Question

3) You measure the length of 5 radish seedlings at 7 days and 10 days and get the following results in mm (do NOT use R except for part (d)). Use a one sided test (why?): seedling #: 5 days: 7 days: 33.8 10.686 1 30 35 -5 2 20 27 -7 3 38 46 -8 4 49 58 -9 5 32 34 -2 40 12.145 difference: -6.2 2.775 (a) Is there a difference in length? (b) Repeat (a), but now use a regular (unpaired) t-test. (c) What happened in (b)? (d) Verify the normal distribution assumption for the paired t-test you did in (a). (You may use R for this part).

Answer 1

we have - given paired data los seeding 1 4 S 5 days 36 33.8 49 / 10.686 32 32 20 7 days 27 / 46 58 34 40 12.145 difference | -5 -7 1-8 -9 -2 -6.2 2.775 a) is there is a difference in length ? It is our claim.

To test i Hoi Mj = 0 vs Hi Mjito where a = difference between . 5 days ¢ 7 days. 4 Mja populahon mean difference between two variables, Test statistic : t= . Tejf takibe to the strengenge 24.09) (=-4.9960 Assume 2=51. 30.05 . talarda= toos, GE 2.476, Decision - we reject Ho at Cl-d) X100% revel of significance if Iti >talar of here It > talpidf hence we reject Ho at xas).

conclusion . There is enough evidence to conclude in that .. there a difference in length. now use unpaired t test. tel 5 days =9, 4 7 days = ₂ s = 10.686, 5=12.145. 5, = 33.8, = 40, ni=5 To Test n₂ =5 & Hoi - M, =la vs H. M, &M

Test statistics : t= 19, -yo) SA+S22 = 133.8 – 40). 1106802, (h. | F= -0-85699). Assume 2=5% taizi df = 2.306 here df = n, tha2 = 8 . Cassume equal variance) 117 tarzidf fail to reject Ho. Decision :- we rejë ct ito if here Ite talido we at dao.o5.

conclusion :- that There is there is not a enough evidence difference in to length conclude 6 . In ® we fail to reject Ho & for ③ we aclept Hi te reject Ho. - for the paired t-test ¢ Normal distribution Assumption renity.

### t test for paired data : let 5 days as x and 7 days as y > xdays = c(30,20,38,49,32) > ydays = c(35,27,46,58, 34) > di= c(-5,-7,-8,-9,-2) > ## 1) fisrt assumption : n < 30 here satisfy > ## 2) each variable should be normal : for this we can test for difference of normality test > shapiro.test(di) Shapiro-wilk normality test data: di W = 0.93855, p-value = 0.6557 > ## here p value is greater than alpha value (0.05) hence here data is not normal