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

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 * (s)/(√(n) )`

`\mu = 180 +/- 1.729 * (30)/(√(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 * (s)/(√(n) )`

`\mu = 180 +/- 2.093 * (30)/(√(20) )`

= 180 +/- 14.0403

= 165.9597 <
`\mu` < 194.0403