Voter |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
11 |
12 |
13 |
14 |
15 |
16 |
Rating (last year) |
60 |
54 |
78 |
84 |
91 |
25 |
50 |
65 |
68 |
81 |
75 |
45 |
62 |
79 |
58 |
63 |
Rating (this year) |
56 |
48 |
70 |
60 |
85 |
40 |
40 |
55 |
80 |
75 |
78 |
50 |
50 |
85 |
53 |
60 |
SPSS Printouts |
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Mean |
N |
Std. Deviation |
Std. Error Mean |
||||||||||||
Pair 1 |
Rating last year |
64.88 |
16 |
16.733 |
4.183 |
||||||||||
Rating this year |
61.56 |
16 |
15.262 |
3.816 |
|||||||||||
Paired Differences |
t |
df |
Sig. (2-tailed) |
||||||||||||
Mean |
Std. Deviation |
Std. Error Mean |
99% Confidence Interval of the Difference |
||||||||||||
Lower |
Upper |
||||||||||||||
Pair 1 |
Rating last year - Rating this year |
3.313 |
9.680 |
2.420 |
-3.818 |
10.443 |
1.369 |
15 |
.191 |
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Using the above scenario and data analysis, answer the following questions.
How do you interpret the results (be complete) __________________________________________________________________________________________________________________________________________
(a)
HA: Research Hypothesis: (Performance rating has changed fro last year to this year)
(b) H0: Null Hypothesis: (Performance rating has not changed fro last year to this year)
(c) Dependent Sample t test (paired t test)
(d) tcrit = 2.9467
EXPLANATION:
= 0.01
ndf = 15
From Table, critical values of t = 2.9467
(e) Conclusion:
Since the calculated value of t = 1.369 is less than critical value of t = 2.9467, the difference is not significant. Fail to reject null hypothesis.
Conclusion:The data do not support the claim that Performance rating has changed fro last year to this year..
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