Exercises questions answers

| Home | | Advanced Mathematics |

Chapter: Biostatistics for the Health Sciences: Tests of Hypotheses

Biostatistics for the Health Sciences: Tests of Hypotheses - Exercises questions answers


EXERCISES

9.1 The following terms were discussed in Chapter 9. Give definitions of them in your own words:

a.     Hypothesis test

b.     Null hypothesis

c.      Alternative hypothesis

d.     Type I error

e.      Type II error

f.       p-value

g.     Critical region

h.     Power of a test

i.       Power function

j.       Test statistic

k.     Significance level

9.2 Chapters 8 and 9 discussed methods for calculating confidence intervals and testing hypotheses, respectively. In what manner are parameter estimation and hypothesis testing similar to one another? In what manner are they different from one another?

9.3 In a factory where he conducted a research study, an occupational medicine physician found that the mean blood lead level of clerical workers was 11.2. State the null and alternative hypotheses for testing that the population mean blood lead level is equal to 11.2. What is the name for this type of hypothesis test?

9.4 Using the data from Exercise 9.3, state the hypothesis set (null and alternative hypotheses) for testing whether the population mean blood lead level exceeds 11.2. What is the name for this type of hypothesis test?

9.5 In the example cited in Exercise 9.3, the physician measures the blood lead levels of smelter workers in the same factory and finds their mean blood lead level to be 15.3. State the hypothesis set (null and alternative hypotheses) for testing whether the mean blood lead level of clerical workers differs from that of smelter workers.

9.6 Using the data from Exercise 9.5, state the hypothesis set (null and alternative hypotheses) for testing whether the mean blood lead level of smelter workers exceeds that of clerical workers.

9.7 The Orange County Public Health Department was concerned that the mean daily fecal coliform level in a particular month at Huntington Beach, Califor-nia, exceeded a safe level. Let us call this level “a.” State the appropriate hy-pothesis set (null and alternative) for testing whether the mean coliform level exceeded a safe standard.

9.8 Suppose we would like to test the hypothesis that mean cholesterol levels of residents of Kalamazoo and Ann Arbor, Michigan, are the same. We know that both populations have the same variance. State the appropriate hypothe-sis set (null and alternative). What test statistic should be used?

9.9 Consider a sample of size 5 from a normal population with a variance of 5 and a mean of zero under the null hypothesis. Find the critical values for a 0.05 two-sided significance test of the mean equals zero versus the mean dif-fers from zero.

9.10 Use the test in Exercise 9.9 (i.e., critical values) to determine the power of the test when the mean is 1.0 under the alternative hypothesis, the variance is 5, and the sample size is 5.

9.11 Again use the test in Exercise 9.9 to determine the power when the mean is 1.5 under the alternative hypothesis and the variance is again 5.

9.12 We suspect that the average fasting blood sugar level of Mexican Americans is 108. A random sample of 225 clinic patients (all Mexican American) yields a mean blood sugar level of 119 (S2 = 100). Test the hypothesis that μ = 108.

a.     What is the hypothesis set for a two-tailed test?

b.     Find the estimated s.e.m.

c.      Find the Z statistic.

d.     What decision should we make, i.e., reject or fail to reject H0 at the α = 0.05 level; reject or fail to reject H0 at the α = 0.01 level?

e.      What type of test is this: exact or approximate?

9.13 In the previous exercise there were two possible outcomes; reject the null hy-pothesis or fail to reject the null hypothesis. Explain in your own words what is meant by these outcomes.

9.14 Test the hypothesis that a normally distributed population has a mean blood glucose level of 100 ( σ2 = 100). Suppose we select a random sample of 30 in-dividuals from this population ( = 98.1, S2 = 126).

a.     What is the hypothesis set (null and alternative) for a two-tailed test?

b.     Find the estimated s.e.m.

c.      Find the Z statistic.

d.     What decision should we make, i.e., reject or fail to reject H0 at the α = 0.05 level; reject or fail to reject H0 at the α = 0.01 level?

e.      What type of test is preferable to run in this situation, exact or approximate? Explain your answer.

9.15 Describe the differences between a one-tailed and a two-tailed test. Give ex-amples of when it would be appropriate to use a two-tailed test and when it would be appropriate to use a one-tailed test.

