Question

The policy of a particular bank branch is that its ATMs must be stocked with enough cash to satisfy customers making withdrawals over an entire weekend. At this branch the expected (i.e., population) average amount of money withdrawn from ATM machines per customer transaction over the weekend is $160 with an expected (i.e., population) standard deviation of $30. Suppose that a random sample of 36 customer transactions is examined and it is observed that the sample mean withdrawal is $172. At α = 0.05 level of significance, is there enough evidence to believe that the population average withdrawal is greater than $160?

(a) Hypothesis: H0 :

Ha :

(b) Test Statistics=

(c) Pvalue=

(d) Decision:

(e) Conclusion:

Answer 1

The provided sample mean is and the known population standard deviation is , and the sample size is

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho:

Ha:

This corresponds to a right-tailed test, for which a z-test for one mean, with known population standard deviation, will be used.

Rejection Region

Based on the information provided, the significance level is α=0.05, and the critical value for a right-tailed test is Zc​=1.64.

The rejection region for this right-tailed test is R={z:z>1.64}

(2) Test Statistics

The z-statistic is computed as follows:

(3)

The P-value is p = P(Z>2.4)= 0.0082

(4) The decision about the null hypothesis

Since it is observed that z=2.4>zc​=1.64, it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is p=0.0082, and since p=0.0082<0.05, it is concluded that the null hypothesis is rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population mean μ is greater than 160, at the 0.05 significance level. Hence there enough evidence to believe that the population average withdrawal is greater than $160.

Graphically

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