Question

In a one-way ANOVA, the following data were collected: SSTT = 211.35, SSE = 280.5, N = 21,I = 6. A.How many samples are there?b. How many degrees of freedom are there for SSTT and SSE?Compute the mean squares MSTr and MSE.d. Compute the value of the test statistic F.Find the critical value for a level of significance of a = 0.025.f. Complete the one-way ANOVA hypothesis test for this data.g. Can you conclude that two or more of the population means are different? Explain why andwrite your conclusion completely. Use an a = 0.025 level of significance.

Answer 1

Write out the given parameters

`\begin{gathered} \text{SSTT}=211.35 \\ \text{SSE}=280.5 \\ N=21 \\ I=6 \end{gathered}`

SSTT is the sum of squares due to test

SSE is the sum squares due to error

N is the total number of observation

I is the number in group

a) The number of samples is N,

`\begin{gathered} N\text{ is given} \\ N=21 \end{gathered}`

Hence, the number of samples is 21

b) (i) The degree of freedom for SSTT is

`\begin{gathered} df_{\text{SSTT}}=I-1 \\ I=6 \\ df_(SSTT)=I-1=6-1=5 \end{gathered}`

Hence the degree of freedom for SSTT is 5

(ii) The degree of freedom for SSE is

`\begin{gathered} df_{\text{SSE}}=N-I \\ =21-6 \\ =15 \end{gathered}`

Hence, the degree of freedom for SSE is 15

(c) Compute the mean squares MST and MSE

MSE is calculated using the formula

`\begin{gathered} \text{MST}=(SSTT)/(I-1)=(211.35)/(6-1) \\ =(211.35)/(5)=42.27 \end{gathered}`

MSE is calculated using the formula

`\begin{gathered} \text{MSE}=(SSE)/(N-I)=(280.5)/(21-6) \\ =(280.5)/(15)=18.7 \end{gathered}`

Hence, MST= 42.7 and MSE= 18.7

(d) Compute the value of the test statistic F

F is calculated using the formula

`\begin{gathered} F=(MST)/(MSE) \\ F=(42.27)/(18.7) \\ F=2.2604 \end{gathered}`

Hence, the test statistics F is 2.2604

(e) Find the critical value for a level of significance of 0.025

Using table, the critical value for a level of significance of 0.025 at df of freedom 5 and 15 is

`F_{\text{critical}}=3.576`

(f)

`\begin{gathered} H_0=\mu_1=\mu_2 \\ H_1=at\text{ least two means differ} \end{gathered}`

(g) We can conclude that two or more of the population means are different. This is because the critical value is greater than the test statistics F.

Hence, we we will reject the null hypothesis and conclude that there is difference in mean population across the six groups