Section 6.2 Solutions

6.32

a. The null hypothesis is a statement about a population, expressed in terms a parameter. Furthermore, the parameter value is a fixed, not a random, quantity. Hence, testing if it will rain tomorrow doesn’t fit our definition of a null hypothesis.

b. The standard deviation of the sample mean is , not .

c. The null hypothesis is a statement about an unknown population parameter (for example, ), not about a sample statistic. Since is a sample statistic, it cannot be used as the basis of a null hypothesis.

6.33

a. The null hypothesis is a statement of “status-quo” or “no difference.” The alternative hypothesis includes the research statement. In this example, these two are turned around.

b. The null hypothesis is a statement about an unknown population parameter (for example, ), not about a sample statistic. Since is a sample statistic, it cannot be used as the basis of a null hypothesis.

c. The smaller the P-value is, the stronger the evidence is against the null hypothesis. For a given significance level , test results are statistically significant if the P-value . Furthermore, typical values of are 0.05 or 0.01 (i.e., small values). Hence, a P-value of 0.95 would never give statistically significant results.

6.37

a.

b.

6.38

I am not including a normal curve picture simply because I can’t create it in Word. A picture should definitely be a part of your solution, though.

Because the alternative hypothesis is two-sided, we need to not only consider the value of our test statistic and more extreme, but we also need to consider the negative value of our test statistic and more extreme (since ahead of time we weren’t sure which direction the difference would go).

P-value =

This is exactly the same as , which is the short-hand notation we use in class.

6.41

Our P-value is 0.032. Since P-value 0.05, our results are statistically significant at the 0.05 level (i.e., we have enough evidence to reject the null hypothesis at the 0.05 level). But since P-value > 0.01, our results are not significant at the 0.01 level (i.e., we do not have enough evidence to reject the null hypothesis at the 0.01 level).

6.50

Assuming the average math scores for all students in DC are the same for 2000 and 2003, there is less than a 5% chance of seeing the results from our sample or more extreme results. Hence, our data are very unlikely assuming the 2000 and 2003 averages are the same. That is, our data are not compatible with the 2000 and 2003 averages being the same, so we have strong evidence that the averages are, in fact, different.

6.54

In order to use the z-test, we must assume the distribution of new words is approximately normal (since the sample size is so small, the CLT will only apply if the original population is close to normal).

The test statistic is and the P-value is (Although a normal curve picture isn’t included with this solution, I encourage you to draw it.)

Assuming the mean number of new words is 8.9, there is a 0.102 chance of observing our sample mean, 10.2, or a greater sample mean. This is not surprising, so we have no reason to doubt the null hypothesis that the mean number of new words is 8.9. That is, we do not have evidence that these sonnets were written by a different author.

6.56

Since the sample size is so large (n = 160), we can use the z-test even though we aren’t told the population is normal (since the Central Limit Theorem tells us the distribution of is approximately normal).

a.

b. The test statistic is and the P-value is (Although a normal curve picture isn’t included with this solution, I encourage you to draw it.) Assuming the average calcium level for rural Guatemalan pregnant women is 9.5 mg/dl, there is only a 0.0272 chance of observing our sample mean, 9.57 mg/dl, or a more extreme sample mean. Because our data are so unlikely, we have evidence that the average calcium level for these women is different from 9.5 mg/dl. (Our results are statistically significant at the 0.05 level.)

c. A 95% confidence interval for the mean calcium level is (9.51 mg/dl, 9.63 mg/dl). Since the test results (of a two-sided test) are significant at the 0.05 level, we already knew this 95% confidence interval would not include the null-hypothesized value of 9.5 mg/dl. This is an example where we have statistical significance, but we may not have practical significance—that is, practically speaking, our sample mean is not significantly different from 9.5 mg/dl.

6.62

Since the sample size is so large (n = 100), we can use the z-test even though we know the population is not normal (since the Central Limit Theorem tells us the distribution of is approximately normal).

The hypotheses are .

The test statistic is and the P-value is (Although a normal curve picture isn’t included with this solution, I encourage you to draw it.)

Our P-value falls between 0.01 and 0.05. Hence, our results are statistically significant at the 0.05 level, but they are not statistically significant at the 0.01 level.

6.63

If results are statistically significant at the 1% level, then we have enough evidence to reject the null hypothesis (that is, P-value 0.01). Because the 1% level is more stringent than the 5% level (more evidence is required), the results will also be significant at the 5% level (P-value 0.01 means that P-value 0.05).

6.64

If test results are statistically significant at the 5% level, then P-value 0.05. But based only on this information, we cannot know if P-value 0.01, too. Hence, we cannot say whether the results are also significant at the 1% level.

6.71

In order to use the z-confidence interval, we must assume the distribution of all readings is close to normal (since the sample size is so small, the CLT will only apply if the original population is close to normal).

From the data, the sample mean is pCi/l.

a. A 95% confidence interval for the mean reading of all detectors is (99.04 pCi/l, 109.22 pCi/l).

b. The hypotheses are . Since the 95% confidence interval contains the null-hypothesized value of 105 pCi/l, we cannot reject the null hypothesis at the 0.05 level. Hence, at the 0.05 significance level, we do not have evidence that the mean reading of all detectors is different from 105 pCi/l.