Question

1. Recall that a bank manager has developed a new system to reduce the time customers spend waiting for teller service during peak hours. The manager hopes the new system will reduce waiting times from the current 9 to 10 minutes to less than 6 minutes. Suppose the manager wishes to use the random sample of 100 waiting times to support the claim that the mean waiting time under the new system is shorter than six minutes.

a. Letting μ represent the mean waiting time under the new system, set up the null and alternative hypotheses needed if we wish to attempt to provide evidence supporting the claim that μ is shorter than six minutes. b. The random sample of 100 waiting times yields a sample mean of 5.46 minutes. Assuming that the population standard deviation equals 2.47 minutes, use critical values to test H0 versus Ha at each of α = .10, .05, .01, and .001.

c. Using the information in part b, calculate the p-value and use it to test H0 versus Ha at each of α = .10, .05, .01, and .001.

d. How much evidence is there that the new system has reduced the mean waiting time to below six minutes?

Please show work for all questions

Answer 1

a. Letting μ represent the mean waiting time under the new system, set up the null and alternative hypotheses needed if we wish to attempt to provide evidence supporting the claim that μ is shorter than six minutes.

a)

The Null and Alternative Hypotheses

HOM= 6 minutes

HA :μ<6 minutes

b)

The t statistic is obtained using the formula,

Where,

Sample mean, X = 5.46

Sample standard deviation, s= 2.47

t= 5.46 - 6 2.47/V100

t = -2.186

The critical values are obtained from t critical value table for degree of freedom = n - 1 = 100 - 1 = 99 and for different value of significance level = $\alpha$.

$\alpha$	t-critical
0.1	-1.29016
0.05	-1.66039
0.01	-2.36461
0.001	-3.1746

Decision Rule: Reject the null hypothesis if t-statistic is greater than t-critical value,

t-statistic		t-critical	Decision
-2.186	>	-1.29016	Reject the null hypothesis
-2.186	>	-1.66039	Reject the null hypothesis
-2.186	<	-2.36461	Do not reject the null hypothesis
-2.186	<	-3.1746	Do not reject the null hypothesis

c)

The p-value is obtained from t distribution table for t-statistic = -2.186 and degree of freedom = n - 1 = 100 - 1 = 99 and significance level = $\alpha$

Decision Rule: Reject the null hypothesis if P-value is less than significance level $\alpha$,

t-statistic	P-value		$\alpha$	Decision
-2.186	0.015586	<	0.1	Reject the null hypothesis
-2.186	0.015586	<	0.05	Reject the null hypothesis
-2.186	0.015586	>	0.01	Do not reject the null hypothesis
-2.186	0.015586	>	0.001	Do not reject the null hypothesis

d)

the minimum value of significance level $\alpha$ at which, the result is significant is 0.05 hence there is statistical evidence at 5% significance level that the new system has reduced the mean waiting time to below six minutes.