Question

Question 20 Not yet answered Points out of 8.00 Here is some output from R Studio. This is data is on mice and a new drug that helped with preventing early death in mice. The researcher was interested in its effects on the weights of the mice. The data is based on 10 mice that received the treatment. Research question: Does the weight of the mice increase significantly after the 3 months of the treatment? Write a conclusion based on the output evaluating the p-value in the context of the study. Source on Save a 1 + Weight of the mice before treatment 2 x<-c(200.1, 190.9, 192.7, 213, 241.4, 196.9, 172.2, 185.5, 205.2, 193.7) 3 # Weight of the mice after treatment 4 y<-c(392.9, 393.2, 345.1, 393, 434, 427.9, 422, 383.9, 392.3, 352.2) 5 resc-t. testy, x, paired=TRUE) 6 res P Flag question 6:4 (Top Level) Console Terminal > # Weight of the mice before treatment > *<-c(200.1, 190.9, 192.7, 213, 241.4, 196.9, 172.2, 185.5, 205.2, 193.7) > # Weight of the mice after treatment >y<-c(392.9, 393.2, 345.1, 393, 434, 427.9, 422, 383.9, 392.3, 352.2) > resc-t.test(y, x, paired=TRUE) > res Paired t-test data: y and x t - 20.883, df - 9, p-value - 6.2e-89 alternative hypothesis: true difference in means is not equal to 95 percent confidence interval: 173.4219 215.5581 sample estimates: mean of the differences 194.49

Conclusion based off p-value?

Since the p-value is 6.2e-09 then we have strong evidence to reject the null hypothesis. Is this correct or do we fail to make a conclusion because the validity conditions are not met?

Also a follow up question is:

If the validity conditions for the two-sample or paired t-tests are not met, what kind of tests can you do?

Answer 1

Since the p-value is 6.2e-09 then we have strong evidence to reject the null hypothesis. This is correct conclusion.

If validity conditions are not meet, that is If the assumption that samples are not independent of each other is failed then we have to use independent sample t test.