Question

A market research consultant hired by a leading soft-drink company wants to determine the proportion of consumers who favor its low-calorie drink over the leading competitor's low-calorie drink in a particular urban location. A random sample of 250 consumers from the market under investigation is provided in the file P08_17.xlsx.

a. Calculate a 95% confidence interval for the proportion of all consumers in this market who prefer this company's drink over the competitor's. Round your answers to three decimal places, if necessary.

Count = 134

n = 250

Answer 1

Solution :

Given that,

n = 250

x = 135

Point estimate = sample proportion = = x / n = 135 /250=0.536

1 - = 1 - 0.536=0.464

At 95% confidence level

= 1 - 95%  

= 1 - 0.95 =0.05

/2 = 0.025

Z/2 = Z_{0.025 = 1.96}

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

= 1.96 (((0.536*0.464) / 250)

= 0.062

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.536 - 0.062< p <0.536 + 0.062

0.474< p < 0.598

The 95% confidence interval for the population proportion p is : 0.474, 0.598