Question

Suppose we pick a simple random sample of 28 Major League Baseball players, and find that their average weight is 212.7 pounds, with a sample standard deviation of 15.6 pounds. Find a 99% confidence interval for the mean weight of all MLB players

Answer 1

Confidence Interval (204.53, 220.87)

Explanation:

Confidence interval is range of values that you can a some % confident that the true mean of the population lies.

The formula for constructing a confidence Interval

X ± z б/√n

Where X is the mean , б is the standard deviation , n is the sample and z is the value for a desired confidence level.

First find z

Degrees of freedom = 28 - 1 = 27

Level of significance = 1-0.99= 0.01

Then we look up this value in the t distribution

z = 2.771

Lower limit = 212.7- 2.771*15.6/√28 =204.53

Upper limit = 212.7+ 2.771*15.6/√28 =220.87

Confidence Interval (204.53, 220.87)

This means that we are 90% confident that the mean weight of Major Legue Baseball players is between 204.53 and 220.87

Answer 2

Explanation:

We would use the t- distribution.

From the information given,

Mean, μ = 212.7 pounds

Standard deviation, σ = 15.6 pounds

number of sample, n = 28

Degree of freedom, (df) = 28 - 1 = 27

Alpha level,α = (1 - confidence level)/2

α = (1 - 0.99)/2 = 0.005

We will look at the t distribution table for values corresponding to (df) = 27 and α = 0.005

The corresponding z score is 2.77

We will apply the formula

Confidence interval

= mean ± z ×standard deviation/√n

It becomes

212.7 ± 2.77 × 15.6/√28

= 212.7 ± 2.77 × 2.948

= 212.7 ± 8.17

The lower end of the confidence interval is 212.7 - 8.17 = 204.53 pounds

The upper end of the confidence interval is 212.7 + 8.17 = 220.87 pounds