Question

One factor in rating a National Hockey League team is the mean weight of its players. A random sample of players from the Detroit Red Wings was obtained. The weight (in pounds) of each player was carefully measured, and the resulting data have a sample size of 18 with a sample mean of 205 pounds and a sample standard deviation of 11.6 pounds. You can assume that all of the assumptions are met.

a) What are the assumptions that are required to perform inference on this data?

b) Should you use a z distribution or a t distribution in this problem. Note that you will only get one try to get this question correct.Please explain the correct answer.

c) Find the 98.9% confidence interval for the true mean weight of the players from the Detroit Red Wings. Just answer, no need solution.

c) i) If this would be a z distribution, what would be the critical value? Please use 4 decimal places.

c) ii)  If this would be a t distribution, what would be the critical value? Please use 4 decimal places.

c) iii) If this would be a t distribution, what would be the degrees of freedom?

c) iv) The 98.9% confidence interval for the true mean weight of the players from the Detroit Red Wings is

( , )

Answer 1

a) Assumptions:

The sample is random and independent.

The distribution is normally distributed.

b) We will use a t-distribution in this problem as the population standard deviation is not given.

And the distribution is normal.

c)

i) If this would be a z distribution, what would be the critical value?

At α = 0.011 two tailed critical value, z_c = ABS(NORM.S.INV(0.011/2)) = 2.5427

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ii)  If this would be a t distribution, what would be the critical value?

At α = 0.011 and df = n-1 = 17, two tailed critical value, t-crit = T.INV.2T(0.011, 17) = 2.8532

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iii) If this would be a t distribution, what would be the degrees of freedom?

df = n-1 = 17

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v) The 98.9% confidence interval for the true mean weight of the players from the Detroit Red Wings is

x̅ = 205, s = 11.6, n = 18

98.9% Confidence interval :

At α = 0.011 and df = n-1 = 17, two tailed critical value, t-crit = T.INV.2T(0.011, 17) = 2.8532

Lower Bound = x̅ - t-crit*s/√n = 205 - 2.853 * 11.6/√18 = 197.1990

Upper Bound = x̅ + t-crit*s/√n = 205 + 2.853 * 11.6/√18 = 212.8010

197.199 < µ < 212.801