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).

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

[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

In which

z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].

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

This means that [tex]n = 180, \pi = \frac{167}{180} = 0.9278[/tex]

95% confidence level

So [tex]\alpha = 0.05[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.05}{2} = 0.975[/tex], so [tex]Z = 1.96[/tex].

The lower limit of this interval is:

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

The upper limit of this interval is:

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

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