Question

Going back to problem 1, in real life you can, without much difficulty, get the mean grade of Prof. Lax’s classes but that is about it; meaning you will have no idea how his grades would be distributed, nor would you have any idea about the standard deviation of these grades. (I doubt Prof. Lax would advertise his laxness on his website. Contrary what you might believe that is academically bad form and might negatively affect his students’ hireability in the job market). However, you have access to Miss Z’s data (which she swears is obtained by a random selection process) and the grades she obtained in her random sample of nine were: 79, 75, 84, 63, 98, 52, 87, 99, 83 a. To help Miss Z with her decision to take this course with Prof. Lax or not, create a 97% confidence interval (CI) for the mean using Miss Z.’s data. Make sure that you do the necessary checks. b. Does your interval capture the rumored population mean of 85? c. Calculate the margin of error (ME or simply E) of your confidence interval. d. Miss Z thinks a margin of error (or E) of 7 points or more will have a significant negative effect on her GPA. How does the ME (or E) of your 97% CI from part (c) compare to what she says her GPA can afford? If your CI’s ME (or E) is different than 7 points she can afford what are the ways you can use to reduce the margin of error down to 7 or smaller. Discuss all that can be done.

Answer 1

From the data, we calculate the mean and standard deviation

Mean (X bar) = Sum of Values /n

and

Mean = 80

n = 9

Std Dev = 15.24

a)

Conditions

• Randomization Condition: The data must be sampled randomly. It is given that Miss Z obtained her sample randomly, hence this conditions is satisfied.
• Independence Assumption: The sample values must be independent of each other. This can be assumed to be true.

Alpha = 0.03

Z_Critical = 2.17

Hence, 97% CI = Mean +/- Z_Critical * Std Dev / n^1/2 = 80 +/- 2.17 *(15.24/9^1/2) = {68.98,91.02}

b)

Since the value 85 lies in the interval {68.98,91.02}, Hence 85 lies in our interval

c)

ME = 2* Z_Critical * Std Dev / n^1/2 = 2*2.17 *(15.24/9^1/2) = 22.05

d)

Since our ME is greater than 7, hence it will have negative impact on GPA as per Mrs Z

To reduce the confidence interval, we can reduce the alpha or reduce the standard deviation ie (if the GPA scores vary less)