service is normally distributed with a mean of 3 minutes and a standard

deviation of 1 minute. The quality assurance dept found in a sample of

50 customers that the mean waiting time was 2.75 minutes. At the .05

significance level, can we conclude that the mean waiting time is less

than 3 minutes?

a) State the null hypothesis adn the the alternate hypothesis.

Let mu be the population mean waiting time.

Test the following hypotheses:

H0: mu = 3

Ha: mu < 3

b) State the decision rule?

Since the population standard deviation is known, we obtain the critical value from the normal table.

At alpha=0.05, for the one-tailed test:

Z(1-alpha)=Z(0.95)=1.64

The critical value is -1.64 for left-tailed test.

Decision rule: reject H0 if test statistic Z0 < -1.64

c) Compute the value of the test statistic?

Let Xbar be the sample mean.

Xbar=2.75

sample size n=50

Let sigma be the population standard deviation.

sigma=1

Let se_Xbar be the standard error of the sample mean.

se_Xbar=sigma/sqrt(n)

=1/sqrt(50)

=0.1414

The test statistic:

Z0=(Xbar-mu)/se_Xbar

=(2.75-3)/0.1414

=-1.77

d) What is your decision regarding H of o?

Z0 < -1.64

Therefore we reject H0 at 0.05 level and conclude that the mean waiting time is less than 3 minutes.

e) What is the p-value? Interpret it.

p-value=P(Z<-1.77)

=1-P(Z<1.77)

=1-0.9616

=0.0384

The p-value is the probability of getting a value of the test statistic as extreme as or more extreme than -1.77 by chance alone, if the null hypothesis H0, is true.