Question

2. A study has been conducted to determine whether the mean spending for internet services differs for residents of three cities. Random samples of people were selected from each city and their spending was recorded. The following ANOVA-single factor computer output was generated.

Anova: Single Factor

ANOVA

Source of Variation SS df MS F

Between Groups ? 2 12.3333333 ?

Within Groups 46.5714285 ? ?

Total 71.2380952 20

a. Fill-in the missing values denoted by ‘?’.

b. Write down the appropriate null and alternative hypotheses.

c. Given the value of the tabulated F = 5.2 (α=.01), test the hypotheses you formulated in ‘b’ above.

d. Clearly state the test’s decision and conclusion. e. Do you need to calculate the LSD and perform pair-wise comparisons? Explain your answer.

Answer 1

a. )

Source of variation	SS	DF	MS	F
Between Groups	24.66667	2	12.33333	4.766871
Within groups	46.57143	18	2.587302
Total	71.2381	20

b. the appropriate null and alternative hypotheses are

Ho:

Ha: at least two means are different

c. Given the value of the tabulated F = 5.2 (α=.01), since F value 4.76 < 5.2 so we don not reject Ho

d. we will conclude that the mean spending for internet services do not differs for residents of three cities.

e. No because LSD and perform pair-wise comparisons are performed when there exist difference but here   all the means are same so we do not need for   LSD and perform pair-wise comparisons. .