Question

This Question: 1 pt 2 of 9 (0 complete) This Quiz: 9 pts possible Consider the data in the table collected from three independent populations. Sample 1 Sample 2 Sample 3 a) Calculate the total sum af squs(SST) and partition the SST into its two components, the sum of squares between (SSB) and the sum of squares within (SSW). b) Use these values to construct a one-way ANOVA table. c) Using α=0.10, what conclusions can be made concerning the population means? 15 Click the icon to view a table of critical F-scores for α-0.10 a) Detemine the values. SST(Type an integer or a decimal.) SSBType an integer or a decimal.) SSW(Type an integer or a decimal.) b) Complete the one-way ANOVA table below. Sum of Degrees of Mean Sum of Source Squares FreedomSquares Between Within Total e integers or decimals. Round to three decimal places as nceded.) c) Let μι μ2, and μ3 be the population means of samples 1, 2, and 3, respectively. What are the correct hypotheses for a one-way ANOVA test? H1 Not all the means are equal. H1: Not all the means are equal What is the critical F-score, Ft? F(Round to three decimal places as needed.) What is the correct conclusion about the population means? H. The data Since the F-statistic in the rejection region, Vsufficient evidence to conclude that the population means are not all the same.

Answer 1

Based on the data given we get the following table :

Sample3 Sample2 Sample 1 15 24 Sum - Average- 266 134 6.377 3.606 St. Dev. - 122 26

The total sample size is N=10. Therefore, the total degrees of freedom are:

df_total​ = 10 −1 = 9

The between-groups degrees of freedom are df_between​ = 3−1 = 2, and

The within-groups degrees of freedom are:

df_within​  = df_total​ − df_between ​= 9 − 2 = 7

a)

The total sum of squares is computed as follows:

$SS_{total }=\sum_{i,j} X_{ij}^{2} - \frac{1}{N}\left ( \sum_{i,j}X_{ij} \right )^{2}$

$=(134+14+266) - \frac{1}{10}(18+6+24)^{2}$

$=414 - \frac{48^{2}}{10} = 183.6$

The within sum of squares is computed as shown in the calculation below:

$SS_{within} = \sum SS_{withingroups}= 26+2+122=150$

The between sum of squares is computed as shown in the calculation below:

$SS_{between} = SS_{total}- SS_{within}$

$=183.6 - 150 = 33.6$

b) Based on the above calculation the ANOVA table is given below :

df MS source 16.800 33.600 0.7840 treatments 150.000 7 21.429 error total 183.600

c) The null and alternative hypothesis is :

Null hypothesis	All means are equal
Alternative hypothesis	Not all means are equal

Answer : C option

The critical F score, at df (2,7) and level of significance 10%

From the f table

$F_{\alpha } (2,7)= 3.2574$

Conclusion :  Since F -statistic is less than crital F value i.e. it lie outside rejection region , so we accept null hypothesis. The data does not have enough evidence to conclude that population means are not all the equal.