Chapter 10
10.44
Does lovastatin reduce the risk of heart attacks? In a Texas study, researchers gave lovastatin to 2,325 people and inactive substitute to 2,081 people. After 5 years 57 of the lovastatin group had sufford a heart attack, compared with the 97 for the inactive pill.
A) State the appropriate hypothesis
B) Obtain statistic and p-value. Interpret the results at a=.01
C) Is normality assured?
D) Is the difference large enough to be important
E) What else would medical researchers need to know before prescribing this drug widely?
(a) H0: p1 = p2 vs. Ha: p1 < p2(b) Hypothesis test for two independent proportions:
p1 / p2 / pc
0.0245 / 0.0466 / 0.035 / p (as decimal)
57/2325 / 97/2081 / 154/4406 / p (as fraction)
57. / 97. / 153.999 / X
2325 / 2081 / 4406 / n
-0.0221 / difference
0. / hypothesized difference
0.0055 / std. error
-3.99 / z
3.35E-05 / p-value (one-tailed, lower)
Conclusion: Since the p-value is less than 0.01, we reject H0.
Therefore, there is sufficient statistical evidence to conclude at 1% level of significance that lovastatin reduces the risk of a heart attack.
(c) Since the sample sizes are large (> 30), the normality assumptions are fulfilled.
(d) Yes (because the value of the z-statistic is quite high). The difference is large enough to be important.
(e) Before prescribing this drug widely, the researchers should also study the possible side effects of this drug.
10.46
To test the hypothesis that students who finished an exam first get better grades. Professor Hardtack kept the mean score of 77.1 with a standard deviation of 19.6 while the last 24 papers handed in showed a mean score of 69.3 with a standard deviation of 24.9. Is this a significant difference at a=.05?
A) State the hypothesis for a right tailed test
B) Obtain a test statistic and p-value assuming equal variances.
C) Is the difference in the mean scores large enough to be important?
D) Is it reasonable to assume equal variances
E) Carry out a formal test for equal variances at a=.05, showing all steps clearly
(a) State the hypotheses for a right-tailed test.Hypotheses: H0: 1 = 2 vs Ha: 1 > 2
(b) Obtain a test statistic and p-value assuming equal variances. Interpret these results.
Hypothesis Test: Independent Groups (t-test, pooled variance)
First Lot / Second Lot
77.1 / 69.3 / Mean
19.6 / 24.9 / std. dev.
25 / 24 / N
47 / Df
7.8000 / difference (First Lot - Second Lot)
499.5760 / pooled variance
22.3512 / pooled std. dev.
6.3874 / standard error of difference
0 / hypothesized difference
1.22 / T
.1141 / p-value (one-tailed, upper)
Conclusion: Since the p-value (0.1141) > 0.05, we fail to reject H0.There is insufficient evidence to conclude at 5% level of significance that students who finish an exam first get better grades.
(c) Is the difference in mean scores large enough to be important?
No. It is not large enough to be important. (Note that we have failed to reject the null hypothesis.)
(d) Is it reasonable to assume equal variances?
The variances appear to vary reasonably with the means and sample sizes. it is reasonable to assume equal variances.
(e) Carry out a formal test for equal variances at α = .05, showing all steps clearly.
Hypotheses: H0: 1^2 = 2^2 vs Ha: 1^2 ≠ 2^2
Level of Significance: a = 5%
Decision Rule: Reject the null hypothesis if p-value < 0.05.
F = s1^2 / s2^2 = 19.6^2 / 24.9^2 = 0.6196
With dof (numerator) = 24 and dof (denominator) = 23, and corresponding to F = 0.6196, the
p-value = 0.8744.
Conclusion: Since the 0.8744 > 0.05, we fail to reject H0. There is insufficient evidence to conclude at 5% level of significance that there is any difference in the variances.
10.56
A sample of 25 concession stand purchases at the October 22 matinee Bride of Chucky showed a mean purchase of $5.29 with a standard deviation of $2.14. The mean appears to be very close but not the variances. At a=.05 is there a difference in variances?
(a) Hypotheses: H0: 1^2 = 2^2 vs Ha: 1^2 ≠ 2^2(b) Level of Significance: = 5%
(c) Decision Rule: Reject the null hypothesis if p-value < 0.05.
(d) Test characteristic: F = s1^2 /s2^2 = 3.02^2 / 2.14^2 = 1.9915
For dof (numerator) = dof(denominator) = 24 and corresponding to F- value = 1.9915,
the p- value = 0.049057
(e) Conclusion: 0.049057 is just less than 0.05, we reject H0. There is just about enough statistical evidence to conclude at 5% level of significance that there is a difference in the variances.