Identify the null hypothesis, alternative hypothesis, test statistic, P-value, conclusion about the null hypothesis, and final conclusion that addresses the original claim.\u00c3\u201a\u00c2
23) The health of employees is monitored by periodically weighing them in. A sample of 54 employees has a mean weight of 183.9 lb. Assuming that CJ is known to be 121.2 lb, use a 0.10 significance level to test the claim that the population mean of all such employees weights is less than 200 lb.\u00c3\u201a\u00c2<\/p>\n\n\n\n
MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.\u00c3\u201a\u00c2
Do one of the following, as appropriate: (a) Find the critical value Za\/2- (b) find the critical value ta\/2- (c) state that\u00c3\u201a\u00c2
neither the normal nor the t distribution applies.\u00c3\u201a\u00c2<\/p>\n\n\n\n
24) 99%; n = 17; CJ is unknown; population appears to be normally distributed.\u00c3\u201a\u00c2
A) Za.\/2 = 2.583 B) Za\/2 = 2.567 C) ta\/2 = 2.898 D) ta.\/2 = 2.921\u00c3\u201a\u00c2
Use the given degree of confidence and sample data to construct a confidence interval for the population mean 1-1\u00c3\u201a\u00c2\u00b7 Assume\u00c3\u201a\u00c2
that the population has a normal distribution.\u00c3\u201a\u00c2<\/p>\n\n\n\n
25) The football coach randomly selected ten players and timed how long each player took to perform a certain drill. The times (in minutes) were:\u00c3\u201a\u00c2
7.0 10.8 9.5 8.0 11.5\u00c3\u201a\u00c2
7.5 6.4 11.3 10.2 12.6\u00c3\u201a\u00c2
Determine a 95% confidence interval for the mean time for all players (include a means plot for the data)\u00c3\u201a\u00c2
A) 7.93 min<>< 11.03=””>
C) 10.93 min< 1-1=””>< 8.03=””>
\u00c3\u201a\u00c2
B) 8.03 min< 1-1=””>< 10.93=””>
D) 11.03 min< 1-1=””>< 7.93=””><\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n Let’s address each part of the problem step by step:<\/p>\n\n\n\n 1. Null Hypothesis (H\u2080) and Alternative Hypothesis (H\u2081):<\/strong><\/p>\n\n\n\n 2. Test Statistic:<\/strong><\/p>\n\n\n\n Since the population standard deviation (\u03c3) is known, we use the z-test for the population mean. The test statistic (z) is calculated as:<\/p>\n\n\n\n [ z = \\frac{\\overline{x} – \\mu}{\\frac{\\sigma}{\\sqrt{n}}} ]<\/p>\n\n\n\n Where:<\/p>\n\n\n\n Calculating the standard error:<\/p>\n\n\n\n [ \\frac{\\sigma}{\\sqrt{n}} = \\frac{121.2}{\\sqrt{54}} \\approx 16.49 \\, \\text{lb} ]<\/p>\n\n\n\n Now, calculating the z-score:<\/p>\n\n\n\n [ z = \\frac{183.9 – 200}{16.49} \\approx -0.976 ]<\/p>\n\n\n\n 3. P-value:<\/strong><\/p>\n\n\n\n The p-value represents the probability of obtaining a test statistic at least as extreme as the one observed, assuming the null hypothesis is true. For a z-score of -0.976, we look up the corresponding cumulative probability in the standard normal distribution table.<\/p>\n\n\n\n Using standard normal distribution tables or a calculator, the cumulative probability for z = -0.976 is approximately 0.164.<\/p>\n\n\n\n Since this is a one-tailed test (\u03bc < 200), the p-value is 0.164.<\/p>\n\n\n\n 4. Conclusion about the Null Hypothesis:<\/strong><\/p>\n\n\n\n The significance level (\u03b1) is 0.10. Since the p-value (0.164) is greater than \u03b1 (0.10), we fail to reject the null hypothesis.<\/p>\n\n\n\n 5. Final Conclusion:<\/strong><\/p>\n\n\n\n There is insufficient evidence at the 0.10 significance level to support the claim that the population mean weight of all employees is less than 200 lb.<\/p>\n\n\n\n Regarding the Multiple Choice Question:<\/strong><\/p>\n\n\n\n The question asks for the critical value for a 99% confidence level with a sample size of 17, assuming the population appears to be normally distributed and the population standard deviation is unknown.<\/p>\n\n\n\n When the population standard deviation is unknown and the sample size is small (n < 30), we use the t-distribution. The degrees of freedom (df) are calculated as:<\/p>\n\n\n\n [ \\text{df} = n – 1 = 17 – 1 = 16 ]<\/p>\n\n\n\n For a 99% confidence level, the critical value corresponds to the 0.005 quantile of the t-distribution with 16 degrees of freedom. Using a t-distribution table or calculator, the critical value is approximately 2.921.<\/p>\n\n\n\n Therefore, the correct answer is:<\/p>\n\n\n\n D) t\u2080.005,16 = 2.921<\/strong><\/p>\n\n\n\n Confidence Interval Calculation:<\/strong><\/p>\n\n\n\n To construct a 99% confidence interval for the population mean, we use the formula:<\/p>\n\n\n\n [ \\overline{x} \\pm t_{\\alpha\/2, \\, \\text{df}} \\times \\frac{s}{\\sqrt{n}} ]<\/p>\n\n\n\n Where:<\/p>\n\n\n\n Since the specific sample data (mean and standard deviation) are not provided in the question, we cannot compute the exact confidence interval. However, the critical value for a 99% confidence level with 16 degrees of freedom is 2.921, as determined above.<\/p>\n\n\n\n Regarding the Football Drill Times:<\/strong><\/p>\n\n\n\n The question provides the following drill times (in minutes): 7.0, 10.8, 9.5, 8.0, 11.5, 7.5, 6.4, 11.3, 10.2, 12.6.<\/p>\n\n\n\n To determine a 95% confidence interval for the mean time for all players, we follow these steps:<\/p>\n\n\n\n Identify the null hypothesis, alternative hypothesis, test statistic, P-value, conclusion about the null hypothesis, and final conclusion that addresses the original claim.\u00c3\u201a\u00c223) The health of employees is monitored by periodically weighing them in. A sample of 54 employees has a mean weight of 183.9 lb. Assuming that CJ is known to be 121.2 lb, use […]<\/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-190599","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/190599","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=190599"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/190599\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=190599"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=190599"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=190599"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}\n
\n
\n
\n
(7.0 – 9.33)^2 = 5.4289, \\quad (10.8 – 9.33)^2 = 2.1769, \\quad (9.5 – 9.33)^2 = 0.0289, \\quad (8.0 – 9.33)^2 = 1.7689, \\quad (11.5 – 9.33)^2 = 4.7089
]
[
(7.5 – 9.33)^2 = 3.4289, \\quad (6.4 – 9.33)^2 = 8.5289, \\quad (11.3 – 9.33)^2 = 3.9289, \\quad (10.2 – 9.33)^2 = 0.7569, \\quad (12.6<\/li>\n<\/ol>\n","protected":false},"excerpt":{"rendered":"