 7.437 and 15.429 weeks<\/strong>.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>The following data represent the age (in weeks) at which babies first crawl based on a survey of 12 mothers. The data are normally distributed and s = 9.950 weeks. Construct and interpret a 90% confidence interval for the population standard deviation of the age (in weeks) at which babies first crawl. Click the icon [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-214574","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/214574","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=214574"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/214574\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=214574"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=214574"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=214574"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}