Question

A new cream that advertises that it can reduce wrinkles and improve skin was subject to a recent study. A sample of 61 women over the age of 50 used the new cream for 6 months. Of those 61 women, 37 of them reported skin improvement (as judged by a dermatologist).

(a) Construct a 98% confidence interval for the percentage of all women who may have skin improvement.

(b) Suppose the confidence interval obtained in part a is too wide. How can the width of this interval be reduced? Discuss all possible alternatives. Which alternative is the best?

PLEASE SHOW DETAILED WORK, thank you!

Answer 1

(a)

n = Sample Size = 61

= Sample Proportion = 37/61 = 0.6066

SE =

= 0.02

ndf = 61 - 1 =60

From Table, critical values of t = 2.3901

Confidence Interval for proportion:

0.6066 (0.0265 X 2.3901)

= 0.6066 0.0633

= ( 0.5433 ,0.6699)

Confidence Interval for proportion:

0.5433< P <,0.6699

So,

A 98% confidence interval for the percentage of all women who may have skin improvement. is given by:

54.33 % < P <, 66.99 %

(b)

the width of the confidence interval be reduced as follows:

(i) Increase the Sample Size from the present 61 women to a higher value

(ii) Reduce the variability of data

(iii) Use one sided Confidence Interval

(iv) Lower the Confidence Level

The best alternative is: (iv) Lower the Confidence Level because it is expensive to increase the sample size. So, lowering the confidence level will shorten the width of confidence interval at the expense of losing some confidence.