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.

A. (0.59, 1.05)

B. (0.44,0.94)C. (0.51, 1.0)

Answer 1

Answer: B. (0.44,0.94)

Step-by-step explanation:

Given : Number of observations : n = 9

Number of successes  : x = 7

Let p be the population proportion of times that the bats would follow the point.

Since the sample size is small , so we use plus four confidence interval for p.

Plus four estimate of p=$\hat{p}=\dfrac{\text{No. of successes}+2}{\text{No. of observations}+4}$

$=\dfrac{7+2}{9+4}\approx0.69$

By z-table , the critical value for 95% confidence level : z* = 1.96

Then, the 95% confidence interval for the population proportion of times that the bats would follow the point. will be :

$\hat{p}\pm z^*\sqrt{\dfrac{\hat{p}(1-\hat{p})}{N}}$ , where N= 13

$0.69\pm (1.96)\sqrt{\dfrac{0.69(1-0.69)}{13}}$

$0.69\pm (1.96)\sqrt{0.0163862084615}$

$0.69\pm (1.96)(0.128008626512)$

$\approx0.69\pm 0.25=(0.69-0.25,\ 0.69+0.25)$

$(0.44,\ 0.94)$

Hence, the  95% confidence interval for the population proportion of times that the bats would follow the point = $(0.44,\ 0.94)$

Thus the correct answer is B. (0.44,0.94)