Question

Leading Businesses The following data show the yearly budgets for leading business sectors in the United States. At α = 0.05, is there a significant difference in the mean budgets of the business sectors? The data are in thousands of dollars.Use the traditional method of hypothesis testing unless otherwise specified.

a. State the hypotheses and identify the claim.

b. Find the critical value.

c. Compute the test value.

d. Make the decision.

e. Summarize the results.

Beverages	Electronics	Food producers	Supportive services
170	46	59	56
128	24	58	37
19	18	33	19
16	14	31	19
12	13	28	17
11	12	22	15
10	10	16	15

Answer 1

ANOVA Test:

(1) Null and Alternative Hypotheses The following null and altermative hypotheses need to be tested Ha: Not all means are equal The above ypoiheses wto ftr a on Aa (2) Rejection Region Based on the information provided, the significance level is α-0.05, and the degrees of freedom are dfi 3 and dfe 3, therefore, the rejection region for this F-test is R-{F: F> F 3.009

(3) Test Statistic: From the given data

Group 1 170 128 19 Group 3 Group 2 Group 4 46 59 37 19 19 17 15 15 14 12 28 12 16 137 19.571 3625 12.541 247 35.286 10419 16.849 Sum- Average 52.286 25.429 5966 15.49 St. Dev.- 67.243 27129.4285714 943.714285714 1703.42857142 1439.71428571 29 29

The total sample size is N 28. Therefore, the total degrees of freedom are dftotal -28-1-27 Also, the between-groups degrees of freedom are dfietueen degrees of freedom are: 4-1-3, and the within-groups dfuithin dftotal - dfbetuween - 27-3- 24 First, we need to compute the total sum of values and the grand mean. The following is obtained Σ xij = 366 + 137 + 247 + 178 = 928 1.3 Also, the sum of squared values is x46266 +3625 +10419+5966 66276 2.J Based on the above calculations, the total sum of squares is computed as follows (yx,) 9 ss,otal= Σ x =66276 35519.429 2.0 1.J The within sum of squares is computed as shown in the calculation below SSwithin- SSuithingroupsー27129.428571429 + 943.71428571429 1703.4285714286 + 1439.7142857143 31216.286

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: Sn 4303.143 M Sbetweendfbetweer Ma51-1043 1434.381 1216.286 -1300.679 24 Now that sum of squares are computed, we can proceed with computing the mean sum of squares: MSbetureendfbetureen 1434.381 MS,ithin-S thin = 31216.286 MSuitheithin 31216.286 1300.679 dfwithin 24 Finally, with having already calculated the mean sum of squares, the F-statistic is computed as F MSitueen1434.381I ithn 1300.679 1.103

Since it is observed that F-1.103 rejected F, 3.009, it is then concluded that the null hypothesis is not Using the P-value approach: The p-value is p 0.3673, and since p concluded that the null hypothesis is not rejected. 0.3673 > 0.05, it is t s concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that not all 4 population means are equal, at the α-0.05 significance level.