Question

A poll of 804 adults aged 18 or older asked about purchases that they intended to make for the upcoming holiday season. One of the questions asked about what kind of gift they intended to buy for the person on whom they would spend the most. Clothing was the first choice of 480 people. Give a 99% confidence interval for the proportion of people in this population who intend to buy clothing as their first choice.

Answer 1

Answer:

The 99% confidence interval for the proportion of people in this population who intend to buy clothing as their first choice is (0.55, 0.64).

Step-by-step explanation:

Let X = number of people who intend to buy clothing as their first choice.

The number of person intending to buy clothing as their first choice in a sample of n = 804 is, x = 480.

Compute the sample proportion of people who intend to buy clothing as their first choice as follows:

$\hat p=\frac{x}{n}=\frac{480}{804}=0.597$

As the sample size is, large, i.e. n = 804 > 30 and is selected from an unknown population, then according to the central limit theorem the sampling distribution of sample proportion follows a Normal distribution.

The mean of this sampling distribution is, $\mu_{\hat p}=\hat p=0.597$.

The standard deviation of this sampling distribution is, $\sigma_{\hat p}=\sqrt{\frac{\hat p(1-\hat p)}{n}}=\sqrt{\frac{0.597(1-0.597)}{804}}=0.0173$

A z-confidence interval will be used to compute the 99% confidence interval for the proportion of people in this population who intend to buy clothing as their first choice.

The critical value of z for 99% confidence level is:

$z_{\alpha/2}=z_{0.01/2}=z_{0.005}=2.58$

*Use a z-table.

Compute the 99% confidence interval for population proportion as follows:

$CI=\hat p\pm z_{\alpha/}\sqrt{\frac{\hat p(1-\hat p)}{n}}\\=0.597\pm 2.58\times 0.0173\\=0.597\pm 0.0446\\=(0.5524, 0.6416)\\\approx(0.55, 0.64)$

Thus, the 99% confidence interval for the proportion of people in this population who intend to buy clothing as their first choice is (0.55, 0.64).