Question

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).

Answer 1

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).