Posted by **Matt** on Wednesday, May 9, 2012 at 8:21pm.

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 -
**MathGuru**, Thursday, May 10, 2012 at 7:29pm
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 -
**Anonymous**, Tuesday, December 15, 2015 at 4:38am
0.065

- Statistics -
**Anonymous**, Monday, May 16, 2016 at 10:17am
Bbccvn

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