Question

A study is being conducted to determine whether a particular exercise program is effective at reducing the mean blood cholesterol level on males between the ages of 35 and 50. Six subjects participated in the program, and their cholesterol levels before and after the program are as follows. 1 Blood Cholesterol Level Subject Before After 265 229 2 240 231 258 227 4 295 240 5 251 238 6 225 226 نیا Using this data, we wish to test the following hypothesis using a level of significance of 0.05, where y, denotes the mean blood cholesterol level before the program, and uz denotes the mean blood cholesterol level after the program. Ho: Mi - Hz = 0 H1:41 - M2 > 0 Answer questions a through e assuming that you have chosen to use a paired t-test. Part a (4 points): State any assumptions that are required to perform this test. Part b (6 points): Identify the test statistic and critical region for the test. State the conclusion for this test. Part c (5 points): What is the P-value for this test? State the conclusion for this test. Part d (5 points): Construct a confidence interval that is relevant to this problem. Show that this confidence interval verifies your conclusion from part c. Part e (5 points): If the null hypothesis is true, what is the probability you conclude that Hy > Hz?

Answer 1

Sol:

Before the program mean =
X
= 255.66667 and after program mean =
2
= 231.83333
Before the program standard deviation = $s_1$ = 23.8635 and after program standard deviation =
52
= 5.84522

sample size = 6 for both after and before program.
Two sample t-test for population mean to test the claim that
17
>
12

Assumption: that the population follows normal distribution and the the groups after and before the program have equall variances. And the samples are independent of each other.

The provided sample means are shown below: X1 = 255.66667 X2 = 231.83333 Also, the provided sample standard deviations are: $i = 23.8635 82 = 5.84522 and the sample sizes are ni = 6 and n2 = 6. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: Mi = 42 Ha: Mi > 2 This corresponds to a right-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 a = 0.05, and the degrees of freedom are df = 10. 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 te = 1.812, for a = 0.05 and df 10. The rejection region for this right-tailed test is R= {t:t > 1.812}. (3) Test Statistics Since it is assumed that the population variances are equal, the t-statistic is computed as follows: t= X1 - X2 (n1 –1)s{+(n2 – 1) s (a + mo) ni+n2-2 2.376 255.66667 – 231.83333 (6-1)23.86352+(6-1)5.845222 6+6-2 (ő + ) (4) Decision about the null hypothesis Since it is observed that t = 2.376 >te = 1.812, it is then concluded that the null hypothesis is rejected. Using the P-value approach: The p-value is p = 0.0194, and since p= 0.0194 < 0.05, it is concluded that the null hypothesis is rejected.


(5) Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that population mean ui is greater than M2, at the 0.05 significance level.

As p value is less than significance level of 0.05 hence we reject null hypothesis and accept the alternative hypothesis that the mean cholesterol level has fallen after the program

d)

Confidence Interval The 95% confidence interval is 1.485 <Mi – 42 < 46.182. Graphically T-Test: t = 2.376, p = 0.0194 0.40 - 0.35 0.30 0.25 0.20 0.15 0.10 0.05 0.00 -4.0 -3.5 -3.0 -2.5 -2.0 -1.5 -1.0 -0.5 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0

As the 95% confidence intervel of difference on the two means doesn't contain 0 hence the interval also verifies the claim made by our test and the conclusion above in part b)

e).

If the null hypothesis is true, what is the probability you conclude that ?1 > ?2?
Answer: this probability of rejecting null hypothesis when its actually true is the Type I error and that for this test is the significance level of the test = 0.05

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