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?

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.  =    ~ N(0,1)

where, = sample proportion of adults who chose chocolate pie = 14%

            n = sample of adults = 500

            p = true proportion

Now, the 99% confidence interval for p =  

Here, = 1% so   = 0.5%. So, the critical value of z at 0.5% significance level is 2.5758.

Also, Margin of error =  = 0.03 for 99% interval.

So, 99% confidence interval for p  =  

                                                        = [0.14 - 0.03 , 0.14 + 0.03]

                                                        = [0.11 , 0.17]

Similarly, 80% confidence interval for p  =  

Here, = 20% so   = 10%. So, the critical value of z at 10% significance level is 1.2816.

Also, Margin of error =  = 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.

it would be 6xy and -16xy.

72 is the answer to this problem