Respuesta :
Answer:
The 96% confidence interval for the proportion of people who say they jog regularly is between 0.1222 and 0.1578. This means that we are 96% sure that the true population proportion of people who jog is between 0.1222 and 0.1578.
With a 99% confidence level, the interval would be wider.
The smallest number of people we could contact to still have a margin of error of 3% at a 96% confidence level is 566.
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence level of [tex]1-\alpha[/tex], we have the following confidence interval of proportions.
[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In which
z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].
The margin of error is:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
A polling organization contacts 1600 people and of these 14% say that they jog regularly.
This means that [tex]n = 1600, \pi = 0.14[/tex]
Construct and interpret a 96% confidence interval for the proportion of people who say they jog regularly.
96% confidence level
So [tex]\alpha = 0.04[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.04}{2} = 0.98[/tex], so [tex]Z = 2.056[/tex].
Margin of error:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]M = 2.056\sqrt{\frac{0.14*0.86}{1600}}[/tex]
[tex]M = 0.0178[/tex]
Lower bound:
Sample proportion subtracted by the margin of error. So
[tex]\pi - M = 0.14 - 0.0178 = 0.1222[/tex]
Upper bound:
Margin of error added to the sample proportion. So
[tex]\pi + M = 0.14 + 0.0178 = 0.1578[/tex]
The 96% confidence interval for the proportion of people who say they jog regularly is between 0.1222 and 0.1578. This means that we are 96% sure that the true population proportion of people who jog is between 0.1222 and 0.1578.
Explain what changes if you instead use a 99% confidence level.
With a 99% confidence level, the value of Z would be higher(2.575), which means that the interval would be wider.
What is the smallest number of people we could contact to still have a margin of error of 3% at a 96% confidence level.
This is n for which [tex]M = 0.03[/tex]
So
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.03 = 2.056\sqrt{\frac{0.14*0.86}{n}}[/tex]
[tex]0.03\sqrt{n} = 2.056\sqrt{0.14*0.86}[/tex]
[tex]\sqrt{n} = \frac{2.056\sqrt{0.14*0.86}}{0.03}[/tex]
[tex](\sqrt{n})^2 = (\frac{2.056\sqrt{0.14*0.86}}{0.03})^2[/tex]
[tex]n = 565.5[/tex]
Rounding up:
The smallest number of people we could contact to still have a margin of error of 3% at a 96% confidence level is 566.
