Question

As an aid for improving students' sleeping habits, nine students were randomly selected to attend a seminar on the importance of sleep in life. The table below shows the number of hours each student slept per week before the seminar.

Students |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |

Before |
9 | 12 | 6 | 15 | 3 | 18 | 10 | 13 | 7 |

After |
9 | 17 | 9 | 20 | 2 | 21 | 15 | 22 | 6 |

a. Construct a 90% confidence interval for the difference of mean number of study hours before and after attending the seminar and INTERPRET

b. At 10% level of significance, did attending the seminar increase the number of hours the students studied per week?

Would gladly appreciate your help! Thank you so much! :)

Answer #1

Ans:

Students | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |

Before | 9 | 12 | 6 | 15 | 3 | 18 | 10 | 13 | 7 |

After | 9 | 17 | 9 | 20 | 2 | 21 | 15 | 22 | 6 |

difference(d) | 0 | 5 | 3 | 5 | -1 | 3 | 5 | 9 | -1 |

mean(d)= |
3.111 |
||||||||

std. dev |
3.333 |

a)df=9-1=8

critical t value=tinv(0.1,8)=1.860

90% confidence interval for mean difference

=3.111+/-1.860*(3.333/sqrt(9))

=3.111+/-2.067

**=(1.044, 5.178)**

b)As both limits of above confidence interval are
**positive**,we can conclude that **attending
the seminar increase the number of hours the students studied per
week.**

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