Question

Mary wants to know if training with her program will improve runners’ times in the mile. She compares 5 runners whom she trains with 5 runners that receive standard instruction (38 points).

Improvement in time with Mary’s method in decrease in seconds	Improvement in time without Mary’s method in decrease in seconds.
30	20
34	25
39	35
50	20
30	25

1. Identify your mean, median, mode, and range for the group that experienced Mary’s method (4 points).

2. Identify your control group (2 points), experimental group (2 points), independent variable (2 points), and dependent variable (2 points).

3. What statistical test should you conduct? (2 points)

4. What are the null and alternative hypotheses? (2 points)

5. What is/are the critical value(s)? (2 points)

6. Calculate the test statistic. (4 points)

7. What is your decision in regards to the null hypothesis? (2 points)

8. What do you conclude about the programs using everyday common language? (2 points)

9. Present your data in graphical form. Use a bar graph and include error bars (4 points).

10. Calculate a 95% confidence interval (3 points).

11. Calculate the effect size (if appropriate) and interpret it (3 points).  

12. Write the results in the correct APA style format. (2 points)

Answer 1

Control group:Improvement in time without Mary’s method in decrease in seconds.

Experimental Group: Improvement in time with Mary’s method in decrease in seconds.

Independent Variable:Type Of method

Dependent Variable: Improvement in time.

The groups are indpendent is Independent t test.

NULL HYPOTHESIS H0:

ALTERNATIVE HYPOTHESIS Ha:

Now I will use here F test to know which test should be used here.

From this test P value=0.2811 > 0.05 therefore WE CAN ASSUME VARIANCES ARE EQUAL.
Now degrees of freedom= n1+n2-2= 5+5-2=8

It is One tailed test

(2) Rejection Region

Based on the information provided, the significance level is \alpha = 0.05α=0.05, and the degrees of freedom are df = 8df=8. 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 right-tailed test is t_c = 1.86tc​=1.86, for \alpha = 0.05α=0.05 and df = 8df=8.

The rejection region for this right-tailed test is R = \{t: t > 1.86\}R={t:t>1.86}.

(3) Test Statistics

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

Decision about the null hypothesis

Since it is observed that t=2.505>tc​=1.86, it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is p = 0.0183, and since p=0.0183<0.05, it is concluded that the null hypothesis is rejected.

Conclusion

It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that population mean μ1​ is greater than μ2​, at the 0.05 significance level.

Confidence Interval

Pooled Variance
s2p = ((df1)(s21) + (df2)(s22)) / (df1 + df2) = 428.71 / 8 = 53.59

Standard Error
s(M1 - M2) = √((s2p/n1) + (s2p/n2)) = √((53.59/5) + (53.59/5)) = 4.63

Confidence Interval
μ1 - μ2 = (M1 - M2) ± ts(M1 - M2) = 11.6 ± (2.31 * 4.63) = 11.6 ± 10.68

95% CI [0.924, 22.276].

You can be 95% confident that the difference between your two population means (μ1 - μ2) lies between 0.92 and 22.28

Effect Size= Cohen's d = (M2 - M1) ⁄ SDpooled

where:

SDpooled = √((SD12 + SD22) ⁄ 2)

Cohen's d = (25 - 36.6) ⁄ 7.320413 = 1.58461.