9.16 Redo Exercise 9.14 but use a one-tailed (left-tail) test.

9.17 Recent advances in DNA testing have helped to confirm guilt or innocence in many well-publicized criminal cases. Let us consider the DNA test results to be the gold standard of guilt or innocence and a jury trial to be the test of a hypothesis. What types of errors are committed in the following two situations?

a. The jury convicts a person of murder who later is found to be innocent by DNA testing.

b. The jury exonerates a person who later is found to be guilty by DNA test-ing.

9.18 Find the area under the t-distribution between zero and the following values:

a.     2.62 with 14 degrees of freedom

b.     –2.85 with 20 degrees of freedom

c.      3.36 with 8 degrees of freedom

d.     2.04 with 30 degrees of freedom

e.      –2.90 with 17 degrees of freedom

f.       2.58 with 1000 degrees of freedom

9.19 Find the critical values for t that correspond to the following:

a.     n = 12, α = 0.05 one-tailed (right)

b.     n = 12, α = 0.01 one-tailed (right)

c.      n = 19, α = 0.05 one-tailed (left)

d.     n = 19, α = 0.05 two-tailed

e.      n = 28, α = 0.05 one-tailed (left)

f.       n = 41, α = 0.05 two-tailed

g.     n = 8, α = 0.10 two-tailed

h.     n = 201, α = 0.001 two-tailed

9.20 Consider the paired t test that was used with the data in Table 9.1, what would the power of the test be if the alternative is that the mean temperature differs by 3 degrees between the cities? What is the power at a difference of 5 degrees? Why does the power depend on the assumed true difference in means?

9.21 Suppose you are planning another experiment like the one in Exercise 9.20. Based on that data: (1) you are willing to assume that the standard deviation of the difference in means is 1.5°F, and (2) you anticipate that the average temperature in New York tends to be 3°F lower than the corresponding tem-perature in Washington on the same day.

a. For such a one-sided paired t-test test, how many test days do you need to obtain 95% power at the specified alternative?

b. How many do you need for 99% power?

c. How many do you need for 80% power?

9.22 What is a meta-analysis? Why are meta-analyses performed?

9.23 What is Bayes’ theorem? Define prior distribution. What is a posterior distri-bution?

9.24 How do Bayesians treat parameters? How do frequentists treat parameters? Are the two approaches different from one another?

9.25 Why can missing data be a problem in data analysis? What is imputation?

9.26 Define sensitivity and specificity. How do they relate to the type I and type II errors in hypothesis testing?

9.27 Here are some questions about hypothesis testing:

a.     Describe the one sample test of a mean when the variance is unknown and when the variance is known.

b.     Describe the use of a two-sample t test (common variance estimate).

c.      Describe when it is appropriate to use a paired t test.

Answers:

9.3 H0: The mean μ = 11.2 versus the alternative HA: The mean μ 11.2. This is a two-sided test.

9.5 H0: The mean difference μ1 μ2 = 0 versus the alternative HA: The mean difference μ1μ2 0.

9.9 The sample size is 5, the population variance is known to be 5, and the data is normally distributed. Under the null hypothesis, the mean is 0. We want to find the critical value C such that P(–C < X < C) = 0.95, where X is the sample mean. Under the null hypothesis, Z = X/(5/5) = X since the standard deviation is 5 and the standard error of the mean is the standard deviation divided by n where the sam-ple size n is in this case 5. Since Z is standard normal and Z = X from the table, we see that C = 1.96.

9.10 In this case, the true mean is 1 and the critical value C is 1.96, as determined in Exercise 9.9. The power of the test is the probability that X > 1.96 or X < –1.96 under the alternative that the mean is 1 instead of 0. Under this alternative, X has a normal distribution with mean equal to 1 and standard error equal to 1. So under the alternative a standard normal Z = X – 1. P(X > 1.96) = P(Z > 0.96) = 0.5 – P(0 < Z < 0.96) = 0.5 – 0.3315 = 0.1685. Now P(X < –1.96) = P(Z < –2.96) = P(Z > 2.96) = 0.5 – P(0 < Z < 2.96) = 0.5 – 0.4985 = 0.0015. So the power of the test is 0.1685 + 0.0015 = 0.17.

9.11 In this case, the true mean is 1.5 and the critical value C is 1.96 as deter-mined in Exercise 9.9. The power of the test is the probability that X > 1.96 or X < –1.96 under the alternative that the mean is 1.5 instead of 0. Under this alternative, X has a normal distribution with mean equal to 1.5 and standard error equal to 1. So under the alternative a standard normal Z = X – 1.5. P(X > 1.96) = P