Question

You work for a large retailer and have been asked to estimate the proportion of your customers that are less than 30 years old. You have sampled a large number of stores and have found that of the 106 customers you have surveyed, 58 are less than 30 years old.

Assuming your sample is valid, what is the upper bound of a 99% confidence interval for the population proportion of customers who are less than 30 years old?
(Report your answer as a decimal and not as a percentage. For example, report 0.05 rather than 5%.)

Answer 1

Here the problem is a population proportion problem. Here first we need to get the sample proportion of people who are less than 30 years old.

out of 106 customers are surveyed, we got 58 customers less than 30 years.

Hence, we can get that 58/106 = 0.547 is the proportion of customers who are less than 30 years old.

Now we will discuss the confidence interval and upper limit:

Finding the margin of error:

We know that a sample proportion is only an estimate for the population proportion. We do not expect the sample proportion to equal the population proportion, so there is some error due to random chance. We use the standard deviation of the sample proportions to describe the amount of error we can expect in random samples. We call this the standard error.

We learned that the standard error of the sample proportion depends on the population proportion and sample size. Here is the formula for the standard error:

When we use a normal model for the sampling distribution, 99% of sample proportions estimate the population proportion within approximately 2.575829 standard errors (Accurate value has been obtained from R, will share the code in appendix section). So the margin of error is the following:

Hence the upper limit will be  

So, upon calculation we get

0.547 + 2.575829 * sqrt ((0.547 * (1 - 0.547)/106)

0.67154

Hence the answer.

Appendix

R-code

qnorm(0.995,0,1)

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