Question

A marketing research company desires to know the mean consumption of milk per week among males over age 25. They believe that the milk consumption has a mean of 2.5 liters, and want to construct a 85% confidence interval with a maximum error of 0.07 liters. Assuming a variance of 1.21 liters, what is the minimum number of males over age 25 they must include in their sample? Round your answer up to the next integer.

Answer 1

Answer: the minimum number of males over age 25 they must include in their sample n = 306

Step-by-step explanation:

Given;

Mean of milk consumption = 2.5litres

Maximum error E = 0.07

Variance S = 1.21 litres

Confidence interval of 85%

Z' = t(0.075)= 1.44

the minimum number of males over age 25 they must include in their sample = n

n = (Z'×S/E)^2

n = ( 1.44 × 0.85/0.07)^2

n = (17.4857)^2

n = 305.75

n = 306

Answer 2

In this exercise we have to use the knowledge of variance to calculate the value of n, so we have that:

the sample is n=306

Organizing the information given in the statement we have that:

• Mean of milk consumption = 2.5litres
• Maximum error E = 0.07
• Variance S = 1.21 litres
• Confidence interval of 85%

So given by the equation we have:

$Z' = t(0.075)= 1.44\\n = (Z'*S/E)^2\\n = ( 1.44 * 0.85/0.07)^2\\n = (17.4857)^2\\n = 305.75\\n = 306$

See more about variance at brainly.com/question/22365883