Question

A survey is to be conducted to determine the average driving in miles by MS dents. The investigator wants to know how are the sample should be taken to confidence interval on the true average within 1.5 miles. A similar study on the past found that the population standard deviation was 82 miles Bonus(5 points): A recent study of 750 Internet users in Europe found that 35% of Internet users were women. What is the 96% confidence interval of the true proportion of women in Europe who use the Internet?

solve both pls

Answer 1

(a)

Z value for 92% confidence interval is 1.75

Margin of error, E = 1.5 miles

$\sigma$ = 8.2 miles

Sample size, n = (z$\sigma$/E)²

= (1.75 * 8.2 / 1.5)²

= 92 (Round to next integer)

(b)

Point estimate, p = 0.35

Standard error of proportion, SE = = 0.0174

VP(1 - p)/n = 0.35* (1 -0.35)/750

Z value of 96% confidence interval is  2.054

96% confidence interval of proportion of women in Europe using the Internet is,

(0.35 - 2.054 * 0.0174 , 0.35 + 2.054 * 0.0174)

(0.3143, 0.3857)