Question

A survey by Genworth Financial Inc., a financial services company, concludes that the cost of long-term care in the United States varies significantly, depending on where an individual lives (The Wall Street Journal, May 16, 2009). An economist collects data from the five states with the highest annual costs (Alaska, Massachusetts, New Jersey, Rhode Island, and Connecticut), in order to determine if his sample data are consistent with the survey's conclusions. The economist provides the following portion of an ANOVA table: a. Specify the competing hypotheses to test whether some differences exist in the mean long-term care costs in these five states. HOMAlaska - Massachusetts = New Jersey = Rhode Island = Connecticut Ha Not all population means are equal. Ho: MAlaska s Massachusetts New Jersey Rhode Island Connecticut Ha Not all population means are equal. HO: Alaska M Massachusetts New Jersey Rhode Island Connecticut Ha Not all population means are equal. b. Complete the ANOVA table. Assume that long-term care costs are normally distributed. (Round intermediate calculations to at least 4 decimal places. Round "MS" to 4 decimal places, and "P' to 3 decimal places.) df MSF ANOVA Source of Variation Between Groups Within Groups Total p-value 0.000 SS 635.0542 253.2192 888.2734

Answer 1

Solution:

Part a) State hypothesis:

Since null hypothesis is always stated as All groups have equal or same means, correct option is first option.

Ho : JlAlaska = | Massachusetts = New Jersey = Jl Rhode Island = Connecticut

H_A: Not all population means are equal

Part b) Complete ANOVA table:

Given:

SSTR = 635.0542

SSW = 253.2192

df_TR = 4

df_W = 20

Thus

MSTR = SSTR / df_TR =

MSTR = 635.0542 / 4

MSTR = 158.7636

and

MSW = SSW / df_W

MSW = 253.2192 / 20

MSW = 12.6610

and

F = MSTR / MSW

F = 158.7636 / 12.6610

F = 12.540

Thus we get:

Source of variation	SS	df	MS	F	p-value
Between Groups	635.0542	4	158.7636	12.540	0.000
Within Groups	253.2192	20	12.6610
Total	888.2734	24

Part c)

Level of significance = 5% = 0.05

Since p-value =0.000 < 0.05 level of significance, we reject null hypothesis H0.

Thus correct answer is:
Yes, since the p-value is less than the significance level.