13.2 We will test the hypotheses
The proportions of the different marital statuses for 25 to 29 year old males in 2000 are the same as for
the general population as given in the table.
At least one of the proportions for the males is different.
The expected counts for the 25-29 year old males is
Note that the expected counts are all greater than 5 and that our sample was random.
The test statistic, X 2=161.77 with 3 df. This is extremely large (well beyond 17.73) in our table. With such a low p-value ( ≈0 ) we have strong statistical evidence against H 0 . That is, we have strong evidence to support the claim that the distribution of proportions of marital status was different for 25-29 yr old males in 2000 than for the general population.
H0 : The 1996 proportions of PhD’s among the various ethnic groups are the same as in 1980.
HA : At least one of the proportions is different.
All expected counts are greater than 1, however, 2 of the 6 = 33% of the expected counts are less than 5. This means that our statistic will not necessarily be close enough to chi-square distributed for our conclusions to be reliable. Therefore, we proceed with caution. X2 = 60.02852. The p-value for the test is less than 0.005 from the table. A p-value this low would mean that we have very significant evidence against the null. So we conclude that we have evidence that the 1996 proportions of PhD’s are different from the 1981 distribution.
(b) We can see that the larges contributor to the test statistic is the difference in the proportion of non-resident aliens being awarded PhD’s. There were 27% awarded in 1996 to this group, but only 12.8% in 1981
(a) The two-way table is shown here. Successes Failures
(b) The proportions of successes in each of the four categories are
Proportions of Successes
It appears that the Patch plus drug had the
highest success rate with the drug alone
(d) The null hypothesis says that there is no difference in the success rates of the various treatments.
(e) Here are the expected counts with the differences of the Observed and Expected Counts
(f) We notice that the O − E values for the successes follow the same sort of pattern as the bar graph
(a) The 8 terms of the X2 statititic are:
(b) From table E, the p-value < 0.0005, which is significiant. This means that we have evidence that the success rates are not the same for all of the various treatments.
(c) The term that contributes most to the statistic is 10.076, the (O − E)2/E for the successes of the patch
plus drug. This is not surprising but consistent with our descriptions in exercise 13.14.
(d) First we conclude that the proportions of success in quitting smoking are not the same for the three treatments. Our analysis shows that people who use both the patch and the new drug have a higher chance for success.
(a) Let p1 = the proportion of the gastric freezing patients that improved, and p2 = the proportion of the placebo group that improved. We test the hypotheses
Treatment Improved Not Improved Treatment Improved Not Improved
(c) There is no significant evidence to support the claim that the gastric freezing treatment has a different proportion of improved patents than a placebo.
We observe from the graphs that there may be a slight, if any, positive association between alcohol consumption and nicotine use in these mothers. It appears that among the mothers who did not use alcohol, we have the highest non-nicotine use rate. The highest nicotine usage occurred in the 0.11-0.99 mg/day group, however, not the 1.00 oz or more group.
H0: There is no association between Alcohol consumption and Nicotine use in pregnant women. HA: There is an association.
Our observed data is summarized in the two-way table
Nicotine (mg/day) Alcohol (oz/day)
Our expected counts (assuming the null hypothesis is true):
Nicotine 16 or more
All expected counts are >5, so we proceed with the chi-square test of association.
The low p-value of our test indicates that there is strong evidence against the null hypothesis. That is, we have strong evidence that there is an association between alcohol consumption and nicotine use in women.
We wish to test if the proportion of rats that develop tumors who can control the shock p1, is less than the proportion of rats that develop tumors who cannot control the shock, p2. The hypotheses are
We can use the chi-square test to get the same result as a 2-proportion z-test when the test is two-sided. Since this test is a one-sided test, we must use the z-test.
The 2-prop z-test gives z = -2.854, p = 0.002. We have significant evidence to conclude that the “happy rats” developed fewer tumors.
We will perform a chi-square goodness of fit test to test the hypotheses:
H0: The proportions of green seeded plants, p1 = ¾, and the proportion of yellow seeded plants, p2 = ¼ . Ha: The proportions are different.
The expected counts from our sample of n = 880 are ¾(880) = 660 green-seeded and ¼(880) = 220 yellow seeded.
Since both expected counts are >5, we proceed. X2 = 2.6727. Our p-value = P(χ 2 > .
Since we would obtain proportions about this size of green and yellow-seeded plants about 10% of the time if the null hypothesis is true, we do not have very significant evidence to doubt (reject) the hypothesized model.
Shown here is the two-way table for the number of strokes per treatment. Also shown is a bar graph comparing the number of strokes for each treatment. It appears that Aspirin and Dipyridamole have about the same “success” rate, but that both treatments together do a better job preventing strokes. Treatment Number of Strokes No Stroke Dipyridamole
We will test whether the differences in the proportions are
statistically significant. That is we test the hypotheses
H0: The proportions of Strokes for each treatment are all the same. Ha: At least one of the proportions is different.
The two-way table of the expected counts, and the corresponding terms of the chi-square statistic:
Treatment Number of Strokes No Stroke Dipyridamole Treatment Number of Strokes No Stroke Dipyridamole
The terms of our statistic that contribute most to the total value are the 9.486 (for the strokes with placebo) and 11.63 (corresponding to strokes with both treatments).
We repeat the analysis without all of the words—just the graphs and test. Treatment Number of Deaths Number of Survivors Dipyridamole Treatment Number of Deaths Number of Survivors Dipyridamole Treatment Number of Deaths Number of Survivors Dipyridamole
We note that there is significant evidence that the proportions of strokes among the treatments is different, but not that the proportions of deaths are different from treatment to treatment. From our bar graphs, and our inferences, we might conclude that there is significant evidence that using aspirin and Dipyridamole may decrease the number of strokes in stroke patients, but has no significant effect on the survival rate of stroke patients.
(a) We will test whether p1, the proportion of men who die in such situations, is greater than p2, the proportion of women that die: H : p = p
We consider the sample from the Titanic a random sample from their peers-- that is, a random
sample of men, and a random sample of women in such situations.
We note that 680, 168, 126 and 317 are all greater than 5, so we proceed.
The TI-84 2-PropZTest gives z = 18.226 and p-value ≈0 . There is strong statistical evidence
that a higher proportion of men die in such situations than women. It is a man's responsibility to protect a woman, even to the point of dying in her place.
(b) Observe a bar graph of the percents of women who died among the three economic classes.
By far, most of the women who died were of
We will now test whether the propotions of
women who died differ among the economic classes. Ha: At least one of these propotions is
The expected counts from the two-way table are:
We note that all of these counts are greater than 5, so we proceed.
The χ2-test gives X2 = 103.767, with a p-value of 2.93×10−23 . This very low p-value is very significant evidence against the null hypothesis. There is strong statistical evidence that there is a difference in the proportions of women who died according to their social status.
(c) The bar graph comparing percents of men by status
that died is shown to the right. It appears that middle
class men fared the worst-- the greatest proportion of
The χ2-test for men gives X2 = 34.62, with a p-value of
×10−8 . This test leads to a similar conclusion
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