In Dallas, some fire trucks were painted yellow (instead of red) to heighten their visibility. During a test period, the fleet of red fire trucks made 153,348 runs and had 20 accidents, while the fleet of yellow fire trucks made 135,035 runs and had 4 accidents. At a = .01, did the yellow fire trucks have a significantly lower accident rate? (a) State the hypotheses. (b) State the decision rule and sketch it. (c) Find the sample proportions and z test statistic. (d) Make a decision. (e) Find the p-value and interpret it. (f ) If statistically significant, do you think the difference is large enough to be important? If so, to whom, and why? (g) Is the normality assumption fulfilled? Explain.
(a) H0: p1 = p2 vs. Ha: p1 > p2(b) Reject the null hypothesis if z > z0.01 = 2.3263
(c) Sample proportions and z test statistic:
p1 / p2 / pc
0.00013 / 0.00003 / 8.32E-05 / p (as decimal)
0.00013 / 0.00003 / 8.32E-05 / p (as fraction)
19.93524 / 4.05105 / 23.98629 / X
153348 / 135035 / 288383 / n
0.0001 / difference
0 / hypothesized difference
3.4E-05 / std. error
2.938318 / z
(d) Since z > z0.01 (2.938 > 2.3263), we reject H0
(e) p-value = 0.0017 Þ The probability that z can be as extreme as the test statistic value is 0.0014.
(f) Yes (because the p-value is very low compared to alpha = 0.01). This means the yellow fire trucks have a significantly lower accident rate than the red fire trucks.
(g) Since the sample sizes are very large, the normality assumptions are fulfilled.
Does lovastatin (a cholesterol-lowering drug) reduce the risk of heart attack? In a Texas study, researchers gave lovastatin to 2,325 people and an inactive substitute to 2,081 people (average age 58). After 5 years, 57 of the lovastatin group had suffered a heart attack, compared with 97 for the inactive pill. (a) State the appropriate hypotheses. (b) Obtain a test 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? (Data are from Science News 153 [May 30, 1998], p. 343.)
(a) H0: p1 = p2 vs. Ha: p1 < p2(b) Hypothesis test for two independent proportions:
p1 / p2 / pc
0.0282 / 0.0466 / 0.0369 / p (as decimal)
65/2325 / 97/2081 / 162/4406 / p (as fraction)
65.449 / 96.995 / 162.444 / X
2325 / 2081 / 4406 / n
-0.0185 / difference
0. / hypothesized difference
0.0057 / std. error
-3.25 / z
.0012 / p-value (two-tailed)
-0.0333 / confidence interval 99.% lower
-0.0036 / confidence interval 99.% upper
0.0148 / half-width
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.
To test the hypothesis that students who finish an exam first get better grades, Professor Hardtack kept track of the order in which papers were handed in. The first 25 papers showed a 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 hypotheses for a right-tailed test. (b) Obtain a test statistic and p-value assuming equal variances. Interpret these results. (c) Is the difference in 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) Hypotheses: H0: m1 = m2 vs Ha: m1 > m2(b)
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)No. It is not large enough to be important. (Note that we have failed to reject the null hypothesis.)
(d) The variances appear to vary reasonably with the means and sample sizes. \ it is reasonable to assume equal variances.
(e)Hypotheses: H0: s1^2 = s2^2 vs Ha: s1^2 ≠ s2^2
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.
A sample of 25 concession stand purchases at the October 22 matinee of Bride of Chucky showed a mean purchase of $5.29 with a standard deviation of $3.02. For the October 26 evening showing of the same movie, for a sample of 25 purchases the mean was $5.12 with a standard deviation of $2.14. The means appear to be very close, but not the variances. At a = .05, is there a difference in variances? Show all steps clearly, including an illustration of the decision rule.
(a) Hypotheses: H0: s1^2 = s2^2 vs Ha: s1^2 ≠ s2^2(b) Level of Significance: a = 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 = 0.0491
(e) Conclusion: 0.0491 is 0.05, we reject H0 and accept Ha. There is just about enough evidence to conclude at 5% level of significance that there is a difference in the variances.