Nine students took the SAT. Their scores are listed below. Later on, they read a book on test preparation and retook the SAT. Their new scores are listed below. Test the claim that the book had no effect on their scores. Use α=0.05. Assume that the distribution is normally distributed. Student 1 2 3 4 5 6 7 8 9 Scores before reading book 72 0 86 0 850 88 0 86 0 710 85 0 1200 95 0 Scores after reading book 74 0 86 0 840 92 0 89 0 720 84 0 1240 97 0 Nine students took the SAT. Their scores are listed below. Later on, they read a book on test preparation and retook the SAT. Their new scores are listed below. Test the claim that the book had no effect on their scores. Use α = 0.05. Assume that the distribution is normally distributed. Student 1 2 3 4 5 6 7 8 9 Scores before reading book 72 0 86 0 850 88 0 86 0 710 85 0 1200 95 0 Scores after reading book 74 0 86 0 840 92 0 89 0 720 84 0 1240 97 0

t= 0.4933

t ≥ t ( 0.025 ,8 ) = 2.306

Since the calculated value of t= 0.4933 is less than t ( 0.025 ,8 ) = 2.306 therefore we accept the null hypothesis at 5 % significance level . On the basis of this we conclude that the book had no effect on their scores.

Step-by-step explanation:

We state our null and alternative hypotheses as

H0: ud= 0 Ha: ud≠0

The significance level is set at ∝= 0.05

The test statistic under H0 is

t= d`/ sd/√n

which has t distribution with n-1 degrees of freedom

The critical region is t ≥ t ( 0.025 ,8 ) = 2.306

Computations

Student       Scores before      Scores after    Difference           d²

                        reading book                    ( after minus before)

1                          720                   740             20                       400

2                        860                   860              0                           0

3                        850                   840             -10                       100

4                        880                   920             40                       1600

5                        860                   890             30                        900

6                        710                    720              10                         100

7                       850                    840              -10                       100

8                      1200                  1240             40                        1600

9                      950                    970              20                           40

∑                    6930                  8020           140                         4840

d`= ∑d/n= 140/9= 15.566

sd²= 1/8( 4840- 140²/9) = 1/8 (4840 - 2177.778) = 2662.22/8= 332.775

sd= 18.2422

t= 3/ 18.2422/ √9

t= 0.4933

Since the calculated value of t= 0.4933 is less than t ( 0.025 ,8 ) = 2.306 therefore we accept the null hypothesis at 5 % significance level . On the basis of this we conclude that the book had no effect on their scores.

2: 10

step-by-step explanation: