8. 0.64/1 points | Previous Answers BBUnderStat11 7.4.014 My Notes Ask Your Teac
ID: 3360382 • Letter: 8
Question
8. 0.64/1 points | Previous Answers BBUnderStat11 7.4.014 My Notes Ask Your Teacher 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 Ib) of pro football players: xii n- 21 246 263 254 251 244 276 240 265 257 252 282 256 250 264 270 275 245 275 253 265 271 Weights (in Ib) of pro basketball players: x2i n2-19 202 200 220 210 193 215 222 216 228 207 225 208 195 191 207 196 182 193 201 (a) Use a calculator with mean and standard deviation keys to calculate x1, s1, x2, and s2. (Round your answers to one decimal place.) X1 = 259.7 s1= 145.3 x2205.8 s2= 168.1 (b) Let 11 be the population mean for X1 and letH2 be the population mean for x2. Find a 99% confidence interval for 1-2-(Round your answers to one decimal place.) ower limit 43.2 upper limit 64.7Explanation / Answer
Solution:
Part a
First of all we have to find mean and standard deviation for given two samples.
Mean = X/n
Var = (X – mean)^2/(n – 1)
SD = sqrt(Var)
Mean and SD for weights of football players is given as below:
No.
X1
(X1 - mean)
(X1 - mean)^2
1
246
-13.7143
188.0820245
2
263
3.2857
10.79582449
3
254
-5.7143
32.65322449
4
251
-8.7143
75.93902449
5
244
-15.7143
246.9392245
6
276
16.2857
265.2240245
7
240
-19.7143
388.6536245
8
265
5.2857
27.93862449
9
257
-2.7143
7.36742449
10
252
-7.7143
59.51042449
11
282
22.2857
496.6524245
12
256
-3.7143
13.79602449
13
250
-9.7143
94.36762449
14
264
4.2857
18.36722449
15
270
10.2857
105.7956245
16
275
15.2857
233.6526245
17
245
-14.7143
216.5106245
18
275
15.2857
233.6526245
19
253
-6.7143
45.08182449
20
265
5.2857
27.93862449
21
271
11.2857
127.3670245
Total
5454
2916.285714
Mean
259.7143
Var
145.8142857
SD
12.07535862
Mean = X/n = 5454/21 = 259.7143
Var = (X – mean)^2/(n – 1) = 2916.285714/(21 – 1) = 145.8142857
SD = sqrt(Var) = sqrt(145.8142857) = 12.07535862
X1bar = 259.7
S1 = 12.1
Now, we have to find mean and SD for weights of basketball players.
No.
X2
(X1 - mean)
(X1 - mean)^2
1
202
-3.8421
14.76173241
2
200
-5.8421
34.13013241
3
220
14.1579
200.4461324
4
210
4.1579
17.28813241
5
193
-12.8421
164.9195324
6
215
9.1579
83.86713241
7
222
16.1579
261.0777324
8
216
10.1579
103.1829324
9
228
22.1579
490.9725324
10
207
1.1579
1.34073241
11
225
19.1579
367.0251324
12
208
2.1579
4.65653241
13
195
-10.8421
117.5511324
14
191
-14.8421
220.2879324
15
207
1.1579
1.34073241
16
196
-9.8421
96.86693241
17
182
-23.8421
568.4457324
18
193
-12.8421
164.9195324
19
201
-4.8421
23.44593241
Total
3911
2936.526316
Mean
205.8421
Var
163.1403509
SD
12.77264072
Mean = X/n = 3911/19 = 205.8421
Var = (X – mean)^2/(n – 1) = 2936.526316/(19 – 1) = 163.1403509
SD = sqrt(Var) = sqrt(163.1403509) = 12.77264072
X2bar = 205.8
S2 = 12.8
Part b
Confidence interval = (X1bar – X2bar) -/+ t*sqrt[(S1^2/n1)+(S2^2/n2)]
Confidence interval = (259.7 – 205.8) -/+ 2.7116*sqrt((12.1^2/21)+(12.8^2/19))
Confidence interval = 53.9 -/+ 2.7116* 3.949058452
Confidence interval = 53.9 -/+ 10.7082669
Lower limit = 53.9 - 10.7082669 = 43.2
Upper limit = 53.9 + 10.7082669 = 64.5
Lower limit = 43.2
Upper limit = 64.7
No.
X1
(X1 - mean)
(X1 - mean)^2
1
246
-13.7143
188.0820245
2
263
3.2857
10.79582449
3
254
-5.7143
32.65322449
4
251
-8.7143
75.93902449
5
244
-15.7143
246.9392245
6
276
16.2857
265.2240245
7
240
-19.7143
388.6536245
8
265
5.2857
27.93862449
9
257
-2.7143
7.36742449
10
252
-7.7143
59.51042449
11
282
22.2857
496.6524245
12
256
-3.7143
13.79602449
13
250
-9.7143
94.36762449
14
264
4.2857
18.36722449
15
270
10.2857
105.7956245
16
275
15.2857
233.6526245
17
245
-14.7143
216.5106245
18
275
15.2857
233.6526245
19
253
-6.7143
45.08182449
20
265
5.2857
27.93862449
21
271
11.2857
127.3670245
Total
5454
2916.285714
Mean
259.7143
Var
145.8142857
SD
12.07535862
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