Independent random samples of professional football and basketball players gave the following information. Assume that the weight distributions are mound-shaped and symmetric.

Question

Weights (in lb) of pro football players: x₁; n₁ = 21

245	263	255	251	244	276	240	265	257	252	282
256	250	264	270	275	245	275	253	265	272

Weights (in lb) of pro basketball players: x₂; n₂ = 19

204	200	220	210	192	215	221	216	228	207
225	208	195	191	207	196	182	193	201

(a) Use a calculator with mean and standard deviation keys to calculate x₁, s₁, x₂, and s₂. (Round to four 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 99% confidence interval for 𝜇₁ − 𝜇₂. (Round to one decimal place.)

lower limit
upper limit

(c) 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 unknown.  

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

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

Answer 1

Answer:

A)

Statistic	Football Players (x1)	Basketball Players (x2)
Sample size (n)	21	19
Mean (x̄)	259.7619	205.8421
Standard deviation (s)	12.1610	12.7378

B)

Statistic	Value
Lower limit	43.2
Upper limit	64.6

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

Work:

A) Sample means and sample standard deviations

FOOTBALL PLAYERS (Group 1)

Data (n₁ = 21):

245, 263, 255, 251, 244, 276, 240, 265, 257, 252, 282, 256, 250, 264, 270, 275, 245, 275, 253, 265, 272

Sum of values = 5455

Mean (x̄₁):

x̄₁ = 5455 / 21 = 259.7619

Sample standard deviation (s₁): Formula: s₁ = sqrt[ Σ(x − x̄₁)² / (n₁ − 1) ]

Sum of squared deviations: Σ(x − x̄₁)² = 2957.8095

s₁ = sqrt(2957.8095 / 20)

s₁ = sqrt(147.8905)

s₁ = 12.1610

BASKETBALL PLAYERS (Group 2)

Data (n₂ = 19):

204, 200, 220, 210, 192, 215, 221, 216, 228, 207, 225, 208, 195, 191, 207, 196, 182, 193, 201

Sum of values = 3911

Mean (x̄₂):

x̄₂ = 3911 / 19 = 205.8421

Sample standard deviation (s₂): Formula: s₂ = sqrt[ Σ(x − x̄₂)² / (n₂ − 1) ]

Sum of squared deviations: Σ(x − x̄₂)² = 2920.5263

s₂ = sqrt(2920.5263 / 18)

s₂ = sqrt(162.2982)

s₂ = 12.7378

Summary:

Football players (group 1):

x̄₁ = 259.7619

s₁ = 12.1610

n₁ = 21

Basketball players (group 2):

x̄₂ = 205.8421

s₂ = 12.7378

n₂ = 19

B) 99% confidence interval for µ₁-µ₂

Point estimate (difference in sample means):

Difference in sample means: x̄₁-x̄₂ = 259.7619 - 205.8421 = 53.9

Standard error of the difference (Welch)

Formula: SE = sqrt( (s₁² / n₁) + (s₂² / n₂) )

s₁² / n₁ = (12.1610²) / 21

= 147.8905 / 21

= 7.0424

s₂² / n₂ = (12.7378²) / 19

= 162.2982 / 19

= 8.5410

SE = sqrt(7.0424 + 8.5410)

SE = sqrt(15.5834)

SE = 3.9496

Degrees of freedom (Welch–Satterthwaite)

Formula: df = [ (s₁²/n₁ + s₂²/n₂)² ] / [ ( (s₁²/n₁)² / (n₁−1) ) + ( (s₂²/n₂)² / (n₂−1) ) ]

Plug in values: df ≈ 37

Critical t-value for 99% confidence

For df ≈ 37 and 99% confidence: t* ≈ 2.715

Margin of error

ME = t* × SE

ME = 2.715 × 3.9496

ME = 10.73

Confidence interval

Lower limit: 53.9198 − 10.73 = 43.19 ≈ 43.2

Upper limit: 53.9198 + 10.73 = 64.65 ≈ 64.6

Therefore, 99% confidence interval for μ₁ − μ₂:

Lower limit = 43.2

Upper limit = 64.6