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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.

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yemmy

Answer:

Step-by-step explanation:

Given Parameters

Mean, [tex]x[/tex] = 180

total samples, n = 20

Standard dev, [tex]\sigma[/tex] = 30

[tex]\alpha[/tex] = 1 - 0.95 = 0.05 at 95% confidence level

Df = n - 1 = 20 - 1 = 19

Critical Value, [tex]t_\alpha[/tex], is given by

[tex]t_{c}=t_{\alpha, df} = t_{0.05,19} = 1.729[/tex]

a).

Confidence Interval, [tex]\mu[/tex], is given by the formula

[tex]\mu = x +/- t_c \times \frac{s}{\sqrt{n} }[/tex]

[tex]\mu = 180 +/- 1.729 \times \frac{30}{\sqrt{20} }[/tex]

[tex]\mu = 180 +/-11.5985[/tex]

[tex]191.5985 > \mu > 168.4015[/tex]

b).

Critical Value, [tex]t_{\alpha/2}[/tex], is given by

[tex]t_{c}=t_{\alpha/2, df} = t_{0.05/2,19} = 2.093[/tex]

Confidence Interval, [tex]\mu[/tex], is given by

[tex]\mu = x +/- t_c \times \frac{s}{\sqrt{n} }[/tex]

[tex]\mu = 180 +/- 2.093 \times \frac{30}{\sqrt{20} }[/tex]

    = 180 +/- 14.0403

    = 165.9597 < [tex]\mu[/tex] < 194.0403