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 inteviews with frail elderly people. In the interview, each person was compared with fictitious others who were worse off. After the interviews, the elderly people reported more satisfaction with their own lives. Following are hypothetical data similar to those obtained in the research study. The scores are measures on a life-satisfaction scale for a sample of n = 9 elderly people who completed the interview. Assume that the population average score on this scale is μ = 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 one-tailed test with α = .05. The life-satisfaction scores for the sample are 18, 23, 24, 44, 19, 27, 23, 26, 25.

we reject H₀

We have enough evidence to claim that people in the sample are significantly  more satisfied than others in general population at 95 % of CI

Step-by-step explanation:

Population mean    μ₀  = 20

Sample data   18  23  24 44  19 27 23 26 25

Then:

Sample mean      μ   =  25,44

Sample standard deviation     s = 7,14

Sample size     n  =  9

Hypoyhesis test:

Null hypothesis                                H₀              μ   =   μ₀

Alternative hypothesis                   Hₐ               μ   >   μ₀

Significance level    α  = 0,05    CI = 95 %

We must develop a  t-student one-tail test  ( to the right ) as follows

t(c)  =  ??

degree of freedom    df = n - 1         df =  8      and    α = 0,05

Then from t-student table    t(c)  =  1, 8595

To calculate   t(s)  =  (  μ  -  μ₀  ) / s/√n

t(s)  =  ( 25,44  -  20 ) * √9 / 7,14

t(s)  = 5,44*3 / 7,14

t(s)  = 2,29

Comparing   t(c) and  t(s)

t(s)  >  t(c)

Then t(s) is in the rejection region   we reject H₀

We have enough evidence to claim that people in the sample are significantly  more satisfied than others in general population at 95 % of CI

x > 1.3

step by step:

8.5−0.9x> 9.8−1.9x

step 1: simplify both sides of the inequality.

−0.9x+8.5> −1.9x+9.8

step 2: add 1.9x to both sides.

−0.9x+8.5+1.9x> −1.9x+9.8+1.9x

1x+8.5> 9.8

step 3: subtract 8.5 from both sides.

1x+8.5−8.5> 9.8−8.5

1x> 1.3

step 4: divide both sides by 1.

1x/1 > 1.3/1