Question

An SAT prep course claims to improve the test score of students. The table below shows the scores for seven students the first two times they took the verbal SAT. Before taking the SAT for the second time, each student took a course to try to improve his or her verbal SAT scores. Do these results support the claim that the SAT prep course improves the students' verbal SAT scores?

Let d=(verbal SAT scores prior to taking the prep course)−(verbal SAT scores after taking the prep course)d=(verbal SAT scores prior to taking the prep course)−(verbal SAT scores after taking the prep course). Use a significance level of α=0.01 for the test. Assume that the verbal SAT scores are normally distributed for the population of students both before and after taking the SAT prep course.

Student | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
---|---|---|---|---|---|---|---|

Score on first SAT | 360 | 530 | 580 | 480 | 550 | 460 | 520 |

Score on second SAT | 400 | 570 | 610 | 550 | 580 | 510 | 550 |

Step 1 of 5: State the null and alternative hypotheses for the test. Ho: μd (=,≠,<,>,≤,≥) 0 Ha: μd (=,≠,<,>,≤,≥) 0

Step 2 of 5: Find the value of the standard deviation of the paired differences. Round your answer to two decimal places.

Step 3 of 5: Compute the value of the test statistic. Round your answer to three decimal places.

Step 4 of 5: Determine the decision rule for rejecting the null hypothesis H0H0. Round the numerical portion of your answer to three decimal places. Reject Ho if (t, I t I) (<,>) _____

Step 5 of 5: Make the decision for the hypothesis test. Reject Null Hypothesis Fail to Reject Null Hypothesis

Answer #1

Pre |
Post |
Difference |

360 | 400 |
-40 |

530 | 570 |
-40 |

580 | 610 |
-30 |

480 | 550 |
-70 |

550 | 580 |
-30 |

460 | 510 |
-50 |

520 | 550 |
-30 |

1).

Null and Alternative hypothesis:

Ho : µd = 0

H1 : µd < 0

2).

∑d = -290

∑d² = 13300

n = 7

Mean , x̅d = Ʃd/n = -290/7 = -41.429

Standard deviation, sd = √[(Ʃd² - (Ʃd)²/n)/(n-1)] = √[(13300-(-290)²/7)/(7-1)] = 14.6385 = 14.64

3)

Test statistic:

t = (x̅d)/(sd/√n) = (-41.4286)/(14.6385/√7) = -7.488

4)

df = n-1 = 6

Critical value, t_c = T.INV(0.01, 6) = -3.143

Reject Ho if t < -3.143

5)

Decision:

Reject the null hypothesis

An SAT prep course claims to improve the test score of students.
The table below shows the scores for seven students the first two
times they took the verbal SAT. Before taking the SAT for the
second time, each student took a course to try to improve his or
her verbal SAT scores. Do these results support the claim that the
SAT prep course improves the students' verbal SAT scores? Let
d=(verbal SAT scores prior to taking the prep course)−(verbal...

An SAT prep course claims to improve the test score of students.
The table below shows the scores for seven students the first two
times they took the verbal SAT. Before taking the SAT for the
second time, each student took a course to try to improve his or
her verbal SAT scores. Do these results support the claim that the
SAT prep course improves the students' verbal SAT scores? Let
d=(verbal SAT scores prior to taking the prep course)−(verbal...

An SAT prep course claims to improve the test score of students.
The table below shows the scores for seven students the first two
times they took the verbal SAT. Before taking the SAT for the
second time, each student took a course to try to improve his or
her verbal SAT scores. Do these results support the claim that the
SAT prep course improves the students' verbal SAT scores?
Let d=(verbal SAT scores prior to taking the prep
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9. An SAT prep course claims to improve the test score of
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the SAT prep course improves the students' verbal SAT scores? Let
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Student
1
2
3
4
5
6
7
Score on first SAT
530
410
380
600
480
440
380
Score on second SAT
560
460
400
620
500
520
430
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times they took the verbal SAT. Before taking the SAT for the
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