Will someone please answer this problem for me. I am struggling with it. I have

ID: 2946798 • Letter: W

Question

Will someone please answer this problem for me. I am struggling with it. I have attached the problem and the data down below.

4. The director of admissions of a small college administered a newly designed entrance test to 20 students selected at random from the new freshman class in a study to determine whether an entrance test score (C8 Test Score) can be predicted from the student's high school grade point average (C7 GPA). This data is provided in your minitab file. Assume that a Least Squares regression model is appropriate and use Minitab to examine this data and answer the questions. Using the data provided, please do the following: A. According to the model created, estimate the test score of a student with a GPA of 3.5 B. Given the first observation (3.1, 5.5) use the regression model to give the residual for this observation, C. Provide a 95% Confidence Interval for ?, the slope parameter.

Explanation / Answer

Solution

Let xi = GPA and yi = test score of the ith student.

Preparatory Work

The estimated least square regression equation is: ycap = 3.1095 + 0.7622x …………. (5)

Now, to work out the answer,

NOTE:

To the point answers are given below. Detailed Excel Calculations and Back-up Theory

follow at the end.

Part (a)

Substituting x = 3.5 in (5),

estimated test score of a student whose GPA is 3.5 = 5.78 ANSWER

Part (b)

Substituting x = 3.1 in (5),

estimated test score of a student whose GPA is 3.1 = 5.47.

Given the actual GPA is 5.5, residual = 5.47 – 5.5 = - 0.03 ANSWER

Part (c)

95% CI for slope parameter, ?1 is: [0.2714, 1.2531] ANSWER

Part (d)

Since the above CI does not contain 0, we conclude that ?1 is not zero. This implies that there is sufficient evidence to suggest that this model provides a sufficient predictor of GPA. ANSWER

Excel Calculations

Xbar

2.50

ybar

5.015

Sxx

9.84

Syy

15.3855

Sxy

7.5

?1cap

0.7622

?0cap

3.1095

s^2

0.5372

sb^2

0.0546

0.05

n-2

tn-2,?/2

2.1009

CIbLB

0.2714

CIbUB

1.2531

Back-up Theory

The linear regression model Y = ?0 + ?1X + ?, ………………………………………..(1)

where ? is the error term, which is assumed to be Normally distributed with mean 0 and variance ?2.

Estimated Regression of Y on X is given by: Y = ?0cap + ?1capX, …………………….(2)

where ?1cap = Sxy/Sxx and ?0cap = Ybar – ?1cap.Xbar..…………………………….…..(3)

Mean X = Xbar = (1/n)sum of xi ………………………………………….……….….(4)

Mean Y = Ybar = (1/n)sum of yi ………………………………………….……….….(5)

Sxx = sum of (xi – Xbar)2 ………………………………………………..…………....(6)

Syy = sum of (yi – Ybar)2 ……………………………………………..………………(7)

Sxy = sum of {(xi – Xbar)(yi – Ybar)} ……………………………………………….(8)

All above sums are over i = 1, 2, …., n,n = sample size ……………………………..(9)

Estimate of ?2 is given by s2 = (Syy – ?1cap2Sxx)/(n - 2)….…………………………..(10)

Standard Error of ?1cap is sb, where sb2 = s2/Sxx

100(1 - ?)% Confidence Interval (CI) for ?1 = ?1cap ± {SE(?1cap) x tn – 2, ?/2}

DONE