Question

A market survey was conducted in a city which numbers 16,000 homemakers. The survey was conducted to estimate the proportion of homemakers who could recognize the brand name of a cleanser based on the shape and color of the container. Of the 1,400 homemakers surveyed, 420 were able to identify the brand name. a. Using the 0.99 degree of confidence, the population proportion lies within what interval? b. An advertising firm claims that at least 30% of all homemakers can recognize the brand name of a cleanser based on the container. Do you agree? Explain.

Answer 1

a) Given that

n=1400

p=420/1400=0.3

q=1-(420/1400)=0.7

To find the 99% CI for population proportion P

The 99% CI is given by the following formula

Where Z is the value of standard normal variate preciely Z0.005

Z0.005=2.57

Hence the confidence interval is

Therefore 99% population proportion lies within the above interval.

b) Now to answer this we need to understand that 30% homemakers in terms of proportion is 0.3.

So to test this claim the hypothesis is

Ho:P=0.3

Vs

H1: P>0.3

The Z statistic to test the above hypothesis is as follows:

Where Z follows standard normal distribution

P=0.3

p=0.3

Q=0.7

So

As the test is one sided the p-value is

Hence as this value of greater than 0.01(1-0.99) accept Ho that is the claim of advertising firm

is not true that is it is not true that atleast 30% of homemakers can recognise the brand name of cleanser based on container.