Question

Suppose that nine bats was used. For each trail, the zoo keeper pointed to one of two "feeders". Suppose that the bats went to the correct feeder (the one that the zoo keeper pointed at) 7 times. Find the 95% confidence interval for the population proportion of times that the bats would follow the point.

Answer 1

The 95% confidence interval for the population proportion of times that the bats would follow the point is (0.5062, 1).

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)`.

For this problem, we have that:

`n = 9, \pi = (7)/(9) = 0.7778`

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)} = 0.7778 - 1.96\sqrt{(0.7778*0.2222)/(9)} = 0.5062`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.7778 + 1.96\sqrt{(0.7778*0.2222)/(9)} = 1.05 = 1`

The proportion cannot be above 1, so 1.05 is 1.

The 95% confidence interval for the population proportion of times that the bats would follow the point is (0.5062, 1).