1. The GPAs of all students at Cal Poly Pomona have a normal distribution. A ran

Question

1.               The GPAs of all students at Cal Poly Pomona have a normal distribution. A random sample of 25 CPP students is taken. Assume n/N .05. (20.5 pts) a) Suppose you do not know the average GPA for all CPP students but you know that distribution has a standard deviation of .29. Construct a 92% confidence interval for µ given that the sample’s average GPA is 3.25 (round to two decimal places). What is the margin of error?

b)     Suppose you do not know the mean or standard deviation of the distribution for all students’ GPAs,but you do know that for the students in the sample, the mean GPA is 3.25 and the standard deviation is .21. Construct the 90% confidence interval for µ (round to two decimal places). What does this interval say about the average GPA over all CPP students?

c)     Can you use the methods used in parts a and b if you did not know the shape of the distribution forCal Poly Pomona students’ GPAs? Explain why or why not.

d)     Suppose you want to make a new random sample of students to estimate the mean GPA of thepopulation, but all you know about the population distribution is that it has a standard deviation of .22. What sample size should you use so that your estimate is within .06 of µ and allows 2% error? What is the confidence level of this interval estimation?

e)     It is found that the average GPA amongst all CPP students is 3.02 and the standard deviation is .29.What is the probability that the mean GPA is between 2.95 and 3.11 for a random sample of 25 students?

Explanation / Answer

Solution

Back-up Theory

Given the population standard deviation, , 100(1 - )% Confidence Interval for population mean µ is:

X bar ± (/n).Z/2, where X bar is the mean of the sample of size n and Z/2 is the upper (/2)% point of Standard Normal Distribution....................................................................... …………………………..(1)

If the population standard deviation, , is not known, 100(1 - )% Confidence Interval for population mean µ is:

X bar ± (s/n).tn – 1, /2, where X bar is the mean of the sample of size n and tn – 1, /2 is the upper (/2)% point of t-Distribution with (n - 1) degrees of freedom………………........................................................(2)

Given the population standard deviation, , to estimate population mean µ within a margin of error of E with % confidence level, sample size, n, required is: (2. Z2/2)/E2 …........................................................(3)

If a random variable X ~ N(µ, 2), i.e., X has Normal Distribution with mean µ and variance 2, and Z is the Standard Normal Variate, i.e., Z ~ N(µ, 2), then X bar ~ N(µ, 2/n) and

P(X bar or t) = P[Z or {(n)(t - µ)/ }] …………………………….……..….......................................(4)

Now, to work out the solution,

Part (a)

Given = 0.29, n = 25, X bar = 3.25, 92% (i.e., = 0.08) Confidence Interval for population mean µ [vide (1) under Back-up Theory] is: 3.25 ± (0.29/5).(1.751)

= 3.25 ± (0.29/5).(1.751) = 3.25 ± 0.102 ANSWER

Part (b)

Since is not known, but X bar = 3.25, and s = 0.21,

90% (i.e., = 0.1) Confidence Interval for population mean µ[vide (2) under Back-up Theory] is: 3.25 ± (0.21/5).(1.711) = 3.25 ± (0.21/5).(1.711) = 3.25 ± 0.072 ANSWER

Part (c)

By Central Limit Theorem, even when the population distribution is not known, Normal approximation for sample average is permissible. Hence, the same methods can be applied. ANSWER

Part (d)

Given = 0.22, E = 0.06, = 0.02, required sample size [vide (3) under Back-up Theory] is:

(0.222)(2.325)2/(0.06)2 = 73 ANSWER

Part (e)

Given µ = 3.02, = 0.29, n = 25, P(2.95 X bar 3.11) [vide (4) under Back-up Theory]

= P(- 1.217 Z 1.552) = 0.9396 – 0.1117 = 0.8279 ANSWER

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