are±zα/2. Use Table II to find the critical values.
Becauseα=0.05, we find from Table II (or Table 9.4) the critical values of ±z0.05/2= ±z0.025= ±1.96, as shown in Fig. 9.17A.
FIGURE 9.17A
−1.96 0 z
1.96 Reject H0 Do not reject H0
0.025 0.025
Reject H0
Step 4 Use Table II to obtain theP-value.
From Step 3, the value of the test statistic isz= −0.88.
The test is two tailed, so the P-value is the proba- bility of observing a value of z of 0.88 or greater in magnitude if the null hypothesis is true. That proba- bility equals the shaded area in Fig. 9.17B, which, by Table II, is 2ã0.1894 or 0.3788. Hence P=0.3788.
FIGURE 9.17B
0 z
z = −0.88 0.88 P- value
CRITICAL-VALUE APPROACH OR P-VALUE APPROACH Step 5 If the value of the test statistic falls in the
rejection region, rejectH0; otherwise, do not rejectH0. The value of the test statistic from Step 3 isz= −0.88.
Figure 9.17A reveals that this value does not fall in the rejection region, so we do not reject H0. The test results are not statistically significant at the 5% level.
Step 5 If P≤α, reject H0; otherwise, do not rejectH0.
From Step 4, P=0.3788. Because the P-value ex- ceeds the specified significance level of 0.05, we do not reject H0. The test results are not statistically significant at the 5% level and (see Table 9.8 on page 408) provide at most weak evidence against the null hypothesis.
Step 6 Interpret the results of the hypothesis test.
Interpretation At the 5% significance level, the (unabridged) data do not provide sufficient evidence to conclude that the mean top speed of all cheetahs differs from 60 mph.
We have now completed the hypothesis test, using all 35 top speeds in Table 9.11. However, recall that the top speed of 75.3 mph is an outlier. Although in this case, we don’t know whether removing this outlier is justified (a common situation), we can still remove it from the sample data and assess the effect on the hypothesis test. With the outlier removed, we determined that the value of the test statistic isz= −1.71.
CRITICAL-VALUE APPROACH OR P-VALUE APPROACH We see from Fig. 9.17A that the value of the test statis-
tic, z= −1.71, for the abridged data does not fall in the rejection region (although it is much closer to the rejection region than the value of the test statistic for the unabridged data, z= −0.88). Hence we do not re- jectH0. The test results are not statistically significant at the 5% level.
For the abridged data, the P-value is the probability of observing a value of z of 1.71 or greater in magnitude if the null hypothesis is true. Referring to Table II, we find that probability to be 2ã0.0436, or 0.0872. Hence
P=0.0872.
Because the P-value exceeds the specified signifi- cance level of 0.05, we do not reject H0. The test re- sults are not statistically significant at the 5% level but, as we see from Table 9.8 on page 408, the abridged data do provide moderate evidence against the null hypothesis.
Interpretation At the 5% significance level, the (abridged) data do not provide sufficient evidence to conclude that the mean top speed of all cheetahs differs from 60 mph. Thus, we see that removing the outlier does not affect the conclusion of this hypothesis test.
Exercise 9.83 on page 419
Statistical Significance Versus Practical Significance
Recall that the results of a hypothesis test are statistically significantif the null hy- pothesis is rejected at the chosen level ofα. Statistical significance means that the data provide sufficient evidence to conclude that the truth is different from the stated null hypothesis. However, it does not necessarily mean that the difference is important in any practical sense.
For example, the manufacturer of a new car, the Orion, claims that a typical car gets 26 miles per gallon. We think that the gas mileage is less. To test our suspicion, we perform the hypothesis test
H0:μ=26 mpg (manufacturer’s claim) Ha:μ <26 mpg (our suspicion), whereμis the mean gas mileage of all Orions.
9.4 Hypothesis Tests for One Population Mean WhenσIs Known 417 We take a random sample of 1000 Orions and find that their mean gas mileage is 25.9 mpg. Assuming σ =1.4 mpg, the value of the test statistic for a z-test is z= −2.26. This result is statistically significant at the 5% level. Thus, at the 5% sig- nificance level, we reject the manufacturer’s claim.
Because the sample size, 1000, is so large, the sample mean, ¯x =25.9 mpg, is probably nearly the same as the population mean. As a result, we rejected the manu- facturer’s claim becauseμis about 25.9 mpg instead of 26 mpg. From a practical point of view, however, the difference between 25.9 mpg and 26 mpg is not important.
? What Does It Mean?
Statistical significance does not necessarily imply practical significance!
The Relation between Hypothesis Tests and Confidence Intervals
Hypothesis tests and confidence intervals are closely related. Consider, for example, a two-tailed hypothesis test for a population mean at the significance levelα. In this case, the null hypothesis will be rejected if and only if the valueμ0given for the mean in the null hypothesis lies outside the (1−α)-level confidence interval forμ. You can examine the relation between hypothesis tests and confidence intervals in greater detail in Exercises 9.95–9.97.
THE TECHNOLOGY CENTER
Most statistical technologies have programs that automatically perform a one-meanz-test.
In this subsection, we present output and step-by-step instructions for such programs.
EXAMPLE 9.14 Using Technology to Conduct a One-Mean z-Test
Poverty and Dietary Calcium Table 9.10 on page 413 shows the daily calcium intakes for a simple random sample of 18 adults with incomes below the poverty level. Use Minitab, Excel, or the TI-83/84 Plus to decide, at the 5% significance level, whether the data provide sufficient evidence to conclude that the mean calcium intake of all adults with incomes below the poverty level is less than the RAI of 1000 mg per day. Assume thatσ =188 mg.
Solution Letμdenote the mean calcium intake (per day) of all adults with incomes below the poverty level. We want to perform the hypothesis test
H0:μ=1000 mg (mean calcium intake is not less than the RAI) Ha:μ <1000 mg (mean calcium intake is less than the RAI)
at the 5% significance level (α=0.05). Note that the hypothesis test is left tailed.
We applied the one-meanz-test programs to the data, resulting in Output 9.1.
Steps for generating that output are presented in Instructions 9.1 on the following page.Note to Excel users:For brevity, we have presented only the essential portions of the actual output.
MINITAB
OUTPUT 9.1 One-meanz-test on the sample of calcium intakes
TI-83/84 PLUS