Question

ATMs must be stocked with enough cash to satisfy customers making withdrawals over an entire weekend. On the other hand, if too much cash is unnecessarily kept in the ATMs, the bank is forgoing the opportunity of investing the money and earning interest. Suppose that at a branch the expected (i.e., population) average amount of money withdrawn from ATM machines per customer transaction over the weekend is $160 with an expected (i.e., population) standard deviation of $30.If a random sample of 36 customer transactions is examined and it is observed that the sample mean withdrawal is $172, is there evidence to believe that the true average withdrawal is more than $160? (Use a .05 level of significance.)

Answer 1

Given that,

Population mean = = 160

Sample mean = = 172

Population standard deviation = = 30

Sample size = n = 36

Level of significance = = 0.05

This a right (One) tailed test.

The null and alternative hypothesis is,

Ho: 160

Ha: 160

The test statistics,

Z = ( - )/ (/)

= ( 172 - 160) / ( 30 /36)

= 2.4

P-value = P(Z > z )

= 1 - P(Z < 2.4)

= 1 - 0.9918

= 0.0082

Therefore, P-value = 0.0082 < = 0.05 , then it is concluded that, reject the null hypothesis.

Conclusion :

It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population mean μ is more than 160, at the 0.05 significance level.