Question

Suppose a consumer advocacy group would like to conduct a survey to find the proportion of consumers who bought the newest generation of an MP3 player were happy with their purchase. Their survey asked consumers if they were happy or unhappy with their purchase. The responses indicated 28 out of 70 customers reported being unhappy with their purchase. Compute a 90% confidence interval for the population proportion of consumers who are happy with their purchase.

(.2492, .5508)

(.4852, .7148)

(.5037, .6963)

(.3037, .4963)

Answer 1

Solution :

Given that,

n = 70

x = 28

Point estimate = sample proportion = = x / n = 28 / 70=0.4

1 - = 1 - 0.4=0.6

At 90% confidence level

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z_{0.05 = 1.645}

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 1.645 (((0.4*0.6) /70 )

=0.0963

A 90% confidence interval for population proportion p is ,

- E < p < + E

0.4 - 0.0963  < p < 0.4 + 0.0963

0.3037< p < 0.4963

The 90% confidence interval for the population proportion p is : (0.3037, 0.4963)