# ANOVA Hypotheses

One Way ANOVA

1. Perform an analysis of variance on these data, and test the following hypotheses
a. Are the differences in mean temperatures for the nine medications the same for all three severities of the blood pressure disorders?
Also produce an interaction plot, and interpret.
b. Is there a significant difference in mean temperatures for the various medications? If there is, use Tukey’s post hoc procedure to identify where the differences are.
c. Is there a significant difference in mean temperatures for the different severity levels? If there is, use Tukey’s post hoc procedure to identify where the differences are.
2. Check to see if the residuals are modeled well by a normal distribution.
3. Rank the data, and repeat (1) using the ranks. State your conclusions.

Solution

Q1

1. a) There is some difference in the mean temp across the medication and severity which can be seen with the interaction plots  Yes there is significant diff which can be checked with ANOVA –

 Anova: Single Factor SUMMARY Groups Count Sum Average Variance A 15 1463 97.53333 0.039524 B 15 1467 97.8 0.027143 C 15 1469.1 97.94 0.066857 D 15 1463.8 97.58667 0.034095 E 15 1467.3 97.82 0.013143 F 15 1467.6 97.84 0.012571 G 15 1462.5 97.5 0.031429 H 15 1468.5 97.9 0.007143 I 15 1468.6 97.90667 0.019238 ANOVA Source of Variation SS df MS F P-value F crit Between Groups 3.511704 8 0.438963 15.73075 6.74E-16 2.012654 Within Groups 3.516 126 0.027905 Total 7.027704 134

So we see that p value is quite low so we reject the null of no difference in the average temp so we say that there is ample diff.

Tukey’sResults-

 Tukey Test 0.055682 0.261707 Mean Diff DIFF (1-YES, 0- N0) AB -0.26667 0 AC -0.40667 0 AD -0.05333 0 AE -0.28667 0 AF -0.30667 0 AG 0.033333 0 AH -0.36667 0 AI -0.37333 0 BC -0.14 0 BD 0.213333 0 BE -0.02 0 BF -0.04 0 BG 0.3 1 BH -0.1 0 BI -0.10667 0 CD 0.353333 1 CE 0.12 0 CF 0.1 0 CG 0.44 1 CH 0.04 0 CI 0.033333 0 DE -0.23333 0 DF -0.25333 0 DG 0.086667 0 DH -0.31333 0 DI -0.32 0 EF -0.02 0 EG 0.32 1 EH -0.08 0 EI -0.08667 0 FG 0.34 1 FH -0.06 0 FI -0.06667 0 GH -0.4 0 GI -0.40667 0 HI -0.00667 0

SO we can see that there are some groups which have significant difference here.

Q2 SOHERE WE CAN SEE THAT RESIDUALS FOR THE SEVERITY 1 ‘ RESIDUALS shows no normal ditribution , rather they follow f

Q3

Now if we rank the data as it was already done , we see whether the medication and severity will have varied impact but we see that there is no change in model and thus there is ample change in temperature.

1. c) ANOVA to compare the mean diff in severity
 SUMMARY Groups Count Sum Average Variance S1 45 4397.9 97.73111 0.064465 S2 45 4397.1 97.71333 0.038455 S3 45 4402.4 97.83111 0.048556 ANOVA Source of Variation SS df MS F P-value F crit Between Groups 0.362815 2 0.181407 3.592825 0.030244 3.064761 Within Groups 6.664889 132 0.050492 Total 7.027704 134

Now we will check with tukey’stest , so the q from table will be 4.48 , MSE = 0.028 from above table and nk will be 3 , that is number of treatments , so we have calculated stat= 4.48*sqrt(0.028/3) = 0.058

 tukey test root value 0.129733 Final Cal value 0.581202 Mean Diff S1,2 0.017778 S1,3 -0.1 S2.,3 -0.11778 all values of difference is less than the stat so we can say that difference is among all three