Question

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

Weights (in lb) of pro football players: x1; n1 = 21 246 261 255 251 244 276 240 265 257 252 282 256 250 264 270 275 245 275 253 265 271

Weights (in lb) of pro basketball players: x2; n2 = 19 203 200 220 210 192 215 223 216 228 207 225 208 195 191 207 196 181 193 201

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

(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 your answers to one decimal place.)

(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 99% level of confidence, do professional football players tend to have a higher population mean weight than professional basketball players?

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

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

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

Answer 1

Answer)

A)

X1 = 248.8, X2 = 196.5

S1 = 52.2, S2 = 43.7

N1 = 22, N2 = 20

B)

As the population standard deviation is unknown, we will use t distribution to estimate the interval

Margin of error = t*standard error

Degrees of freedom is = smaller of n1-1, n2-1

So, df = 19

For degrees of freedom 19, and 99% confidence level, critical value t is = 2.861

t = 2.861

Standard error = √{(s1^2/n1)+(s2^2/n2)}

After substitution

MOE = 42.3718701178

Confidence interval is given by

(X1-x2)-MOE < (u1-u2) < (X1-X2)+MOE

9.92812988218 < (u1-u2) < 94.6718701178

9.9 < (u1-u2) < 94.7

C)

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

D)

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