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: x_{1}; n_{1} = 21
245 | 262 | 254 | 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: x_{2}; n_{2} = 19
202 | 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 x_{1}, s_{1}, x_{2}, and s_{2}. (Round your answers to one decimal place.)
x_{1} = | |
s_{1} = | |
x_{2} = | |
s_{2} = |
(b) Let μ_{1} be the population mean for
x_{1} and let μ_{2} be the
population mean for x_{2}. Find a 99% confidence
interval for μ_{1} − μ_{2}.
(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 standard normal distribution was used because σ_{1} and σ_{2} are unknown.The standard normal distribution was used because σ_{1} and σ_{2} are known. The Student's t-distribution was used because σ_{1} and σ_{2} are known.The Student's t-distribution was used because σ_{1} and σ_{2} are unknown.
a) For Football players :
Sample mean using excel function AVERAGE(), x̅1 = 259.6190 = 259.6
Sample standard deviation using excel function STDEV.S, s1 = 12.1222 = 12.1
Sample size, n1 = 21
For Basketball players :
Sample mean using excel function AVERAGE(), x̅2 = 205.7895 = 205.8
Sample standard deviation using excel function STDEV.S, s2 = 13.0067 = 13.0
Sample size, n2 = 19
b) 99% Confidence interval for the difference:
At α = 0.01 and df = n1+n2-2 = 38, two tailed critical value, t-crit = T.INV.2T(0.01, 38) = 2.712
S²p = ((n1-1)*s1² + (n2-1)*s2² )/(n1+n2-2) = ((21-1)*12.1222² + (19-1)*13.0067²) / (21+19-2) = 157.4766
Lower Bound = (x̅1 - x̅2) - t-crit*√(S²p*(1/n1 +1/n2)) = (259.619 - 205.7895) - 2.712*√(157.4766*(1/21 + 1/19)) = 43.1
Upper Bound = (x̅1 - x̅2) + t-crit*√(S²p*(1/n1 +1/n2)) = (259.619 - 205.7895) + 2.712*√(157.4766*(1/21 + 1/19)) = 64.6
c) The interval consist of numbers that are all positive.
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 σ_{1} and σ_{2} are unknown.
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