The grade point average of four randomly selected college students is recorded at the end of the fall and spring semesters of their senior year, as follows:

Question

Student	1	2	3	4
Fall	1.3	2.2	1.9	2.1
Spring	1.1	1.3	1.2	1.8

Construct a 90% confidence interval for the mean difference in GPA. Round your answers to three decimal places, and use four decimal places for any interim calculations.

Hint: Identify the type of t-procedure, and complete the confidence interval accordingly.

We are 90% confident that the interval -0.914 or 0.136 , -0.136 or 0.914) contains the true mean difference ud in GPA between fall and spring semesters.

How can we calculate this by hand, calculator ti 84, or any other way please

Answer 1

Hi Cynthya,

Are you sure that's a negative and not a positive lower bound 0.136? Here's how I computed that:

CI=Dbar +/- t*(s_d/sqrt(n))

Dbar=mean difference

t*=t-critical value from t-table

s_d=Standard deviation of differences

n=sample size

To compute Dbar, take the difference in each individual measurement ((1.3-1.1), (2.2-1.3)....), add them, and divide by 4.

Dbar=(0.2+0.9+0.7+0.3)/4

Dbar=0.525

Now, use a calculator--in-hand or online--to compute standard deviation. Manual computation is cumbersome.

s_d=0.330

Thus:

Dbar=0.525

s_d=0.330

n=4

Now, to compute t*, we need degrees of freedom. Recall that this is just n-1:

df=4-1=3.

Look at row 3 and the 90% column in the t-table to reach:

t*=2.353

Now:

CI= 0.525 +/- (2.353(0.330/2))

CI=(0.137,0.913)

My bounds are slightly different due to rounding difference, but probably close enough.

All confidence intervals imply that we are x% confident that true measure falls between lower and upper bound. Here:

We are 90% confident that true difference in Fall and Spring GPA is between 0.137 and 0.913 points.

I hope this helps.