The manager of the dairy section of a large supermarket chose a random sample of 250 egg cartons and found that 30 cartons had at least one broken egg. let p denote the proportion of all cartons which have at least one broken egg. Find a point estimate for p and also construct a 90% confidence interval for p.

The point estimate for p is of 0.12.

The 90% confidence interval for p is 0.0862 < p < 0.1538.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of , and a confidence level of , we have the following confidence interval of proportions.

In which

z is the z-score that has a p-value of .

The manager of the dairy section of a large supermarket chose a random sample of 250 egg cartons and found that 30 cartons had at least one broken egg.

This means that

The point estimate for p is of 0.12.

90% confidence level

So , z is the value of Z that has a p-value of , so .

The lower limit of this interval is:

The upper limit of this interval is:

The 90% confidence interval for p is 0.0862 < p < 0.1538.