Answer to Question #140706 in Statistics and Probability for lucy

Question

Suppose that a market research rm is hired to estimate the percent of adults living in a large city who have cell phones. 500 randomly selected adult residents in this city are surveyed to determine whether they have cell phones. Of the 500 people surveyed, 421 responded yes –they own cell phones. 3.1)Using a 90% confidence level, compute a confidence interval estimate for the true proportion of adults residents of this city who have cell phones.(6)3.2)Using a 99% confidence level, compute a confidence interval estimate for the true proportion of adults residents of this city who have cell phones.(4)3.3)Comment on the findings between questions 3.1 & 3.

Answer 1

The sample proportion is computed as follows, based on the sample size $n=500$

and the number of favorable cases $x=421$

$$\hat{p}=\dfrac{x}{n}=\dfrac{421}{500}=0.842$$

The corresponding confidence interval is computed as shown below:

$$CI(Proportion)=$$ $$=\big(\hat{p}-z_c \sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}, \hat{p}+z_c \sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}\big)$$

3.1. The critical value for $\alpha=0.1$ is $z_c=z_{1-\alpha/2}=1.645.$ The corresponding confidence interval is

$$CI(Proportion)=$$

$$=\big(0.842-1.645 \sqrt{\dfrac{0.842(1-0.842)}{500}},$$

$$0.842+1.645 \sqrt{\dfrac{0.842(1-0.842)}{500}}\big)=$$

$$=(0.815, 0.869)$$

Therefore, based on the data provided, the 90% confidence interval for the population proportion is $0.815<p<0.869,$ which indicates that we are 90% confident that the true population proportion $p$ is contained by the interval $(0.815, 0.869).$

3.2. The critical value for $\alpha=0.01$ is $z_c=z_{1-\alpha/2}=2.576.$ The corresponding confidence interval is

$$CI(Proportion)=$$

$$=\big(0.842-2.576\sqrt{\dfrac{0.842(1-0.842)}{500}},$$

$$0.842+2.576 \sqrt{\dfrac{0.842(1-0.842)}{500}}\big)=$$

$$=(0.800, 0.884)$$

Therefore, based on the data provided, the 99% confidence interval for the population proportion is $0.800<p<0.884,$ which indicates that we are 99% confident that the true population proportion $p$ is contained by the interval $(0.800, 0.884).$

Therefore we see that the width of the confidence interval increases as the confidence level increases.

We know that a higher confidence level gives a larger margin of error, so confidence level is also related to precision.

Increasing the confidence in our estimate makes the confidence interval wider and therefore less precise.