 # S O C A survey has been administered to random samples of respondents in each of five nations. Respondents were asked: “How satisfied are you with your life as a whole?” Responses varied from 1 (very dissatisfied) to 10 (very satisfied). Conduct a test for the significance of the difference in mean scores for each nation between men and women. Brazil Males Females X ¯ 1 = 7.7 X ¯ 1 = 7.6 s 1 = 2.0 s 1 = 2.2 N 1 = 712 N 1 = 783 Ukraine Males Females X ¯ 1 = 5.8 X ¯ 1 = 5.5 s 1 = 2.3 s 1 = 2.4 N 1 = 446 N 1 = 549 Vietnam Males Females X ¯ 1 = 7.2 X ¯ 1 = 7.0 s 1 = 1.9 s 1 = 1.8 N 1 = 762 N 1 = 720 South Africa Males Females X ¯ 1 = 6.8 X ¯ 1 = 7.2 s 1 = 2.5 s 1 = 2.3 N 1 = 1492 N 1 = 1482 Egypt Males Females X ¯ 1 = 5.6 X ¯ 1 = 5.9 s 1 = 2.7 s 1 = 2.7 N 1 = 1557 N 1 = 1493 ### Essentials Of Statistics

4th Edition
HEALEY + 1 other
Publisher: Cengage Learning,
ISBN: 9781305093836 ### Essentials Of Statistics

4th Edition
HEALEY + 1 other
Publisher: Cengage Learning,
ISBN: 9781305093836

#### Solutions

Chapter
Section
Chapter 8, Problem 8.9P
Textbook Problem
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## S O C A survey has been administered to random samples of respondents in each of five nations. Respondents were asked: “How satisfied are you with your life as a whole?” Responses varied from 1 (very dissatisfied) to 10 (very satisfied). Conduct a test for the significance of the difference in mean scores for each nation between men and women. Brazil Males Females X ¯ 1 = 7.7 X ¯ 1 = 7.6 s 1 = 2.0 s 1 = 2.2 N 1 = 712 N 1 = 783 Ukraine Males Females X ¯ 1 = 5.8 X ¯ 1 = 5.5 s 1 = 2.3 s 1 = 2.4 N 1 = 446 N 1 = 549 Vietnam Males Females X ¯ 1 = 7.2 X ¯ 1 = 7.0 s 1 = 1.9 s 1 = 1.8 N 1 = 762 N 1 = 720 South Africa Males Females X ¯ 1 = 6.8 X ¯ 1 = 7.2 s 1 = 2.5 s 1 = 2.3 N 1 = 1492 N 1 = 1482 Egypt Males Females X ¯ 1 = 5.6 X ¯ 1 = 5.9 s 1 = 2.7 s 1 = 2.7 N 1 = 1557 N 1 = 1493

Expert Solution
To determine

To find:

The significant difference in the sample statistics of the two samples.

### Explanation of Solution

Given:

The sample statistics of Brazil is given in the table below,

 Sample 1(Males) Sample 2(Female) X¯1=7.7 X¯1=7.6 s1=2.0 s1=2.2 N1=712 N1=783

The sample statistics of Ukraine is given in the table below,

 Sample 1(Males) Sample 2(Female) X¯1=5.8 X¯1=5.5 s1=2.3 s1=2.4 N1=446 N1=549

The sample statistics of Vietnam is given in the table below,

 Sample 1(Males) Sample 2(Female) X¯1=7.2 X¯1=7.0 s1=1.9 s1=1.8 N1=762 N1=720

The sample statistics of South Africa is given in the table below,

 Sample 1(Males) Sample 2(Female) X¯1=6.8 X¯1=7.2 s1=2.5 s1=2.3 N1=1492 N1=1482

The sample statistics of Egypt is given in the table below,

 Sample 1(Males) Sample 2(Female) X¯1=5.6 X¯1=5.9 s1=2.7 s1=2.7 N1=1557 N1=1493

The five step model for hypothesis testing:

Step 1. Making assumptions and meeting test requirements.

Step 2. Stating the null hypothesis.

Step 3. Selecting the sampling distribution and establishing the critical region.

Step 4. Computing test statistics.

Step 5. Making a decision and interpreting the results of the test.

Formula used:

The formula to calculate the sampling distribution of the differences in sample means is given by,

Z(obtained)=(X¯1X¯2)(μ1μ2)σX¯X¯

Where, X¯1 and X¯2 is the mean of first and second sample respectively,

μ1 and μ2 is the mean of first and second population respectively,

σX¯X¯ is the standard deviation and the formula to calculate σX¯X¯ is given by,

σX¯X¯=s21N11+s22N21

Where, N1 and N2 is the number of first and second population respectively.

Calculation:

For Brazil as the significant difference in the sample statistics is to be determined, a two tailed test is applied.

Follow the steps for two-sample testing as,

Step 1. Making assumptions and meeting test requirements.

Model:

Independent random samples.

Level of measurement is interval ratio.

Sampling distribution is Normal.

Step 2. Stating the null hypothesis.

The statement of the null hypothesis is that there is no significant difference in the sample s of the population. Thus, the null and the alternative hypotheses are,

H0:μ1=μ2

H1:μ1μ2

Step 3. Selecting the sampling distribution and establishing the critical region.

Since, the sample size is large, Z distribution can be used.

Thus, the sampling distribution is Z distribution.

The level of significance is,

α=0.05

Area of critical region is,

Z(critical)=±1.96

Step 4. Computing test statistics.

The population standard deviations are unknown.

The formula to calculate σX¯X¯ is given by,

σX¯X¯=s21N11+s22N21

Substitute 2.0 for s1, 2.2 for s2, 712 for N1, and 783 for N2 in the above mentioned formula,

σX¯X¯=(2.0)27121+(2.2)27831=4711+4.84782=0.0056+0.0062=0.0118

Simplify further,

σX¯X¯=0.10860.11(1)

The sampling distribution of the differences in sample means is given by,

Z(obtained)=(X¯1X¯2)(μ1μ2)σX¯X¯

Under the null hypotheses,

μ1μ2=0

Substitute for μ1μ2 in the above mentioned formula,

Z(obtained)=(X¯1X¯2)σX¯X¯

From equation (1) substitute 7.7 for X¯1, 7.6 for X¯2, and 0.11 for σX¯X¯ in the above mentioned formula,

Z(obtained)=(7.77.6)0.11=0.10.11=0.91

Thus, the obtained Z value is 0.91.

Step 5. Making a decision and interpreting the results of the test.

Compare the test statistic with the critical Z value. The Z score falls in the acceptance region. This implies that there is no sufficient evidence to conclude that the mean scores of men and women in Brazil differ in their response towards satisfaction towards life as a whole. The decision to accept the null hypothesis has only 0.05 probability of being incorrect.

For Ukraine as the significant difference in the sample statistics is to be determined, a two tailed test is applied.

Follow the steps for two-sample testing as,

Step 1. Making assumptions and meeting test requirements.

Model:

Independent random samples.

Level of measurement is interval ratio.

Sampling distribution is Normal.

Step 2. Stating the null hypothesis.

The statement of the null hypothesis is that there is no significant difference in the sample s of the population. Thus, the null and the alternative hypotheses are,

H0:μ1=μ2

H1:μ1μ2

Step 3. Selecting the sampling distribution and establishing the critical region.

Since, the sample size is large, Z distribution can be used.

Thus, the sampling distribution is Z distribution.

The level of significance is,

α=0.05

Area of critical region is,

Z(critical)=±1.96

Step 4. Computing test statistics.

The population standard deviations are unknown.

The formula to calculate σX¯X¯ is given by,

σX¯X¯=s21N11+s22N21

Substitute 2.3 for s1, 2.4 for s2, 446 for N1, and 549 for N2 in the above mentioned formula,

σX¯X¯=(2.3)24461+(2.4)25491=5.29445+5.76548=0.0119+0.0105=0.0224

Simplify further,

σX¯X¯=0.14860.15(2)

The sampling distribution of the differences in sample means is given by,

Z(obtained)=(X¯1X¯2)(μ1μ2)σX¯X¯

Under the null hypotheses,

μ1μ2=0

Substitute μ1μ2=0 in the above mentioned formula,

Z(obtained)=(X¯1X¯2)σX¯X¯

From equation (2) substitute 5.8 for X¯1, 5.5 for X¯2, and 0.15 for σX¯X¯ in the above mentioned formula,

Z(obtained)=(5.85.5)0.15=0.30.15=2

Thus, the obtained Z value is 2.

Step 5. Making a decision and interpreting the results of the test.

Compare the test statistic with the critical Z value. The Z score falls in the rejection region. This implies that there is a sufficient evidence to conclude that the mean scores of men and women in Ukraine differ in their response towards satisfaction towards life as a whole. The decision to reject the null hypothesis has only 0.05 probability of being incorrect.

For Vietnam as the significant difference in the sample statistics is to be determined, a two tailed test is applied

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