Question

Suppose that ten bats were used in the experiment. 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) 8 times. Find the 95% confidence interval for the population proportion of times that the bats would follow the point. (0.62, 1.0) (0.477, 0.951) (0.321, 0.831)

Answer 1

Answer: (0.477, 0.951)

Step-by-step explanation:

Given : Number of observations : n = 10

Number of successes  : x = 8

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

Because the number of observation is not enough large , so we use plus four confidence interval for p.

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

$\hat{p}=\dfrac{8+2}{10+4}=\dfrac{10}{14}\approx0.714$

We know that , the critical value for 95% confidence level : z* = 1.96 [By using z-table]

Now, the required confidence interval will be :

$\hat{p}\pm z^*\sqrt{\dfrac{\hat{p}(1-\hat{p})}{N}}$ , where N= 14

$0.714\pm (1.96)\sqrt{\dfrac{0.714(1-0.714)}{14}}$

$0.714\pm (1.96)\sqrt{0.014586}$

$0.714\pm (1.96)(0.120772513429)$

$\approx0.714\pm0.237=(0.714-0.237,\ 0.714+0.237)$

$(0.477,\ 0.951)$

Hence, the 95% confidence interval for the population proportion of times that the bats would follow the point = (0.477, 0.951)