1. A study explored the effect of ethanol on sleep time. Fifteen rats were randomized to one of three treatments. Treatment 1 was water (control). Treatment 2 waas 1g of ethanol per kg of body weight, and Treatment 3 was 2g/kg. The amount of REM sleep (in minutes) in a 24hr period was recorded: Treatment 1: 63, 54, 69, 50, 72 Treatment 2: 45, 60, 40, 56 Treatment 3: 31, 40, 45, 25, 23, 28
(a) Graph the data. Why did you choose the graph that you did and what does it tell you?
(b) Create an ANOVA table for the data using the formulas provided in class. Show your work. You may use R to check your answers.
(c) Evaluate the ANOVA assumptions graphically. Was ANOVA appropriate here?
(d) Based on the ANOVA table, make a conclusion in the context of the problem.
(e) Create 95% CIs for all pairwise comparisons of means using the Tukey method. Do this without R show your work. (You may use R to check your work). Summarize your results using letter codes. What do you conclude?



Answer :

We have to find a problem at the 5% significance level.

Sample size = 10.

We're losing one till test if the Significant value is at five and 10. The critical value of W. C. is 44. When we look at the test, we're supposed to do one tailed. Which tail is it?

We are asked if one is more effective than the other.

Everything is mostly positive here, and we find what our values are. Mhm. 10 patients who used love level high school obtained additional sleep hours by using the following data set High bromide.

That's the control group minus the other one.

The control group is made up of people. I'm not sure if it's a left or right tail, so we'll just calculate the two ends.

The other tail is going to be divided by two and the right tail is going to be 44. 55 minutes 44 is going to be 11. It's very handy. Okay. If it's less than 11 it won't reject and if it's greater than 44 it will reject. Is it sound good? We had 10 students who were at the positive values and they were 1.9 plus 0.8 plus 1.1 plus 0.1. We get 23.4 for our nut. Okay. Critical value leads us straight to the middle. We failed to reject the null hypothesis and this was just a right tail test. If we tried it again at the 1% significance level, it would change our values to be 50. We did 10 tons 11 divided by two. Fifty is going to be five. Again, 23.4 lands us right in there.

We failed to reject the hypothesis again.

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