Question

Researchers are studying stress and want to know if a new drug reduces feelings of chronic stress. They take a sample of 12 chronically stressed individuals and randomly assign them into one of three groups. The control group does not get the drug, the first experimental group gets one dose of the drug, and the second experimental group gets two doses of the drug. Then they record the number of symptoms of stress that the participants display. What can they conclude about the effect of the drug on stress?

Group
Control	One Dose	Two Doses
9	8	4	?X2=	498
8	8	3	G=	74
6	7	5	N=	12
7	5	4	k=	3
T=30	T=28	T=16
SS= 5	SS= 6	SS= 2
n= 4	n= 4	n= 4
M= 7.5	M= 7	M= 4

1) State the null and alternative hypotheses

H₀:

H_A:

2) Identify the two degrees of freedom necessary to find the critical value and then find the critical value

df (on top)=

df (on bottom)=

F_crit =

3) Complete the ANOVA Table:

Source	SS	df	MS	F
Between treatments	___	___	____	___
Within treatments	____	____	____	____
Total	_____	_____	X	____

4) Interpret the results. (state if you reject or fail to reject the null hypothesis, if there is a statistically significant difference, and if the p-value is greater than or less than .05)

PLEASE HELP!!! & SHOW WORK! Thank you!

Answer 1

We can directly use here one way anova by Excel.

Step1) Enter data in Excel.

Step 2) Data >>Data analysis >> One way anova >>Select data >>ok

1) the null and alternative hypotheses

HA: The means are not all equal.

2)

df (on top)=3-1=2

df (on bottom)=12-3=9

Fcrit =4.2565.......................by using =FINV(0.05,2,9)

3)

ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Between Groups	28.66666667	2	14.33333333	9.923076923	0.005292924	4.256494729
Within Groups	13	9	1.444444444
Total	41.66666667	11

4) P value is 0.0053 ......................by using =FDIST(9.923,2,9)

P value 0.0053 < 0.05

We reject H0.

We can say that ,The differences between some of the means are statistically significant

i.e. conclude that not all population means are equal.