Posted by Nadine on Sunday, March 10, 2013 at 1:30pm.
"Working Mom’s Journal" reported that the mean time a mother, with her small children, spends in at the convenience store is 7.3 minutes. A sample of 20 moms is chosen from your neighborhood, and it is found that the mean time they spend in a convenience store was 8.2 minutes with a standard deviation of 1.4 minutes. Using , test the claim that the average amount of time a mom and her children spend in a convenience store is greater than 7.3 minutes.
Determine which test statistic you will use: the standard normal distribution, or the student’s t distribution. Explain why you chose this test statistic.
Establish the null and alternative hypotheses, state the claim.
Test the claim at and discuss your results, should you reject or not reject the null hypothesis, should you reject or except the claim.

Statistics  MathGuru, Monday, March 11, 2013 at 7:30pm
Try a ttest since your sample size is rather small.
Formula:
t = (sample mean  population mean)/(standard deviation divided by the square root of the sample size)
sample mean = 8.2
population mean = 7.3
standard deviation = 1.4
sample size = 20
Plug in the values and calculate the ttest statistic.
Find the critical value for a onetailed test using degrees of freedom (df = n  1). Use a ttable. Compare to your ttest statistic calculated above. If the ttest statistic exceeds the critical value from the table, reject the null. If the ttest statistic does not exceed the critical value from the table, do not reject the null.
I hope this will help get you started.
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