Question

In Major League Baseball, the American League (AL) allows a designated hitter (DH) to bat in place of the pitcher, but in the National League (NL), the pitcher has to bat. However, when an AL team is the visiting team for a game against an NL team, the AL team must abide by the home team's rules, and thus, the pitcher must bat. A researcher is curious if an AL team would score more runs for games in which the DH was used. She samples 20 games for an AL team for which the DH was used, and 20 games for which there was no DH. The data are below. The population standard deviation for runs scored is known to be 2.49 for both groups. Assume the populations are normally distributed.
DH no DH
0 3
6 6
8 2
2 4
2 0
4 5
7 7
7 6
6 1
5 8
1 12
1 4
5 6
4 3
4 4
2 0
7 5
11 2
10 1
0 4

Required:
a. Is there evidence to suggest that more runs are scored in games for which the DH is used? Use α=0.10.
b. Can it be concluded that more runs are scored in games for which the DH is used?

Answer 1

Answer:

Step-by-step explanation:

From the given information:

Let:

$\mu_1$ represent the population mean no. for DH group.

$\mu_2$ represent the population mean no. for no DH group  

$n_1$ represent the sample sizes of DH

$n_2$ represent the sample sizes of  no DH

Sample size: $n_1 = n_2 = 20$

For both groups, the population standard deviation of runs scored = 2.54  

i.e

$\sigma^2_1 = \sigma_2^2 = 2.54$

The null & alternative hypothesis;

$H_o : \mu_1 \le \mu_2 \\ \\ H_a: \mu_1 > \mu_2$

Level of significance:

The test statistics is:

$z = \dfrac{\bar x_1 - \bar x_2}{\sqrt{\dfrac{\sigma^2_1}{n_1} + \dfrac{\sigma_2^2}{n_2} }}$

where;

$\bar x_1$ = sample mean for DH group

$\implies \dfrac{1}{n_1} \sum \limit ^{n_1}_{i=1} x_1 \\ \\ = \dfrac{1}{20}(0+6+8+2+2+4+7+7+6+5+1+1+5+4+4+5+7+11+10+0) \\ \\ = \dfrac{92}{20}$

= 4.6

$\bar x_2$ = sample mean for no DH group

$\implies \dfrac{1}{n_2} \sum \limit ^{n_1}_{i=1} x_1 \\ \\ = \dfrac{1}{20}(3+6+2+4+0+5+7+6+1+8+12+4+6+3+4+0+5+2+1+4) \\ \\ = \dfrac{83}{20}$

= 4.15

Now:

$z = \dfrac{4.60- 4.15}{\sqrt{\dfrac{2.54^2}{20} + \dfrac{2.54^2}{20} }} \\ \\ = \dfrac{0.45}{\sqrt{0.32258 +0.32258 }} \\ \\ = \dfrac{0.45}{0.803219}$

= 0.5602

Since the test is one-tailed by looking at that $H_a:$

The P-value = $P(Z > z)$

$\implies 1 - P(Z \le z) \\ \\ = 1 - P(Z \le 0.5602)$

Using Excel Function " =normdist(z)"; we have":

$= 1- 0.7123$

P-value = 0.2877

Decision rule: To reject $H_o$, if $p-value < \alpha \ at \ 0.10$

Conclusion: SInce P = 0.2877 which is $> \ \alpha \ at \ 0.10.$

We fail to reject the $H_o$ and conclude that there is insufficient evidence to support the given claim that: $\text{more runs are scored in games for which DH is used.}$