Question

Independent random samples of professional football and basketball players gave the following information.

Heights (in ft) of pro football players: x₁; n₁ = 45

6.34	6.50	6.50	6.25	6.50	6.33	6.25	6.17	6.42	6.33
6.42	6.58	6.08	6.58	6.50	6.42	6.25	6.67	5.91	6.00
5.83	6.00	5.83	5.08	6.75	5.83	6.17	5.75	6.00	5.75
6.50	5.83	5.91	5.67	6.00	6.08	6.17	6.58	6.50	6.25
6.33	5.25	6.67	6.50	5.84

Heights (in ft) of pro basketball players: x₂; n₂ = 40

6.05	6.58	6.25	6.58	6.25	5.92	7.00	6.41	6.75	6.25
6.00	6.92	6.83	6.58	6.41	6.67	6.67	5.75	6.25	6.25
6.50	6.00	6.92	6.25	6.42	6.58	6.58	6.08	6.75	6.50
6.83	6.08	6.92	6.00	6.33	6.50	6.58	6.85	6.50	6.58

(a) Use a calculator with mean and standard deviation keys to calculate x₁, s₁, x₂, and s₂. (Round your answers to three decimal places.)

x1 =
s1 =
x2 =
s2 =

(b) Let μ₁ be the population mean for x₁ and let μ₂ be the population mean for x₂. Find a 90% confidence interval for μ₁ – μ₂. (Round your answers to three decimal places.)

lower limit
upper limit

(c) Examine the confidence interval and explain what it means in the context of this problem. Does the interval consist of numbers that are all positive? all negative? of different signs? At the 90% level of confidence, do professional football players tend to have a higher population mean height than professional basketball players?

Because the interval contains only negative numbers, we can say that professional football players have a lower mean height than professional basketball players.

Because the interval contains both positive and negative numbers, we cannot say that professional football players have a higher mean height than professional basketball players.    

Because the interval contains only positive numbers, we can say that professional football players have a higher mean height than professional basketball players.

(d) Which distribution did you use? Why?

The Student's t-distribution was used because σ₁ and σ₂ are unknown.

The standard normal distribution was used because σ₁ and σ₂ are known.     

The Student's t-distribution was used because σ₁ and σ₂ are known.

The standard normal distribution was used because σ₁ and σ₂ are unknown.

Answer 1

	Football ( X )	Σ ( Xi- X̅ )2	Basketball ( Y )	Σ ( Yi- Y̅ )2
	6.34	0.0258	6.05	0.1624
	6.5	0.1028	6.58	0.0161
	6.5	0.1028	6.25	0.0412
	6.25	0.005	6.58	0.0161
	6.5	0.1028	6.25	0.0412
	6.33	0.0227	5.92	0.2841
	6.25	0.005	7	0.2992
	6.17	0.0001	6.41	0.0018
	6.42	0.0579	6.75	0.0882
	6.33	0.0227	6.25	0.0412
	6.42	0.0579	6	0.2052
	6.58	0.1606	6.92	0.2181
	6.1	0.0099	6.83	0.1421
	6.58	0.1606	6.58	0.0161
	6.5	0.1028	6.41	0.0018
	6.42	0.0579	6.67	0.05
	6.25	0.005	6.67	0.0471
	6.67	0.2408	5.75	0.4942
	5.91	0.0725	6.25	0.0412
	6	0.0321	6.25	0.0412
	5.83	0.122	6.5	0.0022
	6	0.0321	6	0.2052
	5.83	0.122	6.92	0.2181
	5.08	1.2085	6.25	0.0412
	6.75	0.3257	6.42	0.0011
	5.83	0.122	6.58	0.0161
	6.17	0.0001	6.58	0.0161
	5.75	0.1843	6.08	0.1391
	6	0.0321	6.75	0.0882
	5.75	0.1843	6.5	0.0022
	6.5	0.1028	6.83	0.1421
	5.83	0.122	6.08	0.1391
	5.91	0.0725	6.92	0.2181
	5.67	0.2594	6	0.2052
	6	0.0321	6.33	0.0151
	6.08	0.0099	6.5	0.0022
	6.17	0.0001	6.58	0.0161
	6.58	0.1606	6.85	0.1576
	6.5	0.1028	6.5	0.0022
	6.25	0.005	6.58	0.0161
	6.33	0.0227
	5.25	0.8636
	6.67	0.2408
	6.5	0.1028
	5.84	0.1151
Total	278.07	5.893	258.12	3.8889

Mean X̅ = Σ Xi / n
X̅ = 278.07 / 45 = 6.179
Sample Standard deviation S_X = √ ( (Xi - X̅ )2 / n - 1 )
S_X = √ ( 5.893 / 45 -1 ) = 0.366

Mean Y̅ = ΣYi / n
Y̅ = 258.12 / 40 = 6.453
Sample Standard deviation S_Y = √ ( (Yi - Y̅ )2 / n - 1 )
S_Y = √ ( 3.8889 / 40 -1) = 0.316

DF = 82

Confidence interval :-

t(α/2, DF) = t(0.1 /2, 82 ) = 1.664

Lower Limit =
Lower Limit = -0.397
Upper Limit =
Upper Limit = -0.151
90% Confidence interval is ( -0.397 , -0.151 )

Part c)

Because the interval contains only negative numbers, we can say that professional football players have a lower mean height than professional basketball players.

Part d)

The Student's t-distribution was used because σ₁ and σ₂ are unknown.