P-value of 0.0101. (a) the in context. (b) What would you make if α = 0.05? If α = 0.01? Justify your answer. Correct Answer (a) If the mean score on the SSHA for older students at this school is really 115, there is a 1.01% chance of finding a sample of 45 older students with a mean score of at least 125.7. (b) Reject H0 if α = 0.05; fail to reject H0 if α = 0.01 since 0.01 < 0.0101 < 0.05. For the study of Jordanian children in , the sample mean hemoglobin level was 11.3 g/dl and the sample standard deviation was 1.6 g/dl. A significance test yields a P-value of 0.0016. (a) the in context. t t value value
drug for treating Alzheimer’s disease will experience nausea. To test this claim, a significance test is carried out of H0:p = 0.10 Ha:p < 0.10 You that the power of this test at the 5% significance level against the alternative p = 0.08 is 0.64. (a) Explain in simple language what “power = 0.64” means in this setting. (b) You could get higher power against the same alternative with the same α by changing the number of you make. Should
alumni to test the hypothesis H0: p = 0.99. Refer to . In Simon’s SRS, 16 of the students were (a) the test (b) Find the using . this result as an area a Correct Answer (a) z = 1.23 (b) 0.2186 to . For Yvonne’s survey, 96 students in the sample said they seldom or value
approve of the new parking arrangement. The principal cites this as evidence that the change was effective. Perform a test of the principal’s claim at the α = 0.05 significance level. Correct Answer State: H0: p = 0.37 versus Ha: p > 0.37, where p is the actual proportion of students who are satisfied with the parking situation. Plan: One-sample z test for p. Random: The sample was randomly selected. Normal: The expected number of successes np0 = 74 and failures (1 − p0) = 126 are both at least 10. Independent: There were 200 in the sample, and since there are 2500 students in the population, the sample is less than 10% of the population. Do: z = 1.32, P-value = 0.0934. Conclude: Since our P-value is greater than 0.05, we fail to reject the null hypothesis. We do not have enough evidence to conclude that the new parking arrangement increased student satisfaction with parking at this school. A drug manufacturer claims that less than 10% of patients who take its new drug for treating Alzheimer’s disease will experience nausea. To test this claim, researchers conduct an experiment. They give the new drug to a random sample of 300 out of 5000 Alzheimer’s patients whose families have
(b) at α = 0.05 but not at α = 0.01. (c) at α = 0.01 but not at α = 0.05. (d) at both α = 0.05 and α = 0.01. (e) because we don’t know the value of ôop. Correct Answer Which of the following 95% confidence intervals would lead us to reject H0: p = 0.30 in favor of Ha: p ≠ 0.30 at the 5% significance level? (a) (0.29, 0.38) (b) (0.19, 0.27) (c) (0.27, 0.31) (d) (0.24, 0.30)
Printed Page 586 SECTION 9.3 Summary Significance tests for the mean μ of a Normal population are based on the sampling distribution of the sample mean X. Due to the central limit theorem, the resulting procedures are approximately correct for other population distributions when the sample is large. If we somehow know σ, we can use a z test statistic and the standard Normal distribution to perform calculations. In practice, we typically do not know σ. Then, we use the one-sample statistic with P-values calculated from the distribution with − 1 degrees of freedom. The is approximately correct when Random The data were produced by random sampling or a randomized experiment. Normal The population distribution is Normal OR the sample size is large (n ≥ 30). Independent Individual observations are independent. When sampling without replacement, check that the population is at least 10 times as large as the sample. Confidence intervals provide additional information that significance do not—namely, a range of plausible values for the parameter μ. A two-sided test of H0: μ = μ0 at significance level α gives the same conclusion as a 100(1 − α)% confidence interval for μ. Analyze by first taking the difference within each pair to produce a single sample. Then use one-sample procedures. Very small differences can be highly significant (small P-value) when a test is based on a large sample. A statistically significant difference need not be practically important. Plot the data to display the difference you are seeking, and use confidence intervals to estimate the actual values of parameters. On the other hand, lack of significance does not imply that H0 is true. Even a large difference can fail to be significant when a test is based on a small sample. Significance tests are not always valid. Faulty data collection, outliers in the data, and other practical problems can invalidate a test. Many tests run at once will probably produce some significant results by chance alone, even if all the null hypotheses are true. 9
part (a) an of Ha: μ ≠ 5. Correct Answer (a) : 0.025 < < 0.05. Technology: the P-value is 0.043. Reject H0 at the 5% significance level. Fail to reject H0 at the 1% significance level. (b) : 0.05 < P-value < 0.10. Technology: the P-value is 0.086. Fail to reject H0 at both levels. The t statistic from a sample of = 25 for the test of H0: μ = 64 Ha: μ ≠ 64 has the value = −1.12. (a) the for this test (i) and (ii)
about their daily water intake. A graph of the data shows a roughly symmetric shape with no outliers. The figure below displays Minitab output for a one-sample t interval for the population mean. Is there convincing evidence at the 10% significance level that the blogger’s claim is incorrect? Use the confidence interval to justify your answer. The P-value for a two-sided test of the null hypothesis H0: μ = 10 is 0.06. (a) Does the 95% confidence interval for μ include 10? Why or why not? (b) Does the 90% confidence interval for μ include 10? Why or why not? Correct Answer (a) Yes, P-value > 0.05. (b) No, since P-value is < 0.10. The P-value for a one-sided test of the null hypothesis H0: μ = 15 is 0.03. (a) Does the 99% confidence interval for μ include 15? Why or why not? (b) Does the 95% confidence interval for μ include 15? Why or why not? Exercises 87 and 88 refer to the following setting. An understanding of cockroach biology may lead to an effective control strategy for these annoying insects. Researchers studying the absorption of sugar by insects feed a random
clearly Show your Exercises 107 and 108 refer to the Case Closed about “normal” body temperature on . (8.3) Check that the conditions are met for performing inference about the mean body temperature in the population of interest. Correct Answer Random: The adults measured were chosen at random. Normal: There were many more than 30 measurements in the sample. Independent: The sample size was 130, which is clearly less than 10% of the healthy 18- to 40-year-olds in the United States. The conditions are met.