In: Statistics and Probability
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 |
9 |
6 |
8 |
2 |
2 |
4 |
3 |
0 |
4 |
5 |
7 |
7 |
7 |
6 |
6 |
1 |
5 |
8 |
1 |
12 |
1 |
4 |
5 |
6 |
4 |
3 |
5 |
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.
z =
Let be the true average number of runs are scored in games for which the DH is used and be the true average number of runs are scored in games for which the DH is not used. We want to test if there evidence to suggest that more runs are scored in games for which the DH is used, that is, we want to test if . This is the alternative hypothesis as the alternative hypothesis needs to be strict inequality such as
The hypotheses are
The level of significance to test the hypothesis is
Using the sample we know the following
We also know the population standard deviations of number of runs scored
The standard error of difference in means is
2 samples of games are collected independent of the other. This is a one tailed test (right tail as the alternative hypothesis has ">") of 2 sample means of independent samples. Since we know the population standard deviation, we can say that the sampling distribution of difference in sample means is normally distributed. Hence we use a 2 sample z test.
The hypothesized value of the difference in population mean is
The test statistic is
ans: the test statistic is 0.8715
This is a right tailed test. The right tail critical value is
Using the standard normal tables, we get for z=1.28, P(Z<1.28) = 0.95
The critical value is 1.28
We will reject the null hypothesis, if the test statistic is more than the critical value.
Here, the test statistic is 0.8715 and it is less than the critical value 1.28. Hence we do not reject the null hypothesis.
we conclude that there is no sufficient evidence to support the claim that more runs are scored in games for which the DH is used.