Posted by Matt on .
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,
Use a onesample ztest.
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 ztable at .01 level of significance for a onetailed test.
If the ztest statistic exceeds the critical value from the ztable, reject the null and conclude µ > 24819. If the ztest statistic does not exceed the critical value from the ztable, do not reject the null.
I hope this will help get you started. 
Statistics 
Anonymous,
0.065

Statistics 
Anonymous,
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