Question

(6 points) A new cream that advertises that it can reduce wrinkles and improve skin was subject to a recent study. A sample of 50 women over the age of 50 used the new cream for 6 months. Of those 50 women, 31 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 a = 0.01. (a) Test statistic: z = (b) Critical Value: z* = (c) The final conclusion is A. We can reject the null hypothesis that p 0.6 and accept that p > 0.6. That is, the cream can improve the skin of more than 60% of women over 50. B. There is not sufficient evidence to reject the null hypothesis that p 0.6. That is, there is not sufficient evidence to reject that the cream can improve the skin of more than 60% of women over 50.

Answer 1

Ans :

(a)

Test statitic is

$z_{\alpha }= 2.33$

(b)

Critical value

z = 0.289

(c)

B. There is not sufficient evidence to reject the null hypothesis that p = 0.6 . That is there is not sufficient evidence to reject that the cream can improve the skin of more than 60 % of women over 50

Here we want to test whether the cream will improve the skin of more than 60% of women over the age of 50 .

Let p be the proportion of women over the age of 50

Therefore p = 60 % ~ 0.6

The null hypothesis is given as

i.e the cream will improve the skin of is significantly different than 60% of women over the age of 50 .

vs the alternative hypothesis

i.e the cream will improve the skin of more than 60% of women over the age of 50 .

From the sample of 50 women, 31 of them reported skin improvement.

n = 50

x = 31

2 p= n

$=\frac{31}{50}$

=0.62

Compute the standard error.

$SE=\sqrt{p*(1-p)/n}$

$=\sqrt{0.6(1-0.6)/50}$

=  0.06928203

Computing test statistic

$z = \frac{\hat{p}-p}{SE}$

$=\frac{0.62 - 0.6}{0.06928203 }$

= 0.2886751
Obtaining the z -critical value for 0.01 level of significance from the normal distribution table

0 1 2 3 4 5 6 7 8 9 0.0 5000 5040 SUSU 5120 5160 5199 523952795319 5359 0.1 5398 5438 5478 5517 5557 5596 5636 5675 5714 5753 0.25793 58325871 5910 5948 5987 6026 60646103 6141 0.36179 6217 6255 6293 6331 6368 6406 6443 6480 6517 0.465546591 6628 6664 6700 6736 67726808 6844 6879 0.5 6915 6950 6985 7019 7054 7088 7123 7157 7190 7224 0.6 7257 7291 7324 7357 7389 74227454 7486 7517 7549 0.7 7580 7611 7642 7673 7704 7734 7764 7794 7823 7852 0.8 7881 7910 7939 7967 7995 8023 8051 8078 8106 8133 0.981598186 8212 8238 82648289 8315 8340 8365 8389 1.0 8113 8438 8461 8485 8508 8531 8554 8577 8599 8621 1.1 8643 86658686 8708 87298749 8770 8790 8810 8830 1.2 8849 8869 8888 8907 8925 8914 896289808997 9015 1.39032 90499066 9082 90999115 9131 9147 9162 9177 1.4 91929207 9222 9236 9251 9265 9279 9292 9306 9319 1.5 9332 9345 9357 9370 93829394 9406 9418 9429 9441 1.6 9452 9463 9474 9484 94959505 9515 9525 9535 9545 1.7 9554 9564 9573 9582 9591 959996089616 9625 9633 1.8 9641 96499656 9664 9671 9678 9686 9693 9699 9706 1.9 9713 9719 9726 9732 9738 9744 9750 9756 9761 9767 2.0 9772 9778 9783 9788 9793 9798 9803 9808 9812 9817 2.1 | 9821 982698309834 9838 9842 98469850 9854 9857 2.29861 9864 9868 9871 9875 9878 9881 9884 9887 9890 2.3 9893 9896 9898 99019904 9906 9909 9911 9913 9916 2.49918 9920 9922925 9927 99299931 9932 99349936 2.5 9938 9940 9941 9943 9945 9946 9948 9949 99519952 2.6 9953 99559956 9957 9959 9960 996199629963 9964 2.79965 9966 9967 9968 99699970 9971 99729973 9974 2.8 9974 9975 9976 9977 9977 9978 997999799980 9981 2.9 9981 998299829983 9984 99849985 998599869986

$z_{\alpha }= 2.33$

Decision rule :

Reject null hypothesis if $z < z_{\alpha }$

i.e z (calculated ) < z (critcal)

Since z (calculated ) = 2.887 < z (critcal) =2.33

We failed to reject the null hypothesis .

There is not sufficient evidence that the cream will improve the skin of more than 60% of women over the age of 50 .