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 3 258 227 4 295 240 5 251 238 6 225 226 Using this data, we wish to test the following hypothesis using a level of significance of 0.05, where ui denotes the mean blood cholesterol level before the program, and uz denotes the mean blood cholesterol level after the program. Ho:M1 – M2 = 0 H1:41 - M2 > 0 Answer questions a through e assuming that you have chosen to use a paired t-test. A) State any assumptions that are required to perform this test. B) Identify the test statistic and critical region for the test. State the conclusion for this test. C) What is the P-value for this test? State the conclusion for this test. D) Construct a confidence interval that is relevant to this problem. Show that this confidence interval verifies your conclusion from part c. E) If the null hypothesis is true, what is the probability you conclude that 41 > uz?

Answer 1

As per Question Hoz M, - Mg = 0 V6 M. 41-42 70 A) we are using paired te test So the required assiemptions for the testis The observations are indepandant of each other 2) The dependant variable should be approximatey normally distributed (3) No significant authers in the difference between the two related groups B) Pest statistics To find ? Subject before mfites d=44-9) d-T (-J) 265 229 -12.17 148.02 260 19.83 227 -31 -7.17 11.36 4 295 240 -53 - 31.17 971.36 251 238 -13 1083 117.36 225 226 26.83 616.69 a = -23.833 Ecd-d) ² = 2124.83 Sy n- -36 2 231 220.02 3 258 6 1 = 5(da) 3 as n 26 20.61 2126.83 S

a atna -2.83 - 23.83 20161/66 tabulated for its at 0.05 level it 2,015 hence we have test statistics -- 2.83 hence wetact 2.83 - 2.83 4 2.83 > which is greater than ttabulated So hear, we have sufficient ovidance to reject Ho. Sl. love is a critical region of the test e) PC to 2.83) = foo, P(+52 2.571) = 0.025 Plts 7 3.365) = 0.00 Using equality formulate ac 2.83 for 2 a = 2.571 froi) = 0.025 3.365 from) 2001 22

n-al from) - f121) ff M2) fla.) by putting val we get fon) 20.020 The p-value for the test is 0.0201, which is less that cos, co-level) hence, we have sufficient evidance for Ho. hence, we conclude that I, -4270 D) we have a I toos, nix Sa have knows -23.83 I 2.015 X 2061 mear (-50.78 -6.87) This does not indlude the value o hence, there is difference in the both & which varifier our result in E) As nuls hypo is true se 4, = 42 When dur p.value is greter than so The prob that these conclude el,7ll is less than 0.05