Question

38. Refer to the Baseball 2016 data, which report information on the 30 Major League Baseball teams for the 2016 season.

League Year Oper Team Sal AttendandVins Team ERA HR 5.090. 261 Arizona Mati onal Atlanta National 190 199870. 76 2036216 4.51 0.255 199687. 62 2020914 BaltimoreAmerican 1992115. 59 2172344 4. 220. 256 Boston American 1912 182. 16 2955434 4 0. 282 208 Chicago (National Chicago SAmerican CincinnatNational ClevelandAmerican Colorado National Detroit American Houston American Kans as C|American 3. 15 1914116. 65 3232420 0. 256 199 168 1991 98. 71 1746293 4. 1 0. 257 2003 116. 73 1894085 4. 910. 256 185 199486. 34 1591667 3. 84 0. 262 0. 275 199598. 26 2602524 4. 91 2000 172. 28 2493859 4. 240. 267 211 4. 060. 247 2000 69. 06 2306623 1973112. 91 2557712 81 4. 210. 261 LA AngelsAmerican LA DodgerNational Miami National 1966146. 45 3016142 156 1962 223. 35 3703312 91 3. 70. 249 201284. 64 1712417 4. 050. 263 128 194 2001 98. 68 2314614 4. 080. 244 MinnesotAmerican 2010 108. 26 1963912 5. 080. 251 200 NY Mets National 200999. 632789602 3. 580. 246 NY YankeeAmerican 2009213. 47 3063405 4. 160. 252 196680. 28 1521506 akland Hmeri can PhiladelrNational 2004 133. 05 1915144 71 Pittsbur National 2001 85. 892249021 4. 210. 257 153 San DiegcNational San FrandNational2000 166. 5 3365256 Seattle American 2004 126. 37 2351426 4. 430. 235 3. 650. 258 130 1999 122. 71 2267928 4 0. 259 223 St. LouisNational 2006 120. 3 3444490 4.080. 255 Tampa BaAmerican 4. 20. 243 199073. 65 1286163 216

Texas	American	1994	144.31	2710402	95	4.37	0.262	215	2016	4.40
Toronto	American	1989	112.90	3392299	89	3.78	0.248	221
Washington	National	2008	166.01	2481938	95	3.51	0.256	203

1. At the .05 significance level, can we conclude that there is a difference in the mean salary of teams in the American League versus teams in the National League?
2. At the .05 significance level, can we conclude that there is a difference in the mean home attendance of teams in the American League versus teams in the National League?
3. Compute the mean and the standard deviation of the number of wins for the 10 teams with the highest salaries. Do the same for the 10 teams with the lowest salaries. At the .05 significance level, is there a difference in the mean attendance for the two groups?

Answer 1

A)

To test the hypothesis for the difference in the mean salary of teams in the American League versus teams in the National League, the data values are arranged for for American and National team salary.

The data values for team salary are,

	American	National
	115.59	70.76
	182.16	87.62
	98.71	116.65
	86.34	116.73
	172.25	98.26
	69.06	223.35
	112.91	84.64
	146.45	98.68
	108.26	99.63
	213.47	133.05
	80.25	85.89
	122.71	126.37
	73.65	166.5
	144.31	120.3
	112.9	166.01
Mean	122.6013	119.6293
N	15	15
Std Dev	41.9987	40.01802

Two sample t test is used to compare the means assuming equal variance. The test is performed in following steps,

Step 1: The Null and Alternate Hypotheses are,

$H_A: \mu_1-\mu_1\neq 0$

Step 2: Select the appropriate test statistic and level of significance.

The t statistic is used to compare the two population means and the significance level is for the test (Generally 5% significance level used to compare two means)

a-0.05

Step 3: State the decision rules.

The decision rules state the conditions that if,

,

P-value < α 0.05, reject the null hypothesis

P-value > α 0.05, accept the null hypothesis

Step 4: Compute the appropriate test statistic and make the decision.

