Question

6) It has been study. One een levels of treatment for the new drug will be used in the 2 study. One group will be administered the new phase of the drug once per day and a second group twice per day. You select two SRSs of 19 and 21 individuals who have been diagnosed as having high n diagnosed adrug twice per administered the new drug once per dayad cholest total cholesterol. Results are summarized in the table below. once per monthtwice per month sample size (n) Average Decrease (T) Standard Deviation (s) 21 143 10.3 19 18.2 a) (3 pts) In order to determine if receiving the drug twice per day results in larger cholesterol decrease than once per month state the null and alternative hypotheses. (10 pts) Assume the data for both groups are normally distributed. What type of test will you conduct? Calculate the t static for the test, round to 3 decimal places. b) c) (4 pts) Using the attached table, determine the p-value for the test and at a level of significance of a 0.01 is there evidence that the decrease in taking the drug twice per day is greater than taking the drug once per day? Use the conservative method to determine the degrees of freedom. Problem 6 continued ANOVA Decrease Mean Square df of 3 63427.667 423840 000 56 Groups 190283.000 8380.400 Within GroupS Total 149.650 59 198663.400 c) (5 pts) At a level of significance of .05 what is your conclusion? Explain your answer. Decrease Tukey HSD Subset for alpha -0.05 Treatment Placebo and no D&E D&E only 1 per day with D&E 15 15 15 15 3.53 24.00 113.73 2 per day with D&E 135.13 1.000 1.000 1.000 1.000 Means for groups in homogeneous subsets are displayed. a. Uses Harmonic Mean Sample Size 15.000. (8 pts) From the Tukey summary explain your findings. If you needed to lower the total cholestero of an induvial by 110 which treatment would you select based on this early trial?

Answer 1

6)

1 - once per month

2- twice per month

(1) Null and Alternative Hypothese The following null and alternative hypotheses need to be tested Ho: 12 This corresponds to a left-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 0.01, and the degrees of freedom are df 27.84. In fact, the degrees of freedom are computed as follows, assuming that the population variances are unequal Hence, it is found that the critical value for this left-tailed test is tc 27.84. 2.468, for α-: 0.01 and df The rejection region for this left-tailed test is R It:t< -2.468) (3)_Test Statisti Since it is assumed that the population variances are unequal, the t-statistic is computed as follows n1 +n2 117 143 --5.483 (19-1)18.22+(21110.32 1 (4)_Decision about the null hypothesis Since it is observed that t-5.483 < te - rejected 2.468, it is then concluded that the null hypothesis is

a)

Null and alternate hypothesis are given in image (1)

b)

we conduct 2-sample independent t-test with unequal variance

TS = −5.483

c)

df = min (19-1 , 21-1) = 18

p-value = P(T < TS) = 0