Question

ATM's must be stocked with enough cash to satisfy customers making withdrawals over an entire weekend. But if too much cash is unnecessarily keep in the ATM's the bank is forgoing the opportunity of investing the money and earning interest. Suppose that at a particular branch, the population mean amount of money withdrawn from ATM's per customer transaction over the weekend is $160 with a population standard deviation of $30.

a) If a random sample of 36 customer transactions indicates that the sample mean withdrawal amount is $172 is there evidence to believe that the population mean withdrawal amount is no longer $160? (use a two-tailed test with 0.05 level of significance, show null hypothesis, alternative hypothesis and test statistic).
b) Using a proper table report the p-value and interpret its meaning
c) Given above answer, which type of error could you be making?
d) What is your answer in b) if you use a 0.01 level of significance (show test statistic)
e) What is your answer in b) if the standard deviation is $24 (use a =0.05) Show test statistic

Answer 1

a) The null and alternative hypothesis are:

`H_0: \mu=160\\\\H_a:\mu\\eq 160`

Test statistic z = 2.4

b) P-value = 0.016

c) Type I error.

d) P-value = 0.003

Explanation:

a) This is a hypothesis test for the population mean.

The claim is that the population mean withdrawal amount is no longer $160.

Then, the null and alternative hypothesis are:

`H_0: \mu=160\\\\H_a:\mu\\eq 160`

The significance level is 0.05.

The sample has a size n=36.

The sample mean is M=172.

The standard deviation of the population is known and has a value of σ=30.

We can calculate the standard error as:

`\sigma_M=(\sigma)/(√(n))=(30)/(√(36))=5`

Then, we can calculate the z-statistic as:

`z=(M-\mu)/(\sigma_M)=(172-160)/(5)=(12)/(5)=2.4`

b) This test is a two-tailed test, so the P-value for this test is calculated as:

`\text{P-value}=2\cdot P(z>2.4)=0.016`

As the P-value (0.016) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that the population mean withdrawal amount is no longer $160.

c) As the null hypothesis is rejected, the error we could be making is rejecting a true null hypothesis. This is a Type I error.

d) Significance level = 0.01.

As the P-value (0.016) is bigger than the significance level (0.01), the effect is not significant.

The null hypothesis failed to be rejected.

e) We can re-calculate the standard error for the new standard deviation as:

`\sigma_M=(\sigma)/(√(n))=(24)/(√(36))=4`

Then, we can calculate the z-statistic as:

`z=(M-\mu)/(\sigma_M)=(172-160)/(4)=(12)/(4)=3`

The recalculated P-value for this test is:

`\text{P-value}=2\cdot P(z>3)=0.003`

As the population standard deviation is smaller, a sample result far from the mean from the null hypothesis becames less probable (given that the null hypothesis is true), what is shown in a lower P-value.