Question

A magazine provided results from a poll of 500500 adults who were asked to identify their favorite pie. Among the 500500 ​respondents, 1414​% chose chocolate​ pie, and the margin of error was given as plus or minus±33 percentage points. Given specific sample​ data, which confidence interval is​ wider: the 9999​% confidence interval or the 8080​% confidence​ interval? Why is it​ wider?

Answer 1

Answer:

99% confidence interval is wider as compared to the 80% confidence interval.

Step-by-step explanation:

We are given that a magazine provided results from a poll of 500 adults who were asked to identify their favorite pie.

Among the 500 respondents, 14​% chose chocolate​ pie, and the margin of error was given as plus or minus ±3 percentage points. 

The pivotal quantity for the confidence interval for the population proportion is given by;

                            P.Q.  =  $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$  ~ N(0,1)

where, $\hat p$ = sample proportion of adults who chose chocolate​ pie = 14%

            n = sample of adults = 500

            p = true proportion

Now, the 99% confidence interval for p =  $\hat p \pm Z_(_\frac{\alpha}{2}_) \times \sqrt{\frac{\hat p(1-\hat p)}{n} }$

Here, $\alpha$ = 1% so  $(\frac{\alpha}{2})$ = 0.5%. So, the critical value of z at 0.5% significance level is 2.5758.

Also, Margin of error = $Z_(_\frac{\alpha}{2}_) \times \sqrt{\frac{\hat p(1-\hat p)}{n} }$  = 0.03 for 99% interval.

So, 99% confidence interval for p  =  $0.14 \pm2.5758 \times \sqrt{\frac{0.14(1-0.14)}{500} }$

                                                        = [0.14 - 0.03 , 0.14 + 0.03]

                                                        = [0.11 , 0.17]

Similarly, 80% confidence interval for p  =  $0.14 \pm 1.2816 \times \sqrt{\frac{0.14(1-0.14)}{500} }$

Here, $\alpha$ = 20% so  $(\frac{\alpha}{2})$ = 10%. So, the critical value of z at 10% significance level is 1.2816.

Also, Margin of error = $Z_(_\frac{\alpha}{2}_) \times \sqrt{\frac{\hat p(1-\hat p)}{n} }$  = 0.02 for 80% interval.

So, 80% confidence interval for p  =  [0.14 - 0.02 , 0.14 + 0.02]

                                                           =  [0.12 , 0.16]

Now, as we can clearly see that 99% confidence interval is wider as compared to 80% confidence interval. This is because more the confidence level wider is the confidence interval and we are more confident about true population parameter.