Beginning Statistics, 2nd Edition
Beginning Statistics, 2nd Edition
2nd Edition
ISBN: 9781932628678
Author: Carolyn Warren; Kimberly Denley; Emily Atchley
Publisher: Hawkes Learning Systems
Question
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Chapter 11.1, Problem 1E
To determine

To find:

The test statistic for a hypothesis test for two population means.

Expert Solution & Answer
Check Mark

Answer to Problem 1E

Solution:

The test statistic is 1.16

Explanation of Solution

Given information:

x¯1=9.21,σ1=2.01,n1=45,x¯2=8.76,σ2=1.77,n2=51,H0:μ1μ20

Where: x¯1 and x¯2 are the two sample means, σ1 and σ2 are the two population standard deviations, n1 and n2 are the two sample sizes and H0:μ1μ20 is the null hypothesis.

Steps:

The difference between the means of two populations i.e. μ1μ2 is estimated by the difference between their respective sample means. This is used for comparing the two populations.

Test statistics is a random variable which is calculated from the sample data and used in hypothesis testing.

Test statistic calculates the degrees of acceptance between sample data and null hypothesis.

Formula used:

The test statistic for a hypothesis test for two population means is:

z=(x¯1x¯2)(μ1μ2)σ12n1+σ22n2

Where: x¯1 and x¯2 are the two sample means, μ1μ2 is the hypothetical value of the difference between the two population means σ1 and σ2 are the two population standard deviations, n1 and n2 are the two sample sizes.

Calculation:

Test statistic:

Substituting x¯1=9.21,σ1=2.01,n1=45,x¯2=8.76,σ2=1.77,n2=51,μ1μ2=0 in the above equation,

z=(9.218.76)0(2.01)245+(1.77)251=0.450.0898+0.0614=0.450.3888=1.15741.16

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Chapter 11 Solutions

Beginning Statistics, 2nd Edition

Ch. 11.1 - Prob. 11ECh. 11.1 - Prob. 12ECh. 11.1 - Prob. 13ECh. 11.1 - Prob. 14ECh. 11.1 - Prob. 15ECh. 11.1 - Prob. 16ECh. 11.1 - Prob. 17ECh. 11.1 - Prob. 18ECh. 11.1 - Prob. 19ECh. 11.1 - Prob. 20ECh. 11.1 - Prob. 21ECh. 11.2 - Prob. 1ECh. 11.2 - Prob. 2ECh. 11.2 - Prob. 3ECh. 11.2 - Prob. 4ECh. 11.2 - Prob. 5ECh. 11.2 - Prob. 6ECh. 11.2 - Prob. 7ECh. 11.2 - Prob. 8ECh. 11.2 - Prob. 9ECh. 11.2 - Prob. 10ECh. 11.2 - Prob. 11ECh. 11.2 - Prob. 12ECh. 11.2 - Prob. 13ECh. 11.2 - Prob. 14ECh. 11.2 - Prob. 15ECh. 11.2 - Prob. 16ECh. 11.2 - Prob. 17ECh. 11.3 - Prob. 1ECh. 11.3 - Prob. 2ECh. 11.3 - Prob. 3ECh. 11.3 - Prob. 4ECh. 11.3 - Prob. 5ECh. 11.3 - Prob. 6ECh. 11.3 - Prob. 7ECh. 11.3 - Prob. 8ECh. 11.3 - Prob. 9ECh. 11.3 - Prob. 10ECh. 11.4 - Prob. 1ECh. 11.4 - Prob. 2ECh. 11.4 - Prob. 3ECh. 11.4 - Prob. 4ECh. 11.4 - Prob. 5ECh. 11.4 - Prob. 6ECh. 11.4 - Prob. 7ECh. 11.4 - Prob. 8ECh. 11.4 - Prob. 9ECh. 11.4 - Prob. 10ECh. 11.4 - Prob. 11ECh. 11.5 - Prob. 1ECh. 11.5 - Prob. 2ECh. 11.5 - Prob. 3ECh. 11.5 - Prob. 4ECh. 11.5 - Prob. 5ECh. 11.5 - Prob. 6ECh. 11.5 - Prob. 7ECh. 11.5 - Prob. 8ECh. 11.5 - Prob. 9ECh. 11.5 - Prob. 10ECh. 11.5 - Prob. 11ECh. 11.5 - Prob. 12ECh. 11.5 - Prob. 13ECh. 11.5 - Prob. 14ECh. 11.5 - Prob. 15ECh. 11.5 - Prob. 16ECh. 11.5 - Prob. 17ECh. 11.5 - Prob. 18ECh. 11.5 - Prob. 19ECh. 11.5 - Prob. 20ECh. 11.5 - Prob. 21ECh. 11.5 - Prob. 22ECh. 11.5 - Prob. 23ECh. 11.5 - Prob. 24ECh. 11.5 - Prob. 25ECh. 11.5 - Prob. 26ECh. 11.6 - Prob. 1ECh. 11.6 - Prob. 2ECh. 11.6 - Prob. 3ECh. 11.6 - Prob. 4ECh. 11.6 - Prob. 5ECh. 11.6 - Prob. 6ECh. 11.6 - Prob. 7ECh. 11.6 - Prob. 8ECh. 11.6 - Prob. 9ECh. 11.6 - Prob. 10ECh. 11.6 - Prob. 11ECh. 11.6 - Prob. 12ECh. 11.6 - Prob. 13ECh. 11.6 - Prob. 14ECh. 11.6 - Prob. 15ECh. 11.6 - Prob. 16ECh. 11.CR - Prob. 1CRCh. 11.CR - Prob. 2CRCh. 11.CR - Prob. 3CRCh. 11.CR - Prob. 4CRCh. 11.CR - Prob. 5CRCh. 11.CR - Prob. 6CRCh. 11.CR - Prob. 7CRCh. 11.CR - Prob. 8CRCh. 11.CR - Prob. 9CRCh. 11.CR - Prob. 10CR
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