Chapter 8.3, Problem 43E

Wilson’s interval: The small-sample method for constructing a confidence interval is a simple, approximation of a more complicated interval known as Wilson’s interval. Let $\hat{p}=x/n.$

Wilson’s confidence interval for p is given by $\frac{\hat{p}+\frac{z_{a/2^2}}{2n}\pm z_{a/2}\sqrt{\frac{\hat{p}\left(1-\hat{p}\right)}{n}+\frac{z_{a/2^2}}{4n^2}}}{1+\frac{z_{a/2^2}}{n}}$

Approximation depends on the level: The small-sample method is a good approximation to Wilson’s method for all confidence levels commonly used in practice, but is best when is close to 2. Refer to Exercise 41.

1. Use Wilson’s method to construct a 90% confidence interval, a 95% confidence interval, and a 99% confidence Interval for the proportion of tenth graders who plan to attend college.
2. Use the small-sample method to construct a 90% confidence interval, a 95% confidence interval, and a 99% confidence interval for the proportion of tenth graders who plan to attend college.
3. For which level is the small-sample method the closest to Wilson’s method? Explain why this is the case.

To determine

Use Wilson’s method to construct the confidence interval.

Answer to Problem 43E

At 90% confidence interval the probability value is

  $\text{0}\text{.3626}<p<\text{0}\text{.7469}$

At 95% confidence interval the probability value is

  $\text{0}\text{.3075}<p<\text{0}\text{.7518}$

At 99% confidence interval the probability value is

  $\text{0}\text{.2026}<p<\text{0}\text{.7486}$

Explanation of Solution

The Wilson’s method formula is given,

  $Wilson’smethod=\frac{p+\frac{z_{\alpha /2}^2}{2n}\pm z_{\alpha /2}\sqrt{\frac{p(1-p)}{n}+\frac{z_{\alpha /2}^2}{4n^2}}}{1+\frac{z_{\alpha /2}^2}{n}}$

Here, p is the probability value and z are critical value

  $p=\frac{9}{15}$

Then compute the Wilson’s confidence interval method

At 90% confidence interval the probability value is

  $\text{0}\text{.3626}<p<\text{0}\text{.7469}$

At 95% confidence interval the probability value is

  $\text{0}\text{.3075}<p<\text{0}\text{.7518}$

At 99% confidence interval the probability value is

  $\text{0}\text{.2026}<p<\text{0}\text{.7486}$

Program:

clc
clear
close all
n=15;
x=9;
p=9/15;
z1=1.645;
z2=1.96;
z3=2.575;
up90=p+(z1/(2*n))+z1*sqrt((p*(1-p)/n)+((z1)^2/(4*n^2)));
up95=p+(z2/(2*n))+z2*sqrt((p*(1-p)/n)+((z2)^2/(4*n^2)));
up99=p+(z3/(2*n))+z3*sqrt((p*(1-p)/n)+((z3)^2/(4*n^2)));
div90=1+(z1^2/n);
div95=1+(z2^2/n);
div99=1+(z3^2/n);
w90p=up90/div90
w95p=up95/div95
w99p=up99/div99
low90=p+(z1/(2*n))-z1*sqrt((p*(1-p)/n)+((z1)^2/(4*n^2)));
low95=p+(z2/(2*n))-z2*sqrt((p*(1-p)/n)+((z2)^2/(4*n^2)));
low99=p+(z3/(2*n))-z3*sqrt((p*(1-p)/n)+((z3)^2/(4*n^2)));
w90n=low90/div90
w95n=low95/div95
w99n=low99/div99

To determine

Using small-sample method to construct the confidence interval.

Answer to Problem 43E

At 90% confidence interval the probability value is

  $\text{0}\text{.4151}<p<\text{0}\text{.7849}$

At 95% confidence interval the probability value is

  $\text{0}\text{.3797}<p<\text{0}\text{.8203}$

At 99% confidence interval the probability value is

  $\text{0}\text{.3106}<p<\text{0}\text{.8894}$

Explanation of Solution

The small-sample method formula is given,

  $small-samplemethod=p\pm z_{\alpha /2}\sqrt{\frac{p(1-p)}{4+n}}$

Here, p is the probability value and z are critical value

  $p=\frac{9}{15}$

Then compute the Wilson’s confidence interval method

At 90% confidence interval the probability value is

  $\text{0}\text{.4151}<p<\text{0}\text{.7849}$

At 95% confidence interval the probability value is

  $\text{0}\text{.3797}<p<\text{0}\text{.8203}$

At 99% confidence interval the probability value is

  $\text{0}\text{.3106}<p<\text{0}\text{.8894}$

Program:

clc
clear
close all
n=15;
x=9;
p=9/15;
z1=1.645;
z2=1.96;
z3=2.575;
up90=p+z1*sqrt((p*(1-p)/(n+4)))
up95=p+z2*sqrt((p*(1-p)/(n+4)))
up99=p+z3*sqrt((p*(1-p)/(n+4)))
low90=p-z1*sqrt((p*(1-p)/(n+4)))
low95=p-z2*sqrt((p*(1-p)/(n+4)))
low99=p-z3*sqrt((p*(1-p)/(n+4)))

To determine

Explain at which interval Wilson’s method close to small-sample method.

Answer to Problem 43E

At the 90 % interval Wilson’s method close to small-sample method.

The Wilson’s method shows

At 90% confidence interval the probability value is

  $\text{0}\text{.3626}<p<\text{0}\text{.7469}$

The Small-sample method shows

At 90% confidence interval the probability value is

  $\text{0}\text{.4151}<p<\text{0}\text{.7849}$

Explanation of Solution

Wilson’s method formula

  $Wilson’smethod=\frac{p+\frac{z_{\alpha /2}^2}{2n}\pm z_{\alpha /2}\sqrt{\frac{p(1-p)}{n}+\frac{z_{\alpha /2}^2}{4n^2}}}{1+\frac{z_{\alpha /2}^2}{n}}$

The small-sample formula

  $small-samplemethod=p\pm z_{\alpha /2}\sqrt{\frac{p(1-p)}{4+n}}$