Respuesta :
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. = [tex]\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] ~ N(0,1)
where, [tex]\hat p[/tex] = sample proportion of adults who chose chocolate pie = 14%
n = sample of adults = 500
p = true proportion
Now, the 99% confidence interval for p = [tex]\hat p \pm Z_(_\frac{\alpha}{2}_) \times \sqrt{\frac{\hat p(1-\hat p)}{n} }[/tex]
Here, [tex]\alpha[/tex] = 1% so [tex](\frac{\alpha}{2})[/tex] = 0.5%. So, the critical value of z at 0.5% significance level is 2.5758.
Also, Margin of error = [tex]Z_(_\frac{\alpha}{2}_) \times \sqrt{\frac{\hat p(1-\hat p)}{n} }[/tex] = 0.03 for 99% interval.
So, 99% confidence interval for p = [tex]0.14 \pm2.5758 \times \sqrt{\frac{0.14(1-0.14)}{500} }[/tex]
= [0.14 - 0.03 , 0.14 + 0.03]
= [0.11 , 0.17]
Similarly, 80% confidence interval for p = [tex]0.14 \pm 1.2816 \times \sqrt{\frac{0.14(1-0.14)}{500} }[/tex]
Here, [tex]\alpha[/tex] = 20% so [tex](\frac{\alpha}{2})[/tex] = 10%. So, the critical value of z at 10% significance level is 1.2816.
Also, Margin of error = [tex]Z_(_\frac{\alpha}{2}_) \times \sqrt{\frac{\hat p(1-\hat p)}{n} }[/tex] = 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.
