That 46 of 100 individuals prefer Cola 2. Complete parts a through c. Do not reject the null hypothesis for the values of p0 between and inclusively. (Type integers or decimals. Do not round.)<\/p>\n\n\n\n
(b) Construct a 95% confidence interval for the proportion of individuals who prefer Cola 2. We are 95% confident the proportion of individuals who prefer Cola 2 is between and (Type integers or decimals rounded to three decimal places as needed.)<\/p>\n\n\n\n
(c) Suppose you changed the level of significance in conducting the hypothesis test to a=0.01. What would happen to the range of values for p0 for which the null hypothesis is not rejected? Why does this make sense? Choose the correct answer below.<\/p>\n\n\n\n
A. The range of values would increase because the corresponding confidence interval would decrease in size.<\/p>\n\n\n\n
B. The range of values would decrease because the corresponding confidence interval would increase in size.<\/p>\n\n\n\n
C. The range of values would increase because the corresponding confidence interval would increase in size.<\/p>\n\n\n\n
D. The range of values would decrease because the corresponding confidence interval would decrease in size.<\/p>\n\n\n\n
The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n Let’s go through each part step by step.<\/p>\n\n\n\n We need to determine the values of ( p_0 ) for which we do not reject the null hypothesis. To do this, we use the confidence interval approach. Since the problem asks for values of ( p_0 ) that are not rejected, this means we look at the confidence interval, which will be calculated in part (b).<\/p>\n\n\n\n We use the formula for a confidence interval for a population proportion:<\/p>\n\n\n\n [ where:<\/p>\n\n\n\n Calculating the standard error (SE):<\/p>\n\n\n\n [ Now, calculating the margin of error (ME):<\/p>\n\n\n\n [ Finally, the confidence interval:<\/p>\n\n\n\n [ Rounded to three decimal places:<\/p>\n\n\n\n [ So, we are 95% confident that the true proportion of individuals who prefer Cola 2 is between 0.362 and 0.558<\/strong>.<\/p>\n\n\n\n If we decrease the level of significance to ( \\alpha = 0.01 ), this means we are using a 99% confidence interval<\/strong> instead of 95%. A higher confidence level means a wider confidence interval<\/strong> because we need to be more certain about our estimate.<\/p>\n\n\n\n Since the confidence interval increases, the range of values for ( p_0 ) for which we do not reject ( H_0 ) also increases<\/strong>. This corresponds to option C<\/strong>:<\/p>\n\n\n\n \u2714 C. The range of values would increase because the corresponding confidence interval would increase in size.<\/strong><\/p>\n\n\n\n In hypothesis testing, the level of significance (( \\alpha )) determines the probability of rejecting a true null hypothesis (Type I error). When ( \\alpha = 0.05 ), the critical values for a two-tailed test are based on a 95% confidence interval. If we reduce ( \\alpha ) to 0.01, we are becoming more cautious and requiring stronger evidence to reject ( H_0 ), meaning our confidence level increases to 99%.<\/p>\n\n\n\n A higher confidence level leads to a wider confidence interval because we need to account for more uncertainty. This increased interval means more values of ( p_0 ) fall within the range where we do not reject<\/strong> the null hypothesis. Consequently, the range of ( p_0 ) values for which ( H_0 ) is not rejected also expands.<\/p>\n\n\n\n This makes sense because a smaller significance level means we are less willing to reject ( H_0 ) unless there is very strong evidence. A wider confidence interval reflects this by including a broader range of plausible values for ( p_0 ). Therefore, as the confidence level increases, the number of values of ( p_0 ) that do not lead to rejection also increases.<\/p>\n\n\n\n Thus, the correct answer is option C: The range of values would increase because the corresponding confidence interval would increase in size<\/strong>.<\/p>\n","protected":false},"excerpt":{"rendered":" That 46 of 100 individuals prefer Cola 2. Complete parts a through c. Do not reject the null hypothesis for the values of p0 between and inclusively. (Type integers or decimals. Do not round.) (b) Construct a 95% confidence interval for the proportion of individuals who prefer Cola 2. We are 95% confident the proportion […]<\/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-203940","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/203940","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=203940"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/203940\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=203940"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=203940"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=203940"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}(a) Finding the Range of ( p_0 ) for Which We Do Not Reject ( H_0 )<\/strong><\/h3>\n\n\n\n
(b) Constructing a 95% Confidence Interval for ( p )<\/strong><\/h3>\n\n\n\n
\\hat{p} \\pm Z_{\\alpha\/2} \\sqrt{\\frac{\\hat{p}(1-\\hat{p})}{n}}
]<\/p>\n\n\n\n\n
SE = \\sqrt{\\frac{0.46(1 – 0.46)}{100}} = \\sqrt{\\frac{0.46 \\times 0.54}{100}} = \\sqrt{0.002484} \\approx 0.0498
]<\/p>\n\n\n\n
ME = 1.96 \\times 0.0498 \\approx 0.0976
]<\/p>\n\n\n\n
0.46 – 0.0976 = 0.3624, \\quad 0.46 + 0.0976 = 0.5576
]<\/p>\n\n\n\n
(0.362, 0.558)
]<\/p>\n\n\n\n(c) Effect of Changing ( \\alpha ) to 0.01<\/strong><\/h3>\n\n\n\n
Explanation (300 Words)<\/strong><\/h3>\n\n\n\n