A consultant for a chain of clothing stores has designed a new inventory control

Question

A consultant for a chain of clothing stores has designed a new inventory control system in order to reduce the time lag between placement of an order by a store and when that order is filled. The old system has a mean time lag of 20 days, and a standard deviation of time lags of 4 days. Consider a test of the null hypothesis that the mean time lag with the new inventory control system is at least 20 days, against the alternative that it is less. Assume that time lags are approximately normally distributed, and that the standard deviation of time lags characterizing the new system is 4 days, the same as that of the old system. In order to test the null hypothesis, a random sample of 16 orders filled under the new system will be taken.

(Continued) A manager for the chain of clothing stores is concerned that the probabilities of Type I and Type II error are unacceptably large. He wants the maximum probability of Type I error to be controlled at 1%, and he also wants the probability of Type II error not to exceed 2% in the event that the population mean is 19. What is the minimum sample size so that these conditions are satisfied?

What sample size should be utilized?

a. n = 306 c. n = 308

b. n = 307 d. n = 309

Explanation / Answer

Here let say smple size = n

population stadnard deviation = 4

standard error of sample mean = /n = 4/ n

H0 : >= 20

H : < 20

Here we will reject the null hypothesis if sample mean will be less than x < 20 - Z99% 4/ n

so

x < 20 - Z99% 4/ n

x < 20 - 2.326 * 4/ n

x < 20 - 9.304/ n

so we want to keep the type II error less than 2% . So type II error means that we will not be able to reject the null hypothesis even if it is false. here the true population mean is 19. So, we will not be able to reject the null hypothesis if x > 20 - 9.304/ n

so Pr(Type II error) ) = 0.02

Pr(Type II error) = Pr(x > 20 - 9.304/ n; 19 ; 4/n) = 0.02

Pr(x < 20 - 9.304/ n; 19 ; 4/n) = 0.98

The z - value

Z = 2.054 = [(20- 9.304/ n - 19)/ 4/n]

2.054 = (1 - 9.304/ n ) / (4/n)

1 = (9.304 + 8.216)/n

n = 9.304 + 8.216 = 17.52

n = 306.95 or say

n = 307

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