Question

A soft-drink manufacturer purchases aluminum cans from an outside vendor. A random sample of 70 cans is selected from a large shipment, and each is tested for strength by applying an increasing load to the side of the can until it punctures. Of the 70 cans, 58 meet the specification for puncture resistance. Find a 95% confidence interval for the proportion of cans in the shipment that meet the specification. Round the answers to three decimal places.

Answer 1

Answer:

The 95% confidence interval for the proportion of cans in the shipment that meet the specification is (0.741, 0.917).

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 zscore that has a pvalue of $1 - \frac{\alpha}{2}$.

For this problem, we have that:

70 cans, 58 meet the specification for puncture resistance. This means that $n = 50, \pi = \frac{58}{70} = 0.829$

95% confidence level

So $\alpha = 0.05$, z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$, so $Z = 1.96$.

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.829 - 1.96\sqrt{\frac{0.829*0.171}{70}} = 0.741$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.829 + 1.96\sqrt{\frac{0.829*0.171}{70}} = 0.917$

The 95% confidence interval for the proportion of cans in the shipment that meet the specification is (0.741, 0.917).