I need all the answers step by step I JSTORInfiniteokhin - Day 68 Difference of Means Practice OInsert Format Tools Add-ons HelpLast edit was seconds ago100%Normal textCalibriBIU A148. Independent random samples of professional football and basketball players gave the followinformation, Assume that the weight distributions are mound-shaped and symmetric.Weights (in Ib.) of pro football players: x,; n, = 21245262255 251244276 240265257252 282256250264270 275 245275 253 265 270Weights (in Ib.) of pro basketball players: x,; n, = 19205 200220210191215221 216228 207 225208195191207196181 193 201a. Determine the mean and sample standard deviation (s) for x,.b. Determine the mean and sample standard deviation (s) for x,.c. Find an 99% confidence interval for the difference of means.d. What does this mean in the context of the problem (do pro football players tend tohave a higher population mean weight than pro basketball players)?MASUS

Answered: 8. Independent random samples of…

A First Course in Probability (10th Edition)

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I JSTOR
Infinite
okhin - Day 68 Difference of Means Practice O
Insert Format Tools Add-ons Help
Last edit was seconds ago
100%
Normal text
Calibri
BIU A
14
8. Independent random samples of professional football and basketball players gave the follow
information, Assume that the weight distributions are mound-shaped and symmetric.
Weights (in Ib.) of pro football players: x,; n, = 21
245
262
255 251
244
276 240
265
257
252 282
256
250
264
270 275 245
275 253 265 270
Weights (in Ib.) of pro basketball players: x,; n, = 19
205 200
220
210
191
215
221 216
228 207 225
208
195
191
207
196
181 193 201
a. Determine the mean and sample standard deviation (s) for x,.
b. Determine the mean and sample standard deviation (s) for x,.
c. Find an 99% confidence interval for the difference of means.
d. What does this mean in the context of the problem (do pro football players tend to
have a higher population mean weight than pro basketball players)?
M
ASUS

Expert Solution

Step 1

Note: Since we only answer up to 3 sub-parts, we'll answer the first 3. Please resubmit the question and specify the other subparts (up to 3) you'd like answered.

Given:

n₁= 21

n₂= 19

Formula Used:

Mean $X$ = $\frac{1}{n} \sum_{}^{} X_{i}$ = $\frac{X_{i} + X_{2} + X_{3} + . . . . . . X_{n}}{n}$

Sample standard deviation S = $\sqrt{\frac{\sum_{}^{} {(X_{i} - X)}^{2}}{n - 1}}$

Confidence Interval for difference of means = $X_{1} - X_{2}$ $\pm$ _{$t_{1 - α / 2, v}$} $\sqrt{\frac{S_{1}^{2}}{n_{1}} + \frac{S_{2}^{2}}{n_{2}}}$

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