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

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explanation

the graph has a slope of 1/3 because each time you move up 1, you go to the right 3. so, slope = rise/run = 1/3. we can cross choice a off the list since we want a slope of 3, rather than 1/3.

the first table shown has a slope of -3. this is because the rise is -6 and the run is 2, so slope = rise/run = -6/2 = -3. note how x increases by 2 each time (run) and y drops by 6 each time (rise). choice b can be eliminated.

the second table has a slope of 3. each time x goes up by 1, the y values increase by 3. slope = rise/run = 3/1 = 3. alternatively you can use the slope formula to prove that the slope is 3 here. use any two rows of values to generate the two points. choice c is one of the answers

the third table also has a slope of 3. let's use the slope formula to prove as such. i'm going to use the first two rows of the table. so i'll use the two points (2,1) and (5,10)

m = slope

m = (y2-y1)/(x2-x1)

m = (10-1)/(5-2)

m = 9/3

m = 3 so we get a slope of 3

choice d is the other answer

3: 2

step-by-step explanation:

shorter piece is 2: 3, so just reverse it, and you get 3: 2