(111-,42)-0 estimate-hypothesized value std error Sp t1

Where pooled standard deviation is obtained using the formula,

(n1-1)s| + (n2-1)s2 /(15-1)41.99872 + (15-1)40.0182 15 15-2

p41.02

122.6013 119.6293)-0

Step 5: The P-value is obtained using the t distribution table for degree of freedom = n1+n2-2=30-2=28

P-value0.8441

Step 6: Conclusion

Since the corresponding P-value is 0.8441 which is greater than 0.05 at the 5% significance level for the two sided alternative hypothesis. Hence, it can be concluded that the null hypothesis is not rejected.

B)

The arranged data values for team attendance are,

	American	National
	2172344	2036216
	2955434	2020914
	1746293	3232420
	1591667	1894085
	2493859	2602524
	2306623	3703312
	2557712	1712417
	3016142	2314614
	1963912	2789602
	3063405	1915144
	1521506	2249021
	2267928	2351426
	1286163	3365256
	2710402	3444490
	3392299	2481938
Mean	2336379	2540892
N	15	15
Std Dev	629257	631793.4

Similarly,

Two sample t test is used to compare the means assuming equal variance.

Step 1: The Null and Alternate Hypotheses are,

$H_A: \mu_1-\mu_1\neq 0$

Step 2: Select the appropriate test statistic and level of significance.

The t statistic is used to compare the two population means and the significance level is for the test (Generally 5% significance level used to compare two means)

a-0.05

Step 3: State the decision rules.

The decision rules state the conditions that if,

,

P-value < α 0.05, reject the null hypothesis

P-value > α 0.05, accept the null hypothesis

Step 4: Compute the appropriate test statistic and make the decision.

(111-,42)-0 estimate-hypothesized value std error Sp t1

Where pooled standard deviation is obtained using the formula,

$s_p=\sqrt{\frac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}}=630526.5$

(2336379- 2540892)-0 =-0.89

Step 5: The P-value is obtained using the t distribution table for degree of freedom = n1+n2-2=30-2=28

$\text{P-value}=0.38196$

Step 6: Conclusion

Since the corresponding P-value is 0.38196 which is greater than 0.05 at the 5% significance level for the two sided alternative hypothesis. Hence, it can be concluded that the null hypothesis is not rejected.

C)

Now, arranging the data values of column wins with respect to the column team salary, the highest 10 and lowest 10 salaries are,

Team Salary	Wins
223.35	91
213.47	84
182.16	93
172.25	86
166.5	87
166.01	95
146.45	74
144.31	95
133.05	71
126.37	68
122.71	86
120.3	86
116.73	68
116.65	103
115.59	89
112.91	81
112.9	89
108.26	59
99.63	87
98.71	78
98.68	73
98.26	75
87.62	68
86.34	94
85.89	78
84.64	79
80.25	69
73.65	68
70.76	69
69.06	84

From these data values, the highest and lowest 10 values are,

	Highest 10	Lowest 10
	91	73
	84	75
	93	68
	86	94
	87	78
	95	79
	74	69
	95	68
	71	69
	68	84
Mean	84.4	75.7
Std dev	10.0466	8.4070

Two sample t test is used to compare the means assuming equal variance.

Step 1: The Null and Alternate Hypotheses are,

$H_A: \mu_1-\mu_1\neq 0$

Step 2: Select the appropriate test statistic and level of significance.

The t statistic is used to compare the two population means and the significance level is for the test (Generally 5% significance level used to compare two means)

a-0.05

Step 3: State the decision rules.

The decision rules state the conditions that if,

,

P-value < α 0.05, reject the null hypothesis

P-value > α 0.05, accept the null hypothesis

Step 4: Compute the appropriate test statistic and make the decision.

(111-,42)-0 estimate-hypothesized value std error Sp t1

Where pooled standard deviation is obtained using the formula,

$s_p=\sqrt{\frac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}}=9.2634$

$t=\frac{(84.4-75.7)-0}{9.2634 \sqrt{\frac{1}{10}+\frac{1}{10}}}=2.1$

Step 5: The P-value is obtained using the t distribution table for degree of freedom = n1+n2-2=20-2=18

$\text{P-value}=0.050077$

Step 6: Conclusion

Since the corresponding P-value is 0.050077 which is greater than 0.05 at the 5% significance level for the two sided alternative hypothesis. Hence, it can be concluded that the null hypothesis is not rejected.