Question

People tend to evaluate the quality of their lives relative to others around them. In a demonstration of this phenomenon, Frieswijk, Buunk, Steverink, and Slaets (2004) conducted fictitious interviews with frail elderly people. In the interview, each person was compared with others who were worse off. After the interviews, the elderly people reported more satisfaction with their own lives (the hypothetical data are reported below). The scores are measures on a life satisfaction scale for a sample of n = 9 elderly people who completed the interview are: 18, 23, 24, 22, 19, 27, 23, 26, 25.

Assume that the average score on this scale is m= 20. Are the data sufficient to conclude that the people in this sample are significantly more satisfied than others in the general population? Use a = .05.

Answer 1

Solution:-

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u < 20
Alternative hypothesis: u > 20

Note that these hypotheses constitute a one-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a one-sample t-test.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = s / sqrt(n)

S.E = 1
DF = n - 1

D.F = 8
t = (x - u) / SE

t = 3.0

where s is the standard deviation of the sample, x is the sample mean, u is the hypothesized population mean, and n is the sample size.

The observed sample mean produced a t statistic test statistic of 3.0.

Thus the P-value in this analysis is 0.009.

Interpret results. Since the P-value (0.009) is less than the significance level (0.05), we have to reject the null hypothesis.

From the above test we have sufficient evidence in the favor of the claim that the people in this sample are significantly more satisfied than others in the general population.