Question

Problem 4 Myeloproliferative neoplasms (MPN) are a type of blood cancer with high inflammation and severe symptoms. A research team from the UCI medical center hypothesized that a Mediterranean dietary intervention, which emphasizes fresh foods with anti-inflammatory properties, may therefore be better than an American diet for reducing symptom severity in MPN patients. Symptom severity in cancer patients can be assessed quantitatively via detailed questionnaires, resulting in a score between 0 and 140 (140=worst). The study started recently. The team has already recruited 28 patients diagnosed with MPN, and randomly assigned them to either a Mediterranean diet or a control American diet (USDA) for ten weeks. The table below summarizes the symptom severity changes for these participants over the course of the study (a larger value indicating a greater reduction). N. Median Treatment Mediterranean USDA 14 Mean 9.13 9.82 SE Mean 2.74 3.44 StDev 10.25 12.86 - 10.5 -8.5 14 Do these findings support the researchers' stated hypothesis? Question 20 1 pts The corresponding null hypothesis relates to which math symbol(s)? OP Med - ÕUSDA or or ēdiff OPMed - PUSDA Одог pdіff OP OH Med — HUSDA ор T Med - TUSDA

Answer 1

Q 20) Here we are interested to test the two population meana.

Therefore, here appropriate hypothesis is

H0 : μ Med - μ USDA

Therefore, Option 6 is right.

The appropriate alternative hypothesis is Directional.

Q 22) Here sample size is small and we have to compare two different means.

Therefore, we can use two sample t test for comparing two means.

Therefore, Option 3rd is right.

Two sample t procedure for two means.

Q 23)

T statistics = -0.157

P value = 0.5618

Q 25)

P Value(0.5618) > Alpha(0.05)

H0 is accepted.

Therefore, researchers claim is false.

(1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: Mi = H2 Ha: Mi > H2 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 = 26. 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 = 1.706, for a = 0.05 and df = 26. The rejection region for this right-tailed test is R= {t:t > 1.706}. (3) Test Statistics Since it is assumed that the population variances are equal, the t-statistic is computed as follows: X-X t= (n-1) s +(n2-1) n1+12-2 (1 + 1) V 9.13 - 9.82 (14-1)10.252+(14-1)12.862/ 1 36 ( 14+14-2 = -0.157 = +) 1 ) (4) Decision about the null hypothesis Since it is observed that t = -0.157 <te = 1.706, it is then concluded that the null hypothesis is not rejected. Using the P-value approach: The p-value is p = 0.5618, and since p= 0.5618 > 0.05, it is concluded that the null hypothesis is not rejected. (5) Conclusion It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean Mi is greater than Mz, at the 0.05 significance level.

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