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 2.54 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
Is there evidence to suggest that more runs are scored in games for which the DH is used? Use α=0.10.
Enter the test statistic - round to 4 decimal places.
Enter the P-Value - round to 4 decimal places.
Can it be concluded that more runs are scored in games for which the DH is used?
Yes or No

Answer 1

Explanation:

The objective is to compute that more runs are scored in games for which DH is used

Let
`\mu_1` denote the population mean number of runs scored for DH group

Let
`\mu_2` denote the population mean number of runs scored for DH group

Let
`n_1` and
`n_2` denote the sample sizes for DH and no DH

From the available information

`n_1=n_2=20`

The population standard deviation of runs score is 2.54 for both the groups.

That is
`\sigma _1^2=\sigma_2^2=2.54`

The null hypothesis is
`H_0:\mu_1\leq \mu_2`

The alternative hypothesis is
`H_1:\mu_1>\mu_2`

Let the level of significance
`\alpha =0.10`

Since, the population standard deviation for both the group is known , even though the sample size is less than 30 , use z test

The test statistic is

`z=\frac{\bar x_1 - \bar x_2}{\sqrt{(\mu_1^2)/(n_1) +(\mu_2^2)/(n_2) } }`

The sample mean for the DH group is computed as

`\bar x_1=(1)/(n_1) \sum_(i=1)^(n_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`

The sample mean of no DH is computed as

`\bar x_2=(1)/(n_2) \sum_(i=1)^(n_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`

The test statistic is

`z=\frac{\bar x_1 - \bar x_2}{\sqrt{(\mu_1^2)/(n_1) +(\mu_2^2)/(n_2) } }`

`=\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`

P-Value

From the alternative hypothesis, it is clear that the test is one tailed test

P-value = P(Z > z)

`1-P(Z\leq z)\\\\1-P(Z\leq 0.5602)`

1 - normsdist (0.5602)

(using excel function "normsdist(z)")

= 1 - 0.7123

= 0.2877

Therefore, the P-Value is 0.2877

Decision Rule

Reject the null hypothesis , if the p-value is less than the level of significance . that is p-value is < 0.10

Here, the p-value is 0.2877 which is greater than the level of significance 0.10

so , fail to reject the null hypothesis and conclude that there is no sufficient evidence to support the claim that more runs scored in games for which DH is used.