# ANOVA Hypotheses

**One Way ANOVA**

- 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.

State your conclusions.

2. Check to see if the residuals are modeled well by a normal distribution.

State your conclusions.

3. Rank the data, and repeat (1) using the ranks. State your conclusions.

**Solution**

**Q1 **

- 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.

- 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 | ||||