Answer:
Step-by-step explanation:
From the given information:
Let:
[tex]\mu_1[/tex] represent the population mean no. for DH group.
[tex]\mu_2[/tex] represent the population mean no. for no DH group
[tex]n_1[/tex] represent the sample sizes of DH
[tex]n_2[/tex] represent the sample sizes of no DH
Sample size: [tex]n_1 = n_2 = 20[/tex]
For both groups, the population standard deviation of runs scored = 2.54
i.e
[tex]\sigma^2_1 = \sigma_2^2 = 2.54[/tex]
The null & alternative hypothesis;
[tex]H_o : \mu_1 \le \mu_2 \\ \\ H_a: \mu_1 > \mu_2[/tex]
Level of significance:
The test statistics is:
[tex]z = \dfrac{\bar x_1 - \bar x_2}{\sqrt{\dfrac{\sigma^2_1}{n_1} + \dfrac{\sigma_2^2}{n_2} }}[/tex]
where;
[tex]\bar x_1[/tex] = sample mean for DH group
[tex]\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}[/tex]
= 4.6
[tex]\bar x_2[/tex] = sample mean for no DH group
[tex]\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}[/tex]
= 4.15
Now:
[tex]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} \\ \\[/tex]
= 0.5602
Since the test is one-tailed by looking at that [tex]H_a:[/tex]
The P-value = [tex]P(Z > z)[/tex]
[tex]\implies 1 - P(Z \le z) \\ \\ = 1 - P(Z \le 0.5602)[/tex]
Using Excel Function " =normdist(z)"; we have":
[tex]= 1- 0.7123[/tex]
P-value = 0.2877
Decision rule: To reject [tex]H_o[/tex], if [tex]p-value < \alpha \ at \ 0.10[/tex]
Conclusion: SInce P = 0.2877 which is [tex]> \ \alpha \ at \ 0.10.[/tex]
We fail to reject the [tex]H_o[/tex] and conclude that there is insufficient evidence to support the given claim that: [tex]\text{more runs are scored in games for which DH is used.}[/tex]
