Question

A survey of 978 online shoppers was conducted In response to the question of what would influence the shopper to spend more money online for a certan year, 26% said tree ship , 19% said offering discounts while shopping, and 7% said product reviews Ca plete parts a through d below a. Construct a 95% confidence interval esin ate of the population propo tin « of online shoppers who would be infue ced to spend more money online with free shipping s1S□(Round to three decimal places as needed) aion proportion x of on ne shoppers who would be inf uenced to spend more Construct a 95% confidence interval estimate of the pop □szsr](Roundtothree decimal places as needed) sRound to three decimal places as neoded) population proportions in (a) through (c) to within t 0.03 money online with discounts offered while shopping e Construct a 95% conkence interval estimate of the population proportion x of online shoppers who would be influenced to spend more money online with product evies d. You have been asked to update te results of this study Determine the sample size necessary to estimate, wih 95% confidence, the For part a), the sample size is F or part b), the sample size i for part c). the sample size is esc z 17 3 4

Answer 1

Solution:- Given that n = 978, free shipping p1 = 0.26, offering discounts p2 = 0.19, reiews = 0.07
Z = 1.96

a. 95% confidence interval for the population proportion : p1 +/- Z*sqrt(p1*q1/n)
= 0.26 +/- 1.96*sqrt(0.26*0.74/978)
= 0.233 , 0.287

b. 95% confidence interval for the population proportion :

= 0.19 +/- 1.96*sqrt(0.19*0.81/978)
= 0.165 , 0.215

c. 95% confidence interval for the population proportion :
= 0.07 +/- 1.96*sqrt(0.07*0.93/978)
= 0.054 , 0.086

d.

n = (Z/E)^2*p*q

for (a) n = (1.96/0.03)^2*0.26*0.74 = 821.248 = 822

for (b) n = (1.96/0.03)^2*0.19*0.81 = 656.913 = 657

for (c) n = (1.96/0.03)^2*0.07*0.93 = 277.876 = 278