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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?

User Irvanjitsingh
by
2.9k points

1 Answer

14 votes
14 votes

Answer:

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 = \frac{\bar x_1 - \bar x_2}{\sqrt{(\sigma^2_1)/(n_1) + (\sigma_2^2)/(n_2) }}

where;


\bar x_1 = sample mean for DH group


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

= 4.6


\bar x_2 = sample mean for no DH group


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

= 4.15

Now:


z = \frac{4.60- 4.15}{\sqrt{(2.54^2)/(20) + (2.54^2)/(20) }} \\ \\ = (0.45)/(√(0.32258 +0.32258 )) \\ \\ = (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.}

User PxDav
by
2.8k points
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