In a study to test whether or not there is a ditference between the average heights ofadult females in two different countries, random samples of size 1 = 120 andn2 = 150 yielded I = 62.7 inches and I = 61. 8 inches. Extensivestudies of a similar kind have shown that it is reasonable to let o1 = 2.5 inchesand o1 = 2. 62 inches. Test at the 0.05 level of significance whether thedifference between these two sample means is significant.Instructions: Assume 8 -0. in addressing the problem obtain the value of the teststatistic and its corresponding p - value up to four decimal places.Oz = 2. 8772; p – value = 0.0040Oz = 2. 8772; p- value = 0. 0020Oz= -2.8772; p- value = 0.9979Oz = 2. 8772;p- value = 0.9980

Answered: Image /qna-images/answer/114b9bc0-f834-4940-9778-58a3e7e05803.jpg

Answered: In a study to test whether or not there…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Kenneth, a competitor in cup stacking, claims that his average stacking time is 8.2 seconds. During a practice session, Kenneth has a sample stacking time mean of 7.8 seconds based on 11 trials. At the 4% significance level, does the data provide sufficient evidence to conclude that Kenneth's mean stacking time is less than 8.2 seconds? Accept or reject the hypothesis given the sample data below. H0:μ=8.2 seconds; Ha:μ<8.2 seconds α=0.04 (significance level) z0=−1.75 p=0.0401 Select the correct answer below: a. Do not reject the null hypothesis because the p-value 0.0401 is greater than the significance level α=0.04. b. Reject the null hypothesis because the p-value 0.0401 is greater than the significance level α=0.04. c. Reject the null hypothesis because the value of z is negative. d. Reject the null hypothesis because |−1.75|>0.04. e. Do not reject the null hypothesis because |−1.75|>0.04.
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