Question

Question 15: (10 points) Suppose a consumer advocacy group would like to conduct a survey to find the proportion p of consumers who bought the newest generation of a PlayStation player were happy with their purchase. a) How large a sample n should they take to estimate p with 2% margin of error and 90% confidence? b) The advocacy group took a random sample of 1000 consumers who recently purchased this PlayStation player and found that 300 were happy with their purchase. Find a 95% confidence interval for p.

Answer 1

Solution Consider all the given values Consider the given values and from the standard normal probability table. p= 0.5 Margin of error (ME)-0.02 Z 1.645 Find the value of n from the general expression. r2 0.5)1-0.5)0.02 r2 1.645 0.25 ( 0.02 1.645

0.25 0.02 1.645 1691.266n n = 1692 Thus, the required value of sample n is 1692. Consider that p" is proportion of consumers who bought the newest play-station player who were happy with purchase. Consider that n= 1000 and X=400. Calculate p 400 1,000 -0.4 For the 95 % confidence table, from the standard probabilities table. Z *-1 . 96

For the 95% confidence table, from the standard probabilities table. Z *-1.96 Find the value of ơ 0.4(1-0.4) 1000 -0.0156 Find the 95% confidence interval for p. (Z* * ơ.) -0.4 t (1.96x0.0156) 0.4 (0.030576) (0.3694,0.4306) Thus, the 95% confidence interval forp is between 0.3694 and 0.4306. 95% o confidence interval-