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ummary statistics are given for independent simple random samples from two populations. Use the pooled t-test to conduct a left-tailed hypothesis test using 1 = 12, s, = 6, n, = 10, x2 = 20, s2 = 4, n2 = 14 O B. Test statistic: t= - 1.526526 Critical value: - 1.717 0.05 0.001 Do not reject Ho O C. Test statistic: t= -3.929 Critical value: - 2.074 P<0.005 O D. Test statistic: t= - 3.929 Critical value: - 1.717 P<0.005 Reiect He

ummary statistics are given for independent simple random samples from two populations. Use the pooled t-test to conduct a left-tailed hypothesis test using 1 = 12, s, = 6, n, = 10, x2 = 20, s2 = 4, n2 = 14 O B. Test statistic: t= - 1.526526 Critical value: - 1.717 0.05 0.001 Do not reject Ho O C. Test statistic: t= -3.929 Critical value: - 2.074 P<0.005 O D. Test statistic: t= - 3.929 Critical value: - 1.717 P<0.005 Reiect He

MATLAB: An Introduction with Applications
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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Summary statistics are given for independent simple random samples from two populations. Use the pooled t-test to conduct a left-tailed hypothesis test using a signiticance level of & =U. X = 12, s, = 6, n = 10, x2 20, s2 = 4, n2 = 14 %3D %3D %3D O A. Test statistic: t= -0.916.916 O B. Test statistic: t= - 1.526526 Critical value: - 2.074 Critical value: - 1.717 0.05 <P<0.10 Do not reject Ho P>0.001 Do not reject Ho C. Test statistic: t= -3.929 O D. Test statistic: t= - 3.929 Critical value: – 1.717 Critical value: - 2.074 P<0.005 P<0.005 Reject Ho Reject Ho
Transcribed Image Text:Summary statistics are given for independent simple random samples from two populations. Use the pooled t-test to conduct a left-tailed hypothesis test using a signiticance level of & =U. X = 12, s, = 6, n = 10, x2 20, s2 = 4, n2 = 14 %3D %3D %3D O A. Test statistic: t= -0.916.916 O B. Test statistic: t= - 1.526526 Critical value: - 2.074 Critical value: - 1.717 0.05 <P<0.10 Do not reject Ho P>0.001 Do not reject Ho C. Test statistic: t= -3.929 O D. Test statistic: t= - 3.929 Critical value: – 1.717 Critical value: - 2.074 P<0.005 P<0.005 Reject Ho Reject Ho
Expert Solution
Step 1

Given : n1=10 , n2=14 , X1-bar=12 , X2-bar=20 , s1=6 , s2=4 ,α=0.05

Therefore , the degrees of freedom is given by,

df=n1+n2-2=10+14-2=22

The pooled estimate of the standard deviation is ,

Sp=(n1-1)s12+(n2-1)s22n1+n2-2=(10-1)62+(14-1)4210+14-2=4.9175

The null and alternative hypothesis is,

Ho : μ1=μ2

H1 : μ1 ˂ μ2

The test left-tailed test.

Our aim is to find the correct option.

 

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