Try a t-test 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 t-test statistic.
Find the critical value for a one-tailed test using degrees of freedom (df = n - 1). Use a t-table. Compare to your t-test statistic calculated above. If the t-test statistic exceeds the critical value from the table, reject the null. If the t-test statistic does not exceed the critical value from the table, do not reject the null.
I hope this will help get you started.
"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.
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