Question

From her purchased bags, Rory counted 110 red candies out of 550 total candies.

Using a 90% confidence interval for the population proportion, what are the lower and upper limit of the interval? Answer choices are rounded to the thousandths place.

Answer 1

Given,

\(\mathrm{n}=550 \ldots \ldots .\) Sample size

\(x=110\)........no. of successes in the sample

Let \(\hat{p}\) denotes the sample proportion.

\(\therefore \hat{p}=\mathrm{x} / \mathrm{n}=110 / 550=0.2\)

Our aim is to construct \(90 \%\) confidence interval.

\(\therefore c=0.90\)

\(\therefore \alpha=1-\mathrm{c}=1-0.90=0.10\)

\(\therefore \alpha / 2=0.10 / 2=0.05\) and \(1-\alpha / 2=0.950\)

\(\therefore Z_{\alpha / 2}=1.645\) (use z table)

Now, the margin of error is given by

\(\mathrm{E}=Z_{\alpha / 2 *} \sqrt{\hat{p}(1-\hat{p}) / n}\)

\(=1.645 * \sqrt{[0.2 *(1-0.2) / 550]}\)

\(=0.028\)

Now the confidence interval is given by

\((\hat{p}-\mathrm{E})

\((0.2-0.028)

\(0.172

Required \(90 \%\) Confidence Interval is \((0.172,0.228)\)

Lower limit : \(0.172\)

Upper limit : \(0.228\)