Question

Suppose the FAA weighed a random sample of 20 airline passengers during the summer and found their weights to have a sample mean of 180 pounds and sample standard deviation of 30 pounds. Assume the weight distribution is approximately normal.
a.) Find a one sided 95% confidence interval with an upper bound for the mean weight of all airline passengers during the summer. Show you work.

b.) Find a 95% prediction interval for the weight of another random selected airline passenger during the summer. Show you work.

Answer 1

Answer:

Step-by-step explanation:

Given Parameters

Mean, $x$ = 180

total samples, n = 20

Standard dev, $\sigma$ = 30

$\alpha$ = 1 - 0.95 = 0.05 at 95% confidence level

Df = n - 1 = 20 - 1 = 19

Critical Value, $t_\alpha$, is given by

$t_{c}=t_{\alpha, df} = t_{0.05,19} = 1.729$

a).

Confidence Interval, $\mu$, is given by the formula

$\mu = x +/- t_c \times \frac{s}{\sqrt{n} }$

$\mu = 180 +/- 1.729 \times \frac{30}{\sqrt{20} }$

$\mu = 180 +/-11.5985$

$191.5985 > \mu > 168.4015$

b).

Critical Value, $t_{\alpha/2}$, is given by

$t_{c}=t_{\alpha/2, df} = t_{0.05/2,19} = 2.093$

Confidence Interval, $\mu$, is given by

$\mu = x +/- t_c \times \frac{s}{\sqrt{n} }$

$\mu = 180 +/- 2.093 \times \frac{30}{\sqrt{20} }$

    = 180 +/- 14.0403

    = 165.9597 < $\mu$ < 194.0403