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).

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

Step-by-step explanation:

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

In which

z is the zscore that has a pvalue of .

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

This means that

95% confidence level

So , z is the value of Z that has a pvalue of , so .

The lower limit of this interval is:

The upper limit of this interval is:

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

the correct answer to your question is a. 4^4

good luck my friend!

hope this !

maybe if i you could me by giving me brainliest?

! < 3

have a splendid day < 3 : )