Perform the indicated one-way ANOVA test. Be sure to do the following: identify
ID: 3323328 • Letter: P
Question
Perform the indicated one-way ANOVA test. Be sure to do the following: identify the claim and state the null and alternative hypotheses. Caleulate the test statistic F and determine the p-value. Decide to reject or to fail to reject the null hypothesis and interpret the decision in the context of the original claim. sume that each population is normally distributed and that the population variances are equal. 12) A researcher wishes to compare times it takes workers to assemble a certain computer component using different machines. Workers are randomly selected and randomly assigned to one of three different machines. The time (in minutes) it takes each worker to assemble the component is recorded. Test the claim that there is no difference in the mean assembly times for the three machines. Use a 0.01. Machine 1 Machine 2 Machine 3 32 32 31 30 40 29 38 31 28 29 25 36 32 35 31 ID: ES6L 10.4.1-14 Decision: Ho : Conclusion: Test statistic: p-value:Explanation / Answer
Solution:- The following table is obtained:
AND nw
The total sample size is N=17. Therefore, the total degrees of freedom(df) are:
df total =17–1=16
Also, the between-groups degrees of freedom are dfbetween=3–1=2, and the within-groups degrees of freedom are:
df_within=df_total – df_between=16–2=14
First, we need to compute the total sum of values and the grand mean. The following is obtained
i,j Xij =189+243+113=545
Also, the sum of squared values is
i,jX2ij = 5959+8519+3211=17689
Based on the above calculations, the total sum of squares is computed as follows
SStotal=i,jX2ij–1N(i,jXij)2=17689–545217=216.941
The within sum of squares is computed as shown in the calculation below:
SS within= SS_withingroups =5.5+83.4285714286+18.75=107.679
The between sum of squares is computed directly as shown in the calculation below:
Now that sum of squares are computed, we can proceed with computing the mean sum of squares:
MS_between=SS_between / df_between=109.263 / 2=54.631
MS_within=SS_within / df_within=107.679 / 14=7.691
Now that sum of squares are computed, we can proceed with computing the mean sum of squares:
MS_between=SS_between / df_between=109.263 / 2=54.631
MS_within=SS_within / df_within=107.679 / 14=7.691
Finally, with having already calculated the mean sum of squares, the F-statistic is computed as follows:
F=MS_between / MS_within=54.6317.691=7.103
(1) Null and Alternative Hypotheses
The following null and alternative hypotheses need to be tested:
Ho: 1 = 2 = 3
Ha: Not all means are equal
The above hypotheses will be tested using an F-ratio for a One-Way ANOVA.
(2) Rejection Region
Based on the information provided, the significance level is =0.01, and the degrees of freedom are df1=2 and df2=2, therefore, the rejection region for this F-test is R={F:F>Fc=6.515}.
(3) Test Statistics
F=MS_between / MS_within=54.6317.691=7.103
(4) Decision about the null hypothesis
Since it is observed that F=7.103 > Fc=6.515, it is then concluded that the null hypothesis is rejected.
Using the P-value approach: The p-value is p=0.0074, and since p=0.0074<0.01, it is concluded that the null hypothesis is rejected.
(5) Conclusion
It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that not all 3 population means are equal, at the =0.01 significance level.
S.no Machine1 Machine2 Machine3 1 32 40 31 2 32 29 28 3 31 38 29 4 30 33 25 5 33 36 6 31 32 7 35