A Web Usage Snapshot indicated a monthly average of 36Internet visits of a particular website per user from home. A random sample of 25 Internet users yielded a sample mean of 36.9visits with a standard deviation of 5.3.

At the 0.05level of significance, can it be concluded that this differs from the national average? Assume the population is normally distributed.

Express your answers correct to 3 decimal places.

a) What test should be performed (z or t)? Type the corresponding letter.

b) Find the critical value(s). If there are 2 values (2 -tail test), separate them with a coma.

c) Calculate the test statistic

d) Make a decision. Type a) if you choose to "Reject the null hypothesis"; Type b) if you choose to "Fail to reject the null hypothesis".

e) Choose the correct conclusion (Type 1 or 2):

1. At 5% level of significance, there is not enough evidence to support the claim that the mean is different from the national average.

2. At 5% level of significance, there is enough evidence to support the claim that the mean is different from the national average.