Question

17 points) The 30 Maior League Baseball teams are divided into two leaques" of 15 teams ench: the American League and the National League. A random sample of 51 American League players and a random sample of 51 National League players were selected. The sample mean and standard deviation of the salaries of the sampled players is shown in the table below. League Mean salary in millions USD SD of salary in millions USD American 3.972 7.067 National 6. 346 6 .304 1. A statistically-minded baseball fan wants to use statistical inference to determine if the salaries of the players in the American League are different from the salaries of the players in the National League. Select the appropriate hypotheses for this research question. A. HONI = M2 VS. HA MI> H2 B. HCM = K2 vs. HM2 C. HX 2 VS. HAHI<M2 2. Calculate the appropriate test statistic. ? = 3. Calculate the p-value. 4. Calculate the effect size. 5. Construct a 95% confidence interval for the difference in American League and National League salaries, American Nation

Answer 1

The provided sample means are shown below:

$\bar X_1 = 3.972$$\bar X_2 = 6.346$

Also, the provided sample standard deviations are:

$s_1 = 7.067$   $s_2 = 6.304$

and the sample sizes are n1 = 51 and n2 = 51.

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: $\mu_1= \mu_2$

Ha: $\mu_1\neq \mu_2$

This corresponds to a two-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used.

(2) Rejection Region

Based on the information provided, the significance level is α = 0.05, and the degrees of freedom are df = 100. In fact, the degrees of freedom are computed as follows, assuming that the population variances are equal:

Hence, it is found that the critical value for this two-tailed test is t_c = 1.984, for α = 0.05 and df = 100.

(3) Test Statistics

Since it is assumed that the population variances are equal, the t-statistic is computed as follows:

$t = \frac{\bar X_1 - \bar X_2}{\sqrt{ \frac{(n_1-1)s_1^2 &plus; (n_2-1)s_2^2}{n_1&plus;n_2-2}(\frac{1}{n_1}&plus;\frac{1}{n_2}) } }$

$t = \frac{3.972-6.346}{\sqrt{ \frac{(51-1)7.0672 &plus; (51-1)6.3042}{51&plus;51-2}(\frac{1}{51}&plus;\frac{1}{51}) } }$

t = −1.79

(4) Decision about the null hypothesis

Since it is observed that |t| = 1.79 < t_c = 1.984, it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is p = 0.0764, and since p = 0.0764 > 0.05, it is concluded that the null hypothesis is not rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean μ1​ is different than μ2​, at the 0.05 significance level.

Confidence Interval

The 95% confidence interval is -5.005 < μ1​−μ2​ < 0.257.