Question

The amount of time it takes a bat to eat a frog was recorded for each bat in a random sample of 12 bats. The resulting sample mean and standard deviation were 21.9 minutes and 7.7 minutes, respectively. Assuming it is reasonable to believe that the population distribution of bat mealtimes of frogs is approximately normal, a. Construct a 95% confidence interval for the mean time for a bat to eat a frog. b. Construct a 95% confidence interval for the variance of the time for a bat to eat a frog.

Answer 1

Answer: a. CI for the mean: 17.327 < μ < 26.473

b. CI for variance: 29.7532 ≤ $\sigma^{2}$ ≤ 170.9093

Step-by-step explanation:

a. To construct a 95% confidence interval for the mean:

The given data are:

mean = 21.9

s = 7.7

n = 12

df = 12 - 1 = 11

1 - α = 0.05

$\frac{\alpha}{2}$ = 0.025

t-score = $t_{0.025,11}$ = 2.2001

Note: since the sample population is less than 30, it is used a t-score.

The formula for interval:

mean ± $t.\frac{s}{\sqrt{n} }$

Substituing values:

21.9 ± 2.200.$\frac{7.7}{\sqrt{12} }$

21.9 ± 4.573

The interval is: 17.327 < μ < 26.473

b. A 95% confidence interval for the variance:

The given values are:

$s^{2}$ = $7.7^{2}$

$s^{2}$ = 59.29

α = 0.05

$\frac{\alpha}{2}$ = 0.025

$1-\frac{\alpha}{2}$ = 0.975

$\chi^{2}_{0.025,11}$ = 21.92

$\chi^{2}_{0.975,11}$ = 3.816

Note: To find the values for $\chi^{2}_{\alpha/2,n-1}$ and $\chi^{2}_{1-\alpha/2,n-1}$, look for them at the chi-square table

The formula to calculate interval:

($\frac{(n-1).s^{2}}{\chi^{2}_{\alpha/2,n-1}} , \frac{(n-1)s^{2}}{\chi^{2}_{1-\alpha/2,n-1}}$)

are the lower and upper limits, respectively.

Substituing values:

($\frac{11.59.29}{21.92} , \frac{11.59.29}{3.816}$)

(29.7532, 170.9093)

The interval for variance is: 29.7532 ≤ $\sigma^{2}$ ≤ 170.9093