Question

A survey of MBA graduates of a business school obtained data on the first-year salary after graduation and years of work experience prior to obtaining their MBA. The data are given in excel.

1. Run the regression analysis (Include all options). Report the least squares regression line. Give the 95% confidence interval for the least squares estimate of the slope. Report the correlation coefficient. Interpret.  Report the coefficient of determination. Interpret. and  Use the ANOVA output and write out the hypothesis being tested, the test statistic, the critical value, p-value, and fully write out the conclusion.

Experience	Salary
8	113.9
5	112.5
5	109
11	125.1
4	111.6
3	112.7
3	104.5
3	100.1
0	101.1
13	126.9
14	97.9
10	113.5
2	98.3
2	97.2
5	111.3
13	124.7
1	105.3
5	107
1	103.8
5	107.4
5	100.2
7	112.8
4	100.7
3	107.3
3	103.7
7	121.8
7	111.7
9	116.2
6	108.9
6	111.9
4	96.1
6	113.5
5	110.4
1	98.7
13	120.1
1	98.9
6	108.4
2	110.6
4	101.8
1	104.4
5	106.6
1	103.9
4	105
1	97.9
2	104.6
7	106.9
5	107.6
1	103.2
1	101.6
0	99.2
1	101.7
6	120.1

Answer 1

Following is the output of excel:

SUMMARY OUTPUT
Regression Statistics
Multiple R	0.694650458
R Square	0.482539258
Adjusted R Square	0.472190044
Standard Error	5.555951324
Observations	52
ANOVA
	df	SS	MS	F	Significance F
Regression	1	1439.269475	1439.26948	46.6256877	1.11404E-08
Residual	50	1543.429756	30.8685951
Total	51	2982.699231
	Coefficients	Standard Error	t Stat	P-value	Lower 95%	Upper 95%
Intercept	100.6157023	1.291837902	77.8857023	7.3154E-54	98.02096955	103.210435
Experience, X	1.490621382	0.218300495	6.8283005	1.114E-08	1.052151941	1.929090822

The least square regression line is

y' = 100.6157 + 1.4906*x

The 95% confidence interval for the least squares estimate of the slope is (1.0522, 1.9291).

Since slope is positive so correlation coefficient is also positive. The correlation coefficient is

r = 0.695

It shows a strong positive relationship between the variables.

The coefficient of determination is

It shows that 48.25% of variation in salary is explained by salary.

---------------------------

Hypotheses are:

The test statistics is

F = 46.63

The p-value is

p-value = 0.0000

Since p-value is less than 0.05 so we reject the null hypothesis. That is model is significant.