A new prep class was designed to improve SAT test scores. Four students were sel

Question

A new prep class was designed to improve SAT test scores. Four students were selected at random. Their exams were recorded, one before the class and one after. The data recorded scores on two practice in table below. Are the scores, on average, higher after the class? Test at a 5% level.

(a) Identify the null hypothesis and alternative hypothesis.

(b) Determine the test statistic. Show all work; writing the correct test statistic, without supporting work, will receive no credit.

(c) Determine the P-value for this test. Show all work; writing the correct P-value, without supporting work, will receive no credit.

(d) Write your conclusion and explain.

SAT Scores Student 1 Student 2 Student 3 Student 4 Scores before class 1840 1960 1920 2150 Score after class 1920 2160 2200 2100

Explanation / Answer

Part a

The null and alternative hypothesis for this test is given as below:

Null hypothesis: H0: Scores after class are same as the scores before class.

Alternative hypothesis: Ha: Scores after class are higher than the scores before class.

H0: µbefore = µafter versus Ha: µbefore < µafter

This is a one tailed test. This is lower tailed test or left tailed test.

Part b

Determine the test statistic. Show all work; writing the correct test statistic, without supporting work, will receive no credit.

We assume 5% level of significance for this test. ( = 0.05)

Test statistic formula is given as below:

Test statistic = t = Dbar / [SD/sqrt(n)]

Calculations for this test are given as below:

X

Y

Di

(Di - DBar)^2

1840

1920

-80

2256.25

1960

2160

-200

5256.25

1920

2200

-280

23256.25

2150

2100

50

31506.25

Dbar = -127.5

SD = 144.0775

Test statistic = t = -127.5/(144.0775/sqrt(4))

Test statistic = t = -127.5/(144.0775/2)

Test statistic = t = -127.5/72.03875

Test statistic = t = -1.769880793

Lower critical value = -2.3534

Part c

= 0.05

DF = n – 1 = 4 – 1 = 3

Test statistic = t = -1.769880793

P-value = 0.0874

By using t-table

Part d

= 0.05

P-value >

So, we do not reject the null hypothesis that Scores after class are same as the scores before class.

There is insufficient evidence to conclude that Scores after class are higher than the scores before class.

X

Y

Di

(Di - DBar)^2

1840

1920

-80

2256.25

1960

2160

-200

5256.25

1920

2200

-280

23256.25

2150

2100

50

31506.25

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