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Information on a packet of seeds claims that 93% of them will germinate. Of the 180 seeds that I planted, only 167 germinated. Compute a 95% two-sided Agresti-Coull CI on the proportion of seeds that germinate. Use as a point estimator p^ the proportion of seeds that germinated during the experiment. Round your answers to two decimal places (e.g. 98.76).

User Lofte
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1 Answer

11 votes

Answer:

The 95% CI on the proportion of seeds that germinate is (0.89, 0.9256).

Explanation:

In a sample with a number n of people surveyed with a probability of a success of
\pi, and a confidence level of
1-\alpha, we have the following confidence interval of proportions.


\pi \pm z\sqrt{(\pi(1-\pi))/(n)}

In which

z is the zscore that has a pvalue of
1 - (\alpha)/(2).

Of the 180 seeds that I planted, only 167 germinated.

This means that
n = 180, \pi = (167)/(180) = 0.9278

95% confidence level

So
\alpha = 0.05, z is the value of Z that has a pvalue of
1 - (0.05)/(2) = 0.975, so
Z = 1.96.

The lower limit of this interval is:


\pi - z\sqrt{(\pi(1-\pi))/(n)} = 0.9278 - 1.96\sqrt{(0.9278*0.0722)/(180)} = 0.89

The upper limit of this interval is:


\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.9278 + 1.96\sqrt{(0.9278*0.0722)/(180)} = 0.9256

The 95% CI on the proportion of seeds that germinate is (0.89, 0.9256).

User KGBird
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