(5)A study determined the number of cavity-free children per 100 children in 16 North American cities before and after public water fluoridation projects. This is paired data since the before and after values come from the same cities. The mean and the standard deviation of thedifferencebetween the data before and after the fluoridation is given below. Test the hypothesis that the number of cavities is significantly lower after the fluoridation. Use 5% level for the significance

(d= the difference between the number of cavities after and before the fluoridation)

Sample size: 16

Mean (diff)= -12.31

Standard Deviation (diff)=13.92

a) Write the appropriate null- and alternative hypotheses,

b) Use Minitab to test the hypothesis for 1-sample t for the difference in the two means. (Copy the Minitab results below)

c) What is the t-value and the p-value?

d) What is the decision regarding the null hypothesis? Use a 5% significance level.

e) What is the conclusion regarding the investigation?

f) With the same mean and the standard deviation, run the 1-sample hypothesis test when the sample size is 5. (Copy the Minitab results below)

g) At a 5% level of significance, is the null hypothesis rejected?

h) How do the answers in parts d and g compare? If there is a difference in the answers for d and g, what might be the reason for that?