Question

Weights of men: 90% confidence; n = 14, x=161.3 lb, s =12.6 lb

Answer 1

The answer is below

Explanation:

Use the given degree of confidence and sample data to find a confidence interval for the population standard deviation . Assume that the population has a normal distribution. Round the confidence interval limits to the same number of decimal places as the sample standard deviation.

Answer: Given that:

Sample size (n) = 14, mean (
`\mu`) = 161.3, standard deviation (
`\sigma`) = 12.6

Confidence(C)= 90% = 0.9

α = 1 - C = 1- 0.9 = 0.1

α/2 = 0.1 / 2 = 0.05

The z score of α/2 correspond to a z score of 0.45 (0.5 - 0.05). This gives:

`z_{(\alpha)/(2) }=1.645`

The margin of error (E) is given by the formula:

`E=z_{(\alpha)/(2) }(\sigma)/(√(n) ) =1.645*(12.6)/(√(14) )=5.5`

The confidence interval = μ ± E = 161.3 ± 5.5 = (155.8, 166.8)

The confidence interval is between 155.8 lb and 166.8 lb. There is a 90% confidence that the mean is between 155.8 lb and 166.8 lb.