A recent study by a Midwestern university has concluded that

A recent study by a Midwestern university has concluded that the time adults spend watching television each week averages 14.6 hours, with a standard deviation of 4.3 hours. Assuming these values are the population parameters, determine the probability that a sample of 100 adults from the population will average more than 15 hours of television per week. Does it seem likely that the university

Solution

Let mu be the population mean

The test hypothesis:

Ho: mu=15 (i.e. null hypothesis)

Ha: mu>15 (i.e. alternative hypothesis)

The test statisitc is

Z=(xbar-mu)/(s/vn)

=(14.6-15)/(4.3/sqrt(100))

=-0.93

It is a right-tailed test.

Assume that the significant level a=0.05

The critical value is Z(0.05) = 1.645 (from standard normal table)

The rejection region is if Z>1.645, we reject the null hypothesis.

Since Z=-0.93 is less than 1.645, we do not reject the null hypothesis.

So we can not conclude that the university