Post a New Question


posted by .

A large furniture store has begun a new ad campaign on local television. Before the campaign, the long term average daily sales were $24,819. A random sample of 40 days during the new ad campaign gave a sample mean daily sale of x(bar)=$25,910 with sample standard deviation s=$1917. Does this indicate that the population mean daily sales is now more than $24,819? Use a 1% level of significance.

  • Statistics -

    Use a one-sample z-test.

    Ho: µ = 24819 ---> null hypothesis
    Ha: µ > 24819 -->alternate hypothesis

    z = (sample mean - population mean)/(standard deviation divided by the square root of the sample size)

    With your data:
    z = (25910 - 24819)/(1917/√40) = ?

    Finish the calculation.

    Check a z-table at .01 level of significance for a one-tailed test.
    If the z-test statistic exceeds the critical value from the z-table, reject the null and conclude µ > 24819. If the z-test statistic does not exceed the critical value from the z-table, do not reject the null.

    I hope this will help get you started.

  • Statistics -


  • Statistics -


Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

More Related Questions

Post a New Question