Respuesta :
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:
[tex]\hat p=\frac{x}{n}=\frac{480}{804}=0.597[/tex]
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, [tex]\mu_{\hat p}=\hat p=0.597[/tex].
The standard deviation of this sampling distribution is, [tex]\sigma_{\hat p}=\sqrt{\frac{\hat p(1-\hat p)}{n}}=\sqrt{\frac{0.597(1-0.597)}{804}}=0.0173[/tex]
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:
[tex]z_{\alpha/2}=z_{0.01/2}=z_{0.005}=2.58[/tex]
*Use a z-table.
Compute the 99% confidence interval for population proportion as follows:
[tex]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)[/tex]
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).
