EQuestion HelpAssume that both populations are normally distributed.(a) Test whether u, #, at the a= 0.01 level of significance for the given sample data.(b) Construct a 99% confidence interval about u, -.Population 1Population 2151517.724.4P- U.UUT (ROUnu LU uree uuciidl Places as TieeQuu.)5.15.3Should the null hypothesis be rejected?O A. Do not reject H, there is sufficient evidence to conclude that the two populations have different means.OB. Do not reject H, there is not sufficient evidence to conclude that the two populations have different means.OC. Reject Ho, there is not sufficient evidence to conclude that the two populations have different means.D. Reject H, there is sufficient evidence to conclude that the two populations have different means.(b) Construct a 99% confidence interval about u, -H2:We are 99% confident that the mean difference is between(Round to two decimal places as needed. Use ascending order.)and

Name: EQuestion HelpAssume that both populations are normally distributed.(
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Description: EQuestion HelpAssume that both populations are normally distributed.(a) Test whether u, #, at the a= 0.01 level of significance for the given sample data.(b) Construct a 99% confidence interval about u, -.Population 1Population 2151517.724.4P- U.UUT

a) Null Hypothesis: H0:μ1=μ2 Alternative Hypothesis: H1:μ1≠μ2 Test statistic: Consider pooled variance: spooled2=(n1-1)s12+(n2-1)s22n1+n2-2 =(15-1)5.12+(15-1)5.3215+15-2 =27.05 t=x¯1-x¯2spooled21n1+1n2 =17.7-24.427.05115+115 =-3.5266p value: df=n1+n2-2=15+15-2=28 p=2P(t<test statistic) =2*P(t<-3.5266) =2*0.0007, from the excel function, =T.DIST(-3.5266,28,TRUE) =0.0015 Decision Rule: IF p value is less than or equal to level of significance, then reject the null hypothesis. Conclusion: Here, the level of significance is 0.01 The p value is less than the level of significance. From the decision rule, the null hypothesis is rejected. Correct option: option D. b) 99% confidence interval about μ1-μ2: t critical value=2.7633, from the excel function, =T.INV.2T(0.01,28) x¯1-x¯2±tspooled21n1+1n2=(17.7-24.4)±2.7633*27.05*115+115 =-6.7±5.2498 =(-11.95, -1.45) rounded to 2 decimals Thus, 99% confidence interval about μ1-μ2 is -11.95 and -1.45.

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Assume that both populations are normally distributed. (a) Test whether u, #, at the a =0.01 level of significance for the given sample data. (b) Construct a 99% confidence interval about u, -H. Should the null hypothesis be rejected? OA. Do not reject Ho, there is sufficient evidence to conclude that the two populations have differont means. O B. Do not reject Hy. there is not sufficient evidence to conclude that the two populations have different means. O C. Reject Ho, there is not sufficient evidence to conclude that the two populations have different means. D. Reject Ho, there is sufficient evidence to conclude that the two populations have different means. (b) Construct a 99% confidence interval about 4, - H.

Assume that both populations are normally distributed. (a) Test whether u, #, at the a =0.01 level of significance for the given sample data. (b) Construct a 99% confidence interval about u, -H. Should the null hypothesis be rejected? OA. Do not reject Ho, there is sufficient evidence to conclude that the two populations have differont means. O B. Do not reject Hy. there is not sufficient evidence to conclude that the two populations have different means. O C. Reject Ho, there is not sufficient evidence to conclude that the two populations have different means. D. Reject Ho, there is sufficient evidence to conclude that the two populations have different means. (b) Construct a 99% confidence interval about 4, - H.

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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