(8) Obesity in adult males is associated with lower levels of sex hormone. A stu
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Question
(8) Obesity in adult males is associated with lower levels of sex hormone. A study investigated a possible
link between obesity and plasma testosterone concentrations in adolescent males between the ages of 14
and 20 years. Plasma testosterone level is measured in nanomoles per liter of blood (nmol/l) for a sample
of 25 obese adolescent males. The sample mean is 0.26 nmol/l, and the population standard deviation is
0.12 nmol/l.
a) (5) Perform a hypothesis test on the claim that the mean testosterone level in obese adolescent males is
greater than 0.24 nmol/l., with a Significant level of 0.1.
You should consider ____________(T/Z) method
b) (3) suppose there is a sample of 25 subjects with a sample mean of 0.26 nmol/l. Consider the following
hypothesis test at a 0.1 significance level:
Ho: =0.24 Ha: >0.24
What is the power of the test if the population mean is actually 0.27 nmol/l?
Explanation / Answer
Solution:-
a) (5) Perform a hypothesis test on the claim that the mean testosterone level in obese adolescent males is
greater than 0.24 nmol/l., with a Significant level of 0.1.
You should consider ____________(T/Z) method
The general rule of thumb for when to use a t score is when your sample: Has a sample size below 30.
And here as the sample size is 25. Therefore, we use T test.
b) (3) suppose there is a sample of 25 subjects with a sample mean of 0.26 nmol/l. Consider the following
hypothesis test at a 0.1 significance level:
Ho: =0.24 Ha: >0.24
The solution to this problem takes four steps: (1) state the hypotheses, (2) formulate an analysis plan, (3) analyze sample data, and (4) interpret results. We work through those steps below:
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: = 0.24
Alternative hypothesis: > 0.24
Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the sample mean is too small.
Formulate an analysis plan. For this analysis, the significance level is 0.10. The test method is a one-sample t-test.
Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).
SE = s / sqrt(n) = 10 / sqrt(25) = 2
DF = n - 1 = 25 - 1 = 24
t = (x - ) / SE = (0.26 - 0.24)/2 = 0.01
where s is the standard deviation of the sample, x is the sample mean, is the hypothesized population mean, and n is the sample size.
The observed sample mean produced a t statistic test statistic of 0.01. We use the t Distribution Calculator to find P(t > 0.01).
The P-Value is 0.496052.
The result is not significant at p < .010
Interpret results. Since the P-value (0.496052) is greater than the significance level (0.10), we cannot reject the null hypothesis.
Conclusion. Fail to reject null hypothesis. We have insufficient evidence to prove the claim that the mean testosterone level in obese adolescent males is greater than 0.24 nmol/l.
What is the power of the test if the population mean is actually 0.27 nmol/l?
Then, t = (x - ) / SE = (0.27 - 0.24)/2 = 0.015
The P-Value is .0494078.
The result is not significant at p < .10.
Since the P-value (0.494078) is greater than the significance level (0.10), we cannot reject the null hypothesis.
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