Question

A state legislator wants to determine whether her performance rating (0-100) has changed from last year...

  1. A state legislator wants to determine whether her performance rating (0-100) has changed from last year to this year. The following table shows the legislator’s performance rating from the same 16 randomly selected voters for last year and this year. At α = 0.01, is there enough evidence to conclude that the legislator’s performance rating has changed? Assume the performance ratings are normally distributed.

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

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

Using the above scenario and data analysis, answer the following questions.

  1. What is the research hypothesis? ­­­­­­­­­­­­­­­­­­___________________________________________________________________
  2. What is the null hypothesis? ___________________________________________________________________
  3. Which t test was used for this problem? (be specific) ________________________
  4. What is tcrit? _____________

How do you interpret the results (be complete) __________________________________________________________________________________________________________________________________________

Homework Answers

Answer #1

(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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