9. High school students can be categorized into two groups by the amount of acti

Question

9. High school students can be categorized into two groups by the amount of activities they are involved in. Let group 1 consist of all high school students who are very involved in sports and other activitics and group 2 consist of all high school students who aren't. The distributions of GPAs in both groups are approximately noral. The mean and standard deviation for group 1 are 2.9 and 0.4, respectively. The mcan and standard deviation for group 2 are 2.7 and 0.5, respectively. Independent random samples of 50 high school students are to be selected from both groups (for a total of 100 students) (a) If the sample mean GPA is to be calculated for both groups and we calculate the difference as involved in activities -not so involved in activities, what is the expected value for the difference in sample means? (b) If the sample mean GPA is to be calculated for both groups and we calculate the difference as involved in activities -not so involved in activities, what is the standard deviation of the sampling distribution of the difference in sample means? (c) What is the probability that the average GPA in the sample of students who are not so involved is higher than the average GPA in the sample of students who are very involved? (d) What is the probability that the average GPAs in the two samples differ by no more than 0.1?

Explanation / Answer

Solution:

We are given

Group 1 (Involved in activities)

Mean = X1bar = 2.9

SD = S1 = 0.4

Sample size = N1 = 50

Group 2 (Not involved in activities)

Mean = X2bar = 2.7

SD = S2 = 0.5

Sample size = N2 = 50

Part a

Expected value for difference in sample means = (X1bar – X2bar)

Expected value for difference in sample means = 2.9 – 2.7

Expected value for difference in sample means = 0.2

Part b

Standard deviation for sampling distribution is given as below:

SD for sampling distribution = sqrt[(S1^2/N1)+(S2^2/N2)]           

SD for sampling distribution = sqrt[(0.4^2/50)+(0.5^2/50)]           

SD for sampling distribution = sqrt[0.0082]

SD for sampling distribution = 0.090554

Part c

Here, we compare 2nd group with 1st group, so assume 1st group parameters as population parameters.

Here, we have to find P(Xbar > 2.9)

Z = (Xbar - µ)/[/sqrt(n)]

Z = (2.7 – 2.9) / [ 0.4/sqrt(50)]

Z = -0.2/ 0.056569

Z = 3.535534

P(Z<3.535534) = 0.999797(by using z-table or excel)

P(Xbar > 2.9) = 1 – P(Xbar<2.9) = 1 - 0.999797 = 0.000203

Required probability = 0.000203

Part d

We have to find P(Difference < 0.1)

For difference in the sample means, we have

Expected value for difference in sample means = Mean = 0.2

SD for sampling distribution of difference = 0.090554

Z = (X – mean) / SD

Z = (0.1 – 0.2) / 0.090554

Z = -0.1/0.090554

Z = -1.10431

P(Z<-1.10431) = 0.134729

(by using z-table or excel)

Required probability = 0.134729

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