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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)

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

243, 261, 255, 251, 244, 276, 240, 265, 257, 252, 282

256, 250, 264, 270, 275, 245, 275, 253, 265, 270

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

205, 200, 220, 210, 192, 215, 221, 216, 228, 207

225, 208, 195, 191, 207, 196, 181, 193, 201

(a) Use a calculator with mean and standard deviation keys to calculatex1,s1,x2, ands2. (Round your answers to four decimal places.)x1=

s1=

x2=

s2=

(b) Let1be the population mean forx1and let2be the population mean forx2. Find a 99% confidence interval for12.(Round your answers to one decimal place.)

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 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'st-distribution was used because1and2are unknown.

The Student'st-distribution was used because1and2are known.

The standard normal distribution was used because1and2are known.

The standard normal distribution was used because1and2are unknown.

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