Question

The following data are the monthly salaries y and the grade point averages x for students who obtained a bachelor's degree in business administration. GPA 2.6 3.4 3.7 3.1 3.5 2.9 Monthly Salary (S) 3,600 3,800 4,200 3,700 4,100 2,500 The estimated regression equation for these data is y- 564.3 +964.3x and MSE-268,482. a. Develop a point estimate of the starting salary for a student with a GPA of 3.0 (to 1 decimal). b. Develop a 95% confidence interval for the mean starting salary for all students with a 3.0 GPA (to 2 decimals). C. Develop a 95% prediction interval for Ryan Dailey, a student with a GPA of 3.0 (to 2 decimals).

Answer 1

:

	GPA (x)	Monthly salary (y)	GPA*Monthly salary	GPA2	Monthly salary2
	2.6	3600	9360	6.76	12960000
	3.4	3800	12920	11.56	14440000
	3.7	4200	15540	13.69	17640000
	3.1	3700	11470	9.61	13690000
	3.5	4100	14350	12.25	16810000
	2.9	2500	7250	8.41	6250000
Sum =	19.2	21900	70890	62.28	81790000

$\bar x = \Sigma x/n = 19.2/6 = \textbf{3.2}$

$SS_{XX} = \Sigma x^2 - \frac{(\Sigma x)^2}{n} = 62.28 - 19.2^2/6 = 0.84$

Regression Equation:

$\hat y = 564.3 + 964.3 x$

MSE = 268482

a) Point estimate of the starting salary for a student with a GPA of 3.0

$\hat y^* = 564.3 + 964.3 *3 = \$\textbf{ 3457.2}$

b) At $\alpha$ = 0.05 and df = 6-2 = 4, _{$t_{\alpha/2}$} = 2.776

95% Confidence interval for the mean starting salary for all student with a GPA of 3.0

$\left (\hat y^* \pm t_{\alpha/2}*\sqrt{MSE* \left (\frac{1}{n} + \frac{(x^*-\bar x)^2}{SS_{XX}} \right )}\ \right )$

$\left (3457.2 \pm 2.776*\sqrt{(268482)*\left ( \frac{1}{6} + \frac{(3-3.2)^2}{0.84} \right )}\ \right )$

$\left (3457.2 \pm 2.776*239.858\ \right )$

$\left (\textbf{ 2791.35, 4123.05}\ \right )$

c) 95% Prediction interval for Ryan dailey, a student with a GPA of 3.0

$\left (\hat y^* \pm t_{\alpha/2}*\sqrt{MSE* \left (1+\frac{1}{n} + \frac{(x^*-\bar x)^2}{SS_{XX}} \right )}\ \right )$

$\left (3457.2 \pm 2.776*\sqrt{(268482)*\left ( 1+\frac{1}{6} + \frac{(3-3.2)^2}{0.84} \right )}\ \right )$

$\left (3457.2 \pm 2.776*570.9762\ \right )$

$\left (\textbf{ 1872.17, 5042.23}\ \right )$