In: Statistics and Probability
People tend to evaluate the quality of their lives relative to others around them. To demonstrate this, researchers conducted interviews with twenty-five (25) frail elderly people. In the interview, each person was compared with fictitious others who were worse off (downward social comparison). After the interviews, the elderly people reported their level of satisfaction with their lives. The scores are measures on a life-satisfaction scale which has a mean of 20 and standard deviation of 4. The 25 elderly had a mean score of 21.65.
Can we conclude that the people in the sample are significantly more satisfied than others in the general population?
Test this with an alpha level of .05.
As we are testing here whether the people in the sample are significantly more satisfied than others in the general population, therefore the null and the alternative hypothesis here are given as:
The test statistic now is computed here as:
As this is a one tailed test, we get the p-value from the standard normal tables as:
p = P(Z > 2.0625) = 0.0196
Therefore 0.0196 is the required p-value here.
As the p-value here is 0.0196 < 0.05 which is the level of significance, therefore the test is significant and we can reject the null hypothesis here and conclude that we have sufficient evidence here that the people in the sample are significantly more satisfied than others in the general population