In: Math
Some Alzheimer’s caregivers were asked to respond to the following statement: “Caregiving enabled me to develop a more positive attitude toward life.” The responses are reflected in the following table.
Strongly Disagree |
Somewhat Disagree |
No Opinion |
Somewhat Agree |
Strongly Agree |
166 |
116 |
171 |
234 |
542 |
a)
from graph, data indicate the true distribution is not uniform
b)
Ho: data indicate the true distribution is uniform
H1: data indicate the true distribution is not uniform
Chi square test for Goodness of fit
expected frequncy,E = expected proportions*total
frequency
total frequency= 1229
observed frequencey, O | expected proportion | expected frequency,E | (O-E) | (O-E)² | (O-E)²/E |
166 | 0.2000 | 245.800 | -79.80 | 6368.04 | 25.907 |
116 | 0.2000 | 245.800 | -129.80 | 16848.04 | 68.544 |
171 | 0.2000 | 245.800 | -74.80 | 5595.04 | 22.763 |
234 | 0.2000 | 245.800 | -11.80 | 139.24 | 0.566 |
542 | 0.200 | 245.800 | 296.20 | 87734.44 | 356.934 |
chi square test statistic,X² = Σ(O-E)²/E =
474.7144
level of significance, α= 0.05
Degree of freedom=k-1= 5 -
1 = 4
P value = 0.0000 [ excel function:
=chisq.dist.rt(test-stat,df) ]
Decision: P value < α, Reject Ho
conclusion: there is enough evidence to conclude that data indicate the true distribution is not uniform