**Name**

**Module ****4**** Homework Assignment**

1. Determine whether the samples are independent or dependent.

The effectiveness of a new headache medicine is tested by measuring the amount of time before the headache is cured for patients who use the medicine and another group of patients who use a placebo drug. Explain.

Solution:Independent samples | Instructor Comments: |

2. Determine whether the samples are independent or dependent.

The effectiveness of a headache medicine is tested by measuring the intensity of a headache in patients before and after drug treatment. The data consist of before and after intensities for each patient. Explain.

Solution:Matched pairs | Instructor Comments: |

3. In a random sample of 500 people aged 20-24, 22% were smokers. In a random sample of 450 people aged 25-29, 14% were smokers. Test the claim that the proportion of smokers in the two age groups is the same. Use a significance level of 0.01. Show all steps of the hypothesis test and all calculations.

Solution:P1=0.22, x1=110; p2=0.145, x2=63; p=173/950 0.182;claim: p1=p2Alternative= p1≠p2Hypothesis test: H0: p1 = p2 H1: p1 p2The critical value is z=±2.575We reject h0 because there is sufficient evidence to reject the claim that the proportion of smokers in the two age groups is the same. | Instructor Comments: |

4. A researcher wishes to determine whether people with high blood pressure can reduce their blood pressure by following a particular diet. Use the sample data below to construct a 99% confidence interval for µ1 – µ2 where µ 1 and µ2 represent the mean for the treatment group and the control group respectively. Show all steps of the hypothesis test and all calculations. Use the following t-distribution table – http://www.itl.nist.gov/div898/handbook/eda/section3/eda3672.htm

**Treatment Group** **Control Group**

n1 = 85 n2 = 75

189.1 = 203.7

s1 = 38.7 s2 = 39.2

Solution: I claim that the treatment population mean is smaller than Control population mean.µ 1= Treatment population meanµ2= Control population meanHo: µ1 = µ2Ha: µ1 < µ2Test- statistic t-value= -2.37P=0.0096Df= 155.01p-value 0.0096 is less that 0.01, this means reject ho equal means. There is enough evidence to support the claim that Treatment population mean is smaller that control population mean. This means that diet to reduce blood pressure works. | Instructor Comments: |

**For questions 5 – 9 below**, use the traditional method of hypothesis testing to test the given claim about the means of two populations. Assume that two dependent samples have been randomly selected from normally distributed populations.** **

Five students took a math test before and after tutoring. Their scores were as follows.

Student |
A |
B |
C |
D |
E |

Before |
71 |
66 |
67 |
77 |
75 |

After |
75 |
75 |
65 |
80 |
87 |

Using a 0.01 level of significance, test the claim that the tutoring has an effect on the math scores.

5. State the null and alternative hypotheses.

Solution:H0: µd = 0. H1: µd ≠ 0. | Instructor Comments: |

6. Find *d*, , and *s**d**,** *

Solution:n = 5 d-bar = 5.20 s (of d) = 5.45 | Instructor Comments: |

7. Find the value of the test statistic, *t*.

Solution:SE = s/√n = 5.44977063737548/√5 = 2.437211521 t = d-bar/SE = 5.2/2.43721152139079 = 2.133585844 p- value = 0.0998 | Instructor Comments: |

8. Find the critical value(s) and state the decision about the null hypothesis.

Solution:Critical values: t = 4.604, -4.604. | Instructor Comments: |

9. State the conclusion in non-technical terms.

Solution:Fail to reject H0. There is not sufficient evidence to support the claim that the tutoring has an effect | Instructor Comments: |