A random sample of five observations from three normally distributed populations

Question

A random sample of five observations from three normally distributed populations produced the following data: Use Table 4.

a.Calculate the grand mean. (Round your answer to 2 decimal places.)

b.Calculate SSTR and MSTR. (Round your final answers to 2 decimal places.)

c.Calculate SSE and MSE. (Round your final answers to 2 decimal places.)

d.Specify the competing hypotheses in order to determine whether some differences exist between the population means.
H0: A = B = C; HA: Not all population means are equal.
H0: A B C; HA: Not all population means are equal.
H0: A B C; HA: Not all population means are equal.

e.Calculate the value of the F(df1,df2) test statistic

f.Using the critical value approach at the 1% significance level, what is the conclusion to the test?
Reject H0 since the test statistic exceeds 5.52.
Reject H0 since the test statistic exceeds 6.93.
Do not reject H0 since the test statistic does not exceed 5.52.
Do not reject H0 since the test statistic does not exceed 6.93.

Treatments 17 15 22 23 28 29 24 31 18 32 15 23 23 25 27 XA 24.8 RB-21.6 RC 24.0 SC 21.0 B 27.3 A 48.2

Explanation / Answer

Step 1                          
   Null Hypothesis Ho :µ1 =µ2 =µ3           
   Alternative Hypothesis : µ1 µ2 µ3           
Step 2                          
   Degrees of freedom between = k - 1 = 3 - 1 = 2                      
   Degrees of freedom Within = n - k = 15 - 3 = 12                      
                          
   Degrees of freedom Total F( k-1,n - k,) at 0.01 is = F Crit = 6.927                      
                          
Step 3                          
   Grand Mean = G / N = 24.8+21.6+24 / 3 = 23.467                      
    SST = ( Xi - GrandMean)^2 = (15-23.467)^2 + (23-23.467)^2 + (31-23.467)^2 + ……..& so on = 413.733                      
   SS Within = (Xi - Mean of Xi ) ^2 =,(15-24.8)^2 + (23-24.8)^2 + (31-24.8)^2 + ……..& so on = 386                      
                          
   SS Between = SST - SS Within = 413.733 - 386 = 27.733                      
Step 4                          
   Mean Square Between = SS Between / df Between = 27.733/2 = 13.867                      
   Mean Square Within = SS Within / df Within = 386/12 = 32.167                      
                          
Step 5                          
   F Cal = MS Between / Ms Within = 13.867/32.167 = 0.431                      
   We got |F cal| = 0.431 & |F Crit| =6.927                      
                          
MAKE DECISION                          
   Hence Value of |F cal| < |F Crit|and Here We Accept Ho                      

[ANSWERS]
a.Calculate the grand mean.
Grand Mean = 23.467

b.Calculate SSTR and MSTR. (Round your final answers to 2 decimal places.)
SST = 413.733   , MSTR=13.867

c.Calculate SSE and MSE. (Round your final answers to 2 decimal places.)
SSE = 386, MSE 32.167

d.Specify the competing hypotheses in order to determine whether some differences exist between the population means.
H0: A = B = C; HA: Not all population means are equal.

e.Calculate the value of the F(df1,df2) test statistic
F Crit = 6.927

f.Using the critical value approach at the 1% significance level, what is the conclusion to the test?
Do not reject H0 since the test statistic does not exceed 6.93

ONE WAY ANOVA Treatments Mean = X /n A 15 23 31 32 23 24.8 B 22 28 18 15 25 21.6 C 17 29 24 23 27 24

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