Question

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.

Answer 1

Answer:

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 $\pi$, and a confidence level of $1-\alpha$, we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the z-score that has a p-value of $1 - \frac{\alpha}{2}$.

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 $n = 250, \pi = \frac{30}{250} = 0.12$

The point estimate for p is of 0.12.

90% confidence level

So $\alpha = 0.1$, z is the value of Z that has a p-value of $1 - \frac{0.1}{2} = 0.95$, so $Z = 1.645$.

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.12 - 1.645\sqrt{\frac{0.12*0.88}{250}} = 0.0862$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.12 + 1.645\sqrt{\frac{0.12*0.88}{250}} = 0.1538$

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