A new cream advertises that it can reduce wrinkles and improve skin. In a recent study, a sample of 64 women over the age of 50 used the new cream for 6 months. Of those 64 women, 58 of them reported skin improvement (as judged by a dermatologist). Is this evidence that the cream will improve the skin of more than 60% of women over the age of 50? Test using α=0.01.

Answer:There is evidence that the cream will improve the skin of more than 60% of women over the age of 50 at the significance level 0.01Step-by-step explanation:We want to test the next hypothesis [tex]H_{0}: p = 0.6[/tex] vs [tex]H_{1}: p > 0.6[/tex] (upper-tail alternative) We have a large sample size n = 64 women over the age of 50. Therefore, the test statistic is  [tex]Z = \frac{\hat{p}-0.6}{\sqrt{0.6(0.4)/64}}[/tex] which is normally distributed. The observed value is   [tex]z = \frac{58/64-0.6}{\sqrt{0.6(0.4)/64}} = 5.001[/tex]. The rejection region is given by RR = {z | z > 2.3263} where 2.3263 is the 99th quantile of the standard normal distribution. Because the observed value 5.001 falls inside RR, we reject the null hypothesis and conclude that there is evidence that the cream will improve the skin of more than 60% of women over the age of 50 at the significance level 0.01