Stats Homework...

Question

Suppose that a car company wants to determine the importance to consumers for service like OnStar, which offers drivers a wifi connection (among other things). Small/preliminary studies have concluded that about 58% of people would like an wifi connection in their car.

(a) Use the given preliminary estimate to determine the sample size required to estimate the proportion of adult Americans who would like wifi capabilities in their car to within .02 with 95% confidence.

(b) How would the interval be affected if 99% confidence interval was desired? Select an answer The interval would get larger The interval would get smaller The population proportion would change  

(c) Use the given preliminary estimate to determine the sample size required to estimate the proportion of adult Americans who would like wifi capabilities in their car to within .02 with 99% confidence. 

Answer 1

Hello Hayley,

a) Since the margin of error is given by Z_∝/2√(p^(1-p^)/n); in other words, the product of the critical value corresponding (1-∝)% confidence and the standard error; solving for n yields:

n = p^(1-p^)*(Z_∝/2/E)², where we may use the preliminary value for p^ of 0.58 and E = 0.02 and Z_∝/2 = 1.96

n = (0.58)*(1-0.58)*(1.96/0.02)

n = (0.58)*(0.42)*(98) ≈ 24 (when rounded up to largest integer)

b) If 99% confidence was desired the critical value Z_∝/2 would be larger thus resulting in a larger margin of error which would produce a larger interval.

c) Using the preliminary estimate with a margin of error, E = 0.02, and 99% level of confidence, the sample size required would be:

n = (0.58)(1-0.58)(2.575/0.02) ≈ 32