Question

Internet Visits

A U.S. Web Usage Snapshot indicate a monthly average of 36 internet visits per user from home. A random sample of 24 Internet user yielded a sample mean of 42.1 visitswith a standard deviation of 5.3. At the .01 level of significance can it be concluded that this differs from the national average?
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Answer #2

The test hypothesis is
Ho:μ=36
Ha:μ not equal to 36

The test statistic is

Z=(xbar-μ)/(s/√n)

=(42.1-36)/(5.3/sqrt(24))

=5.64

Given a=0.01, the critical value is |Z(0.005)|=2.58 (check standard normal table)

Since Z=5.64 is larger than 2.58, we reject Ho.

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