Question

A bank claims that the average cash withdrawal prior to the holidays by customers was $4,100. However, when auditors randomly examined the statements of 50 random bank customers their average withdrawal for that period turned out to be $4,360. Use a value of 5% for the probability of a Type 1 error. If the standard deviation for withdrawals is $1,200, determine the following:

a. Is there enough evidence from the sample of 50 customers to refute the bank’s claim? Explain the reason for your answer. b. What is the p-value based the information given above? Show how you determined the answer. c. Repeat part ‘a’ if the $4,360 average withdrawal was based on 100 random bank customers’ statements.

Answer 1

NullHypothesis : H₀ : average cash withdrawal prior to the holidays by customers was $ 4100. = $ 4100

Alternative Hypothesis : H_a: average cash withdrawal prior to the holidays by customers was $ 4100. $ 4100

Here the type I error probability = 0.05

Standard deviation of withdrawls = = $ 1200  

Average Withdrawal = = $ 4360

sample size n = 50

(a) Standard error of sample mean (se₀) = 1200/sqrt(50) = $ 169.706

Test statistic

Z = (4360 - 4100)/169.706 = 1.532

Here critical value Z_critical = 1.96

so here Z < Z_critical so we fail to reject the null hypothesis and say that there is not  evidence from the sample of 50 customers to refute the bank’s claim

(b) p- value = 2 * Pr(Z > 1.532) = 2 * 0.0628 = 0.1255

(c) Here n = 100

Standard error of sample mean (se₀) = 1200/sqrt(100) = $ 120

Test statistic

Z = (4360 - 4100)/120 = 2.16667

Here critical value Z_critical = 1.96

p- value = 2 * Pr(Z > 2.1667) = 0.0303

so here Z > Z_critical so we reject the null hypothesis and say that there is enough evidence from the sample of 50 customers to refute the bank’s claim