Question

July 29: Problem 1 Previous Problem Problem List Next Problem (10 points) 3 Researchers are comparing two diets, Diet Avs, Diet B. They believe that people tend to lose more weight with Diet A than with Diet B. Note that a higher weight loss is better than a lower weight loss and a negative value for the weight loss variable means that the person gained weight. :-( Weight loss is recorded in kg. Here are the data from a random selection of subjects. Weight Loss/kg) Subject Diet A Diet B 1 0 2 3 4 5 6 7 8 6 4 2 -1 3 7 5 7 0 1 -4 1 7 0 6 7 8 3 7 5 1 7 0 Weight Loss(kg) Subject 1 2 3 4 5 Diet A Diet B 0701 4 Create a csv data set and upload it to JASP. Set preferences in JASP to 3 places after the decimal point Double-check your data set to make sure that you have not mistyped any values Test the research hypothesis at the level of significance.

Answer 1

step 1 - state your null and alternate hypothesis

D= DIET A -DIET B (WEIGHTS)

H0: mean difference between diet A and diet B is null ( $\mu_{d}$ =0) - NULL HYPOTHESIS

H1:diet A is better than diet B ($\mu_{d}>0$ )

STEP 2

value of test statistic = 2.96( working shown below)

STEP3

p value = P(T>|calculated t statistic|) = P(T>2.96) = 0.010

STEP4

p value < 0.05 , reject the null hypothesis(H0) ( working shown below)

Sample 1 Sample 2 Difference = Sample 1 - Sample 2 5 0 5 6 7 -1 4 4 2 1 -4 3 3 1 2 7 7 0 5 5 Average 3.875 1.5 2.375 St. Dev. 2.532 3.742 2.264 n 8 8 From the sample data, it is found that the corresponding sample means are: I1 = 3.875 X = 1.5

Also, the provided sample standard deviations are: 81 = = 2.532 89 = 3.742 and the sample size is n = 8. For the score differences we have D = 2.375 8p = 2.264 (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Hopp = 0 Ha: up > 0 This corresponds to a right-tailed test, for which a t-test for two paired samples be used. (2) Rejection Region Based on the information provided the significance level is a = 0.05, and the degrees of freedom are df = 7. Hence, it is found that the critical value for this right-tailed test is te = 1.895, for a = 0.05 and df = 7. The rejection region for this right-tailed test is R=t:t > 1.895. (3) Test Statistics The t-statistic is computed as shown in the following formula: D t = 2.375 2.264/18 = 2.967 8D/

(4) Decision about the null hypothesis Since it is observed that t = 2.967 >te = 1.895, it is then concluded that the null hypothesis is rejected. Using the P-value approach: The p-value is p=0.0104, and since p = 0.0104 <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 Mi is greater than H2, at the 0.05 significance level. Confidence interval The 95% confidence interval is 0.482 <UD < 4.268. Graphically p-value = Test 0.010443768281464 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 -10 -0.5 D.D 0.5 1.0 1.5 20 2.5 3.0 3.5 4.0