Question

1) The following data represent the results from an independent-measures study comparing two treatment conditions.

Treatment One	Treatment Two
3	3.1
4.5	4.5
4.8	4.5
4.5	3.9
6.5	3.9
4.3	3.8
3.7	3.4

Run the single-factor ANOVA for this data:
F-ratio:   
p-value:   

Now, run the t test on the same data:
t-statistic:   
p-value:   

2) Google Sheets can be used to find the critical value for an F distribution for a given significance level. For example, to find the critical value (the value above which you would reject the null hypothesis) for α=0.05 with dfbetween=3 and dfwithin=56, enter =FINV(0.05,3,56)

Enter this into Google Sheets to confirm you obtain the value 2.769.

You conduct a one-factor ANOVA with 3 groups and 10 subjects in each group (a balanced design). Use Google Sheets to find the critical values for α=0.1 and α=0.05 (report accurate to 3 decimal places).
F0.1=
F0.05=

Answer 1

Anova: Single Factor
SUMMARY
Groups	Count	Sum	Average	Variance	std dev
A	7	31.3	4.471	1	1
B	7	27.1	3.871	0	1
					#VALUE!
ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Between Groups	1.26	1	1.26	1.752	0.210	4.75
Within Groups	8.63	12	0.72
Total	9.89	13

F ratio = 1.752

p value = 0.210

=================

t test

Sample #1   ---->   1                  
mean of sample 1,    x̅1=   4.471                  
standard deviation of sample 1,   s1 =    1.0812                  
size of sample 1,    n1=   7                  
                          
Sample #2   ---->   2                  
mean of sample 2,    x̅2=   3.871                  
standard deviation of sample 2,   s2 =    0.5187                  
size of sample 2,    n2=   7                  
                          
difference in sample means =    x̅1-x̅2 =    4.4714   -   3.9   =   0.60  
                          
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    0.8480                  
std error , SE =    Sp*√(1/n1+1/n2) =    0.4533                  
                          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   0.6000   -   0   ) /    0.45   =   1.324
                          
Degree of freedom, DF=   n1+n2-2 =    12                  

p-value =        0.2102   (excel function: =T.DIST.2T(t stat,df) )              

t statistic =1.324

p value = 0.2102

2)

F(0.1) = F(0.1,2,27) = 2.511

F(0.05,2,27) = 3.354