Dolly Robinson

2021-08-23

Mr. Kram is considering running for mayor of the land of the dawn. Before completing the petitions, he decided to conduct a survey of voters in the land of dawn. A sample of 400 voters reveals that 300 would support him in the May Election. Develop a 99 percent confidence interval for the population proportion.

SchulzD

Let a random sample of size n=400 voters reveals that m=300 would support him n the May Election. Then, m/n gives an estimate of the unknown proportion p. Denote the estimator of p by $\stackrel{^}{p},i.e.,\stackrel{^}{p}=\frac{m}{n}$
Therefore the value of hatp is
$\stackrel{^}{p}=\frac{m}{n}=\frac{3}{4}=0.75$
Now, we know that for large samples, the estimator hatp is approximately normal with mean p and standard deviation $\sqrt{\frac{pq}{n}}$ where $q=1-p$.
In practice p is unknown and we are concerned with estimation of p. The standard deviation
$\sqrt{\frac{pq}{n}}$ is unknown then and has to be estimated.
The statistic
Therefore, we have:
$\sqrt{\frac{\stackrel{^}{p}\stackrel{^}{q}}{n}}=\sqrt{\frac{0.75\cdot 0.25}{400}}=0.022$ Thus, 99% confidence limits for p are $\sqrt{\frac{\stackrel{^}{p}\stackrel{^}{q}}{n}}=\sqrt{\frac{0.75\cdot 0.25}{400}}=0.022$
Therefore, the 99% confidence interval for the population
The confidence interval is (69%,80%).

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