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The frequency distribution shows the results of 200 test scores. Are the test sc

ID: 3229334 • Letter: T

Question

The frequency distribution shows the results of 200 test scores. Are the test scores nomnaly distributed? Use o 30.01 Complete parts (a) through (dL 495-58.5 58 567 67.5-76.5 76 5-85.5 85 5-94.5 20 Using a chi square goodness-of-fittest, you can decide, with some degree of certainty. whether a variable is normally distributed. In all chi-square tests for normality, the null and alternative hypotheses are as follows. Ho: The test scores have a normal distribution. Ha The test scores do not have a normal distribution. a. Find the expected frequencies. Class boundaries 495 585 58.5675 675 7615 5.855 B55945 requency f pected frequency (Round to the nearest integer as needed.) b. Determine the critical value, X0, and the rejection region. xo- Round to three decimal places as needed. Click to select your answers).

Explanation / Answer

Part-a-From Following results we found expected frequencies as Ei

Class Boundary

f

Mid Point x

fx

fx^2

Zi=(Upper Bound-69.39)/8.40

P(Z<zi)

p=P(In the class interval)

Expected Frequency=200*p

(Oi-Ei)^2/Ei

49.5-58.5

20

54

1080

58320

-1.30

0.10

0.10

19.48

0.01

58.5-67.5

60

63

3780

238140

-0.23

0.41

0.31

62.72

0.12

67.5-76.5

82

72

5904

425088

0.85

0.80

0.39

78.07

0.20

76.5-85.5

34

81

2754

223074

1.92

0.97

0.17

34.22

0.00

85.5-94.5

4

90

360

32400

2.99

1.00

0.03

5.23

0.29

Total

200

360

13878

977022

4.23

3.28

1.00

199.72

0.621

Mean=13878/200

69.39

SD

8.40

Part-b

Degree of freedom=k-1=5-1=4

Critical values 02= 13.277 using excel function =CHIINV(0.01,4)

Critical Region is values 2 > 02-option-A

Part-c

From above table test statistic 2=0.621

Part-d

Do not reject H0. At 1% significance level there is no enough evidence to reject the the claim that test scores are normally distributed

Class Boundary

f

Mid Point x

fx

fx^2

Zi=(Upper Bound-69.39)/8.40

P(Z<zi)

p=P(In the class interval)

Expected Frequency=200*p

(Oi-Ei)^2/Ei

49.5-58.5

20

54

1080

58320

-1.30

0.10

0.10

19.48

0.01

58.5-67.5

60

63

3780

238140

-0.23

0.41

0.31

62.72

0.12

67.5-76.5

82

72

5904

425088

0.85

0.80

0.39

78.07

0.20

76.5-85.5

34

81

2754

223074

1.92

0.97

0.17

34.22

0.00

85.5-94.5

4

90

360

32400

2.99

1.00

0.03

5.23

0.29

Total

200

360

13878

977022

4.23

3.28

1.00

199.72

0.621

Mean=13878/200

69.39

SD

8.40

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