Question

SE-MaritalStatus	SE-Income
Not Married	94867
Not Married	97912
Not Married	96653
Not Married	96928
Not Married	96697
Not Married	96522
Not Married	96621
Not Married	98717
Not Married	95744
Not Married	96727
Not Married	96244
Not Married	95432
Not Married	97681
Not Married	95366
Not Married	96572
Married	100947
Married	100837
Married	97303
Married	103144
Married	95706
Married	95385
Married	93901
Married	95994
Married	97663
Married	95865
Married	100964
Married	106627
Married	111478
Married	114932
Married	108781

10. Divide the sample members into 2 groups based on marital status of the head of household.

11. Find xbar1, s1, n1 using Excel

12. Find xbar2, s2, n2 using Excel

13. Find s

Determine whether there is a difference in the mean of the quantitative variable between married and not-married households using the 2 independent samples hypothesis test. Include a screen shot of any applet you used in your calculations.

14. Choose a significance level α.

15. What is the null hypothesis Ho?

16. What is the alternative hypothesis Ha?

17. What is the test statistic t?

18. What is the corresponding p value derived from the t statistic? (Show a scree shot of how you derived the p value.)

19. What is your conclusion on the null and alternative hypothesis?

20. Explain your results in everyday language.

Answer 1

10) The given data can be entered in Excel as follows-

ANOVA in excel

I performed the analysis using basic commands in MS Excel and I am giving the steps to be followed-

1. Select the Data tab and choose the Data Analysis in the top right-hand corner

2. In the Data Analysis menu choose ANOVA: Single Factor and click OK

3. In the ‘Input Range’ box, select all the data in the columns you created, including the variable names

4. Check the ‘Labels in First Row’ box

5. In the ‘Output Range’ box, enter a cell range where Excel will place the output and click OK

6. If the p-value were less than 0.05, you would reject the null hypothesis that says the means of all categories are equal. If the p-value were greater than 0.05, then you would fail to reject the null.

So the desired output is given as-

The answers to questions 11 & 12 are given is the screenshot added above.

We know that the sample standard deviation (s1) can be obtained by sqrt(variance).

n= observation

xbar= Mean

13) s= pooled standard deviation = sqrt(pooled variance) =  

4628.319

14) Choosing a significant level is subjective. However one can choose the universally accepted significant level at 5%.

15) The null hypothesis H0: There is no difference between the mean income of the two groups

16) The alternative hypothesis Ha: The is a significant difference between the mean income of both groups

17 ) the t-statistic is calculated to be =  -2.795

18) Since this is a two-tailed test so the p-value is 0.0093

19) Since the p-value is less than 0.05 (at 5%) so we have enough evidence to reject the null hypothesis.

20) Based on the decision made on the basis of the given sample, we found that there is a significant difference in the average income of two groups (Married and Not Married).