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Answer:
There is a significant difference between marathon runners and men who don't exercise.
Step-by-step explanation:
Null hypothesis: There is no significant difference between marathon runners and men who don't exercise.
Alternate hypothesis: There is a significant difference between marathon runners and men who don't exercise.
Sample 1 (marathon runners)
mean = 51.3 mg/dL
sd = 14.2 mg/dL
n = 19
Sample 2 (men who don't exercise)
mean = 44 mg/dL
sd = 15 mg/dL
n = 32
Pooled SE = 2.43 mg/L
Test statistic (t) = (mean 1 - mean 2) ÷ sqrt[pooled SE(1/n1 + 1/n2)] = (51.3 - 44) ÷ sqrt[2.43(1/19 + 1/32) = 7.3 ÷ sqrt(0.204) = 7.3 ÷ 0.452 = 16.2
The test is a two-tailed test. The critical value is given as 2.009. For a two-tailed test, the region of no rejection of the null hypothesis lies between -2.009 and +2.009.
Conclusion:
Reject the null hypothesis because the test statistic 16.2 falls outside the region bounded by -2.009 and 2.009.
Therefore, there is a significant difference between marathon runners and men who don't exercise.
Using the t-distribution, as we have the standard deviation for the sample, it is found that there is no significant difference between marathon runners and men who don’t exercise.
What are the hypotheses tested?
At the null hypothesis, it is tested if there is no difference, that is:
[tex]H_0: \mu_M - \mu_D = 0[/tex]
At the alternative hypothesis, it is tested if there is a difference, that is:
[tex]H_0: \mu_M - \mu_D \neq 0[/tex]
What is the mean and the standard error for the distribution of the difference?
For each sample, they are given by:
[tex]\mu_M = 51.3, s_M = \frac{14.2}{\sqrt{19}} = 3.26[/tex]
[tex]\mu_D = 44, s_D = \frac{15}{\sqrt{32}} = 2.65[/tex]
Hence, for the distribution of differences:
[tex]\overline{x} = \mu_M - \mu_D = 51.3 - 44 = 7.3[/tex]
[tex]s = \sqrt{s_M^2 + s_D^2} = \sqrt{3.26^2 + 2.65^2} = 4.2[/tex]
What is the test statistic?
It is given by:
[tex]t = \frac{\overline{x} - \mu}{s}[/tex]
In which [tex]\mu = 0[/tex] is the value tested at the null hypothesis.
Hence:
[tex]t = \frac{7.3 - 0}{4.2}[/tex]
[tex]t = 1.74[/tex]
What is the decision?
As stated in this problem, a critical value of [tex]t^{\ast} = \pm 2.009[/tex] is considered.
Since the absolute value of the test statistic is less than the critical value, it is found that there is no significant difference between marathon runners and men who don’t exercise.
To learn more about the t-distribution, you can check https://brainly.com/question/16313918