Is a student\'s GPA affected by whether the student uses John or Mary. The GPA\'

Question

Is a student's GPA affected by whether the student uses John or Mary. The GPA's of 12 randomly selected John users and 13 randomly selected Mary users were calculated: We found the following:

John: 2.22, 0.08, 2.07, 2.69, 2.14, 3.18, 2.02, 4.00, 3.34, 2.02, 3.71, 2.19
Maryl: 3.02, 2.69, 1.87, 2.21, 1.88, 2.77, 2.49, 3.07, 1.47, 3.03, 1.81, 2.58, 3.48

It is known that John GPA's have a standard deviation of 1 and that Mary GPA's have a standard deviation of 0.7. Suppose that the John GPA's are

X1,X2, ... ,X12

and the JJ Email GPA's are

Y1,Y2, ... ,Y13

.
The unknown means are x and y. We are interested in examining x - y. Call the sample means of X and Y, xbar and ybar respectively. Assume that all distributions are normal. Use R for computations.

a)Calculate xbar 1

b) Calculate the variance of xbar 2

c) Calculate ybar 3

d) Calculate the variance of ybar. 4

e) Calculate the variance of xbar - ybar. 5

f) What is the critical value used for a 96% confidence interval for x - y? 6

g) Create a 96% confidence interval for x - y.

7 , 8

i) What is the length of your 96% confidence interval for x - y? 9

j) What would the p value have been if we used this data to test H0:x - y=0
against the alternative Ha:x - y0? 10

Explanation / Answer

Here xbar = 2.47 and ybar = 2.49

a)Calculate xbar = sum of all x / 12 = 2.47

(b) Variance of xbar =   x2/n1 = 1/12= 0.083

(c) ybar = sum of all ys/13 = 2.49

(d) Variance of ybar = y2/n2 = 0.72 /13 = 0.037

(e) Variance of 2xbar - ybar = x2/n1 + y2/n2 = 0.083+0.037 = 0.121

(f) critical value used for a 96% confidence interval for x - y

by seeing Z - table we find critical value Zcritical = 2.055

(g) 96% confidence interval for x - y = (xbar - ybar ) +- Zcritical * sqrt [2xbar - ybar]

= (2.47 - 2.49) +- 2.055 * sqrt [0.121]

= -0.02 +- 0.715 = (-0.735, 0.695)

(i)  length of your 96% confidence interval = 0.695 - (-0.735) = 1.43

(j) P - value = Pr(xbar - ybar <= -0.02; 0 ; 0.348)

Z = (-0.02 -0)/ 0.348 =-0.057

so P- value = 0.48

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