Question

Assume that a sample is used to estimate a population proportion p. Find the 80% confidence interval for a sample of size 362 with 54 successes. Enter your answer as a tri-linear inequality using decimals (not percents) accurate to three decimal places.

Answer 1

We have to find the 80% confidence interval for a population proportion.

The sample size is n = 362 and the number of successes is X = 54.

Then, the sample proportion is p = 0.149171.

`p=(X)/(n)=(54)/(362)\approx0.149171`

The standard error of the proportion is:

`\begin{gathered} \sigma_s=\sqrt{(p(1-p))/(n)} \\ \sigma_s=\sqrt{(0.149171*0.850829)/(362)} \\ \sigma_s=√(0.000351) \\ \sigma_s=0.018724 \end{gathered}`

The critical z-value for a 80% confidence interval is z = 1.281552.

Then, the lower and upper bounds of the confidence interval are:

`LL=p-z\cdot\sigma_s=0.149171-1.281552\cdot0.018724\approx0.1492-0.0240=0.1252`
`UL=p+z\cdot\sigma_s=0.1492+0.0240=0.1732`

As the we need to express it as a trilinear inequality, we can write the 80% confidence interval for the population proportion (π) as:

`0.125<\pi<0.173`

Answer: 0.125 < π < 0.173