Suppose we aredoing a hypothesis test concerning the height of BU students. We d

Question

Suppose we aredoing a hypothesis test concerning the height of BU students. We decide on a level of significance = 0.05 and sample 100 students to find a sample mean of 65.34 with a (sample) standard deviation of 2.53 inches. But suppose for some bizarre reason, we know that the actual correct value for the mean height of BU students is 66.1 inches with a (population) standard deviation of 2.76 inches. (Note that you don't have to worry about the difference of sample and population standard deviations here, since they are given to you, but you DO need to account for the sample size.... ) This problem explores the probability of Type I and Type II errors, and the overall situation can be illustrated as follows (where = the probability of a Type II error). Note that you don't have to know the hypothesis H0 to do this analysis.

(a) What is the probability of a Type I error?

(b) What is the value x?

(c) What is the probability of a Type II error?

(d) How big would we need to make the sample size in order to bring the probability of a Type II error to no more than 0.5% = 0.005? Would the Type I error probability change with the change in sample size?

Explanation / Answer

(a) Here sample mean x = 65.34 inches

sample standard deviation s= 2,53 inches

population mean or true mean = 66.1 inches

population standard deviation = 2.76 inches

Probability of type I error = 0.05

(b) Here x = x + t99,0.05 (s/ sqrt(n)

x = 65.34 + 1.9842 * (2.53/sqrt 100)

x = 65.34 + 1.9842 * 0.253 = 65.842

(c) Pr(Type II error) = Pr(x < 65.842)

where = 66.1 inches

se = / sqrt(n) = 2.76/ sqrt(100) = 0.276

Pr(X < 65.842 ; 66.1; 0.276) = ?

Z = (65.842 - 66.1)/ 0.276 = -0.9348

Pr (Type II error) = Pr(Z < -0.9348) = 0.1749

(d) Here changing the sample sie will not affect the probability of type I error.

here Pr(Type II error) = 0.005

so Z - value for the given probability is -2.575

-2.575 = Pr(X < x ; 66.1 inch ; 2.76/ sqrt(n))

-2.575 = (x - 66.1)/ (2.76/ sqrt(n))

x = 66.1 - 7.107/ sqrt(n)

Here as probability of type I error is same so the value x above which we will reject the null hypotheis is

x = 65.34 + Z95% (2.76)/ sqrt(n) = 65.34 + 1.96 * 2.76/ sqrt(n)

so

66.1 - 7.107/ sqrt(n) = 65.34 + 5.4096/ sqrt(n)

66.1 - 65.34 = 12.0658/ sqrt(n)

sqrt(n) = 12.5166/0.76

n = 271.23 or 272

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