STATISTICS 5

Statistics

Q1

Thesamples are independent. This is because the test constitutesmeasurements comprising of two items (Triola, 2010) the first itemis the new headache medicine and the second item is the placebo drug.

Q2

Thesamples are dependent because test constitutes paired measurementscomprising of one item (Triola, 2010). In this case, theeffectiveness of a headache medicine alone is being examined.

Q3

Nullhypothesis: p1 = p2

Alternativehypothesis: p1≠ p2

Fromthe null hypothesis, p1 – p2 = 0

P1= 22% * 500 = 110 smokers

P2= 14%*450 = 63 smokers

P*= (110 + 63) / (500 + 450) = 0.18

Teststatistic: Z = (0.22 – 0.14) / sqt [(0.18)(0.82) (0.002 + 0.002)]

=0.08 /0.02

=4

Significancelevel = 0.01 critical value two-tailed test = 2.58

Zstatistic is greater than z critical, which is sufficient evidence atthe 0.01 significance level, to reject the null hypothesis and make aconclusion that the two proportions are not equal.

Q4

Confidenceinterval

Df= 157

*α*= 0.01

*1– *α/2

=0.995

Tvalue = 16.3

(µ1- µ2) = 189.1 – 203.7 = -14.6

Interval= -14.6 – 16.3 = -30.9 and -14.6 + 16.3 = 1.7

-30.9< (µ1 – µ2) < 1.7

Q5

Nullhypothesis: d-bar ≤ 0

Alternativehypothesis: d-bar > 0

Q6

Before |
After |
D = after – before |

71 |
75 |
4 |

66 |
75 |
9 |

67 |
65 |
-2 |

77 |
80 |
3 |

75 |
87 |
12 |

Sum = 26 |

d-bar= 26/5 = 5.20

s= 5.45

Q8

SE= s/sqt (n) = 5.45/sqt (5) = 2.44

t= d-bar/SE = 5.2 / 2.44 = 2.13

pvalue = 0.0998

Decision:p value > significance level then null hypothesis is notrejected.

Q9

Thereis no adequate evidence to support the idea that tutoring improvesMath scores.

References

Triola,M. F. (2010). *Elementarystatistics*.Boston: Addison-Wesley.