Answer to Question #234793 in Statistics and Probability for R.K

Question

A random sample of 200 male shoppers in Pavilion shopping mall were interviewed to identify theiron reasons for coming to this particular mall. The factor of 'I prefer Knox brand in the mall' was the most important reason for 120 of those interviewed. Estimate the likely percentage of all male shoppers who frequent this shopping mall primarily because of the Knox branding the mall, using 97.5% confidence limits.

Answer 1

The sample proportion is computed as follows, based on the sample size $n=200$ and the number of favorable cases $X=120$

$$\hat{p}=\dfrac{X}{n}=\dfrac{120}{200}=0.6$$

The critical value for $\alpha=0.025$ is $z_c=z_{1-\alpha/2}=2.2414.$

The corresponding confidence interval is computed as shown below:

$$CI(proportion)=\bigg(\hat{p}-z_c\times\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}},$$

$$\hat{p}+z_c\times\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}\bigg)$$

$$=\bigg(0.6-2.2414\times\sqrt{\dfrac{0.6(1-0.6)}{200}},$$

$$0.6+2.2414\times\sqrt{\dfrac{0.6(1-0.6)}{200}}\bigg)$$

$$=(0.5224, 0.6776)$$

Therefore, based on the data provided, the 97.5% confidence interval for the population proportion is $0.5224<p<0.6776,$ which indicates that we are 97.5% confident that the true population proportion $p$ is contained by the interval $(0.5224, 0.6776).$