Question

A random sample of size n = 121 yielded $\hat{\mathit{p}}$ = .88.

a. Is the sample size large enough to use the methods of this section to construct a confidence interval for p? Explain.

b. Construct a 90% confidence interval for p.

c. What assumption is necessary to ensure the validity of this confidence interval?

Short answer

a. The sample size is not large enough

b. The 90% confidence interval is$\left(0.8314,0.9286\right)$

c.The samples are random samples,and the sample size is large enough, it must be$n\hat{p}⩾15,n\left(1-\hat{p}\right)⩾15$

Step-by-step solution

Given information

A random sample of size n = 121 yielded $\hat{p}$= .88.

They need to compute the following

a. If the sample size is large enough

b. 90% confidence interval for p

c. Assumptions are necessary to ensure the validity of this confidence interval.

(a) Explanation: If the sample size is large enough or not

We know that the sample size is large enough only if $n\hat{p}⩾15,n\left(1-\hat{p}\right)⩾15$

Here

$\begin{array}{rcl}n\hat{p}& =& 121\left(0.88\right)=106.48⩾15\\ n\left(1-\hat{p}\right)& =& 121\left(1-0.88\right)=14.52<15\\ & & \end{array}$

The second condition is not satisfied.

Hence the sample is not large

(b) Calculation: 90% confidence interval for p

Here the confidence coefficient is 90%, hence $\alpha =10\%$.Now from the standard normal distribution table,$z_{\frac{\alpha }{2}}=1.645$

The margin of error

$\begin{array}{rcl}E& =& z_{\frac{\alpha }{2}}\times \sqrt{\frac{\hat{p}\left(1-\hat{p}\right)}{n}}\\ & =& 1.645\times \sqrt{\frac{0.88\times \left(1-0.88\right)}{121}}\\ & =& 0.0486\\ & & \end{array}$

Hence 90% lower limit:$\hat{p}-E=0.88-0.0486=0.8314$

90% upper limit:$\hat{p}+E=0.88+0.0486=0.9286$

The 90% confidence interval is$\left(0.8314,0.9286\right)$

(c) The validity of assumptions

The following conditions are to be satisfied in order to calculate the construction of the confidence interval:

1. The samples are a random sample

2. The sample size is large enough; it must be$n\hat{p}⩾15,n\left(1-\hat{p}\right)⩾15$