Hypothesis Testing Problem Set

¶ … SE = 2.1820749866

Hypothesis Testing

Answer each question completely to recieve full credit

There is a new drug that is used to treat leukemia. The following data represents the remission time in weeks for a random sample of 21 patients using the drug.

State the Null Hypothesis

Let XA represent the mean remission time of the new drug and XB the mean remission time of the previous drug, then the null hypothesis states that the eukemia remission times are not significantly different (Ho: XA = XB) between the two drugs.

State the Alternatice Hypothesis

The alternative hypothesis is that the time to remission is significantly different (H1: XA ? XB) between the two drugs.

State the Level of significance

= .01, two-tailed

State the Test Statistic

Z scores

Perform Calculations

Since ? = .01 and this is a two-tailed test, then the value for each tail would be .005. This corresponds to a Zcrit value of ±2.58.

XA = 17.095; SDA = 10.000; Std. Error (SEA) = 10/?21 = 2.182

Z = (XB - XA)/SEA = (12.5-17.095)/2.182 = -2.11

The probability of getting a Z score of -2.11 is p = .0174, or a Z score of ±2.11 is p = .0348, two-tailed.

Statistical Conclusion

Siince Zcrit > Z, then the null hypothesis cannot be rejected. This conclusion is also supported by the fact that approximately 1.7% of patients using the new drug would be expected to have a remission time equal to or less than the mean remission time obtained using the previous drug.

Experimental Conclusion

At the confidence level selected for this comparison, the mean remission times between the two drugs are not significantly different.

Let X be a random variable representing the remission time in weeks for all patients using the new drug. Assume that the distribution of x is normal. A previously used drug treatment has a mean remission time of 12.5 weeks. Does the data indicate that the mean remission time using the new drug is different from 12.5-week at a level of significance of 0.01?

Problem 2

Men Women

20 29

37 28

46 20

23 28

20 42

23 45

21 19

15 45

20 16

28 32

27 38

20 45

30 41

22 34

27 28

38 21

29 42

20 21

16 30

27 28

42 30

37 43

39 40

39 16

32 44

16 15

21 16

26 20

17 41

39 16

Count 30-30 t-Test: Two-Sample Assuming Equal Variances

Means 27.2-30.4

Variance 77.151-112.323 Variable 1 Variable 2

Std. Dev. 8.784 10.598 Mean 27.2

Combined Std. Error 2.513 Variance 77.1505747126 112.3229885057

t statistic -1.273 Observations 30

Pooled Variance 94.7367816092

Hypothesized Mean Difference 0

df 58

t Stat -1.2733162878

P (T

t Critical one-tail 1.6715527629

P (T

t Critical two-tail 2.001717468

Hypothesis Testing

Answer each question completely to recieve full credit.

2. We wish to test the claim that the mean body mass index (BMI) of men is equal to the mean BMI of women. Use the data to the right to test this claim.

State the Null Hypothesis

The null hypothesis is that there is no difference between the BMI of men and women: Ho: XA = XB, if XA and XB represent mean BMI for men and women, respectively.

State the Alternatice Hypothesis

The alternative hypothesis is that there is a significant difference between the BMI for men and women (H1: XA ? XB).