Shoplifting costs retail businesses a great deal of money every year. In spite o

Question

Shoplifting costs retail businesses a great deal of money every year. In spite of this the historical evidence suggests that only 50% of all shoplifters are turned over to the police. A random survey of 40 retailers revealed that 24 of them had turned their most recent shoplifter in to the police. Use alpha = 0.05 for all tests. 6) Test to see if this sample data indicates that more shoplifters are being turned in than in the past? b. (2) What assumptions must hold for this test to be performed? c. (2) Describe both Type I and Type II errors in the context of this problem. d. (5) Calculate the probability of Type II error if the true proportion of shoplifters turned over to police is 55%. e. (5) Calculate the probability of Type II error if the true proportion of shoplifters turned over to police is 55%, if the sample size is increased to 100. f. (2) What does this tell you about the relationship between alpha, beta and sample size?

Explanation / Answer

(a) Null Hypothesis : H0: p = 0.50

Alternative Hypothesis : Ha : p > 0.50

where p is the proportion of shoplifters turned over to police.

(b) Assumptions are

(i) Randomly sampled

(ii) Data is normally distributed.

(c) Type I error is we reject the null hypothesis even it is True that means we will say that proportion of shoplifters turned over to police is more than 50% but in truth, it is equal to 50%.

Type II error means we don't reject the null hypothesis even it is false that means we will say that proportions of shoplifters turned over to police is equal to 50% but in truth, it is more than 50% .

(d) p0= 0.55

First we will calculate rejection region.

Standard error of proportion se0= sqrt [ p(1-p)/N] = sqrt [0.5 * 0.5/40] = 0.079

Rejection Region is p > p' where p' = 0.50 + Z0.05 se0 = 0.50 + 1.645 * 0.079 = 0.63

Type II error is the probability of rejecting the null hypothesis when it is true.

Pr (p < 0.63; 0.55; 0.079)

Z = (0.63 - 0.55)/ (0.079) = 1.013

so Pr (p < 0.63; 0.55; 0.079) = 0.8438

(e) Now sample size = 100

so standard error of proportion = sqrt [ p(1-p)/N] = sqrt [ 0.5* 0.5/100] = 0.05

so Rejection region p < pc where pc = 0.50 + 1.645 * 0.05 = 0.5823

so Type II error = Pr (p < 0.5823; 0.55; 0.05) by normal distribution

Z = (0.5823 - 0.55)/ 0.05 = 0.646

so  Type II error = Pr (p < 0.5823; 0.55; 0.05) = 0.7405

(2) When we increase sample size then type II error reduces and when we decrease alpha, type II error increases.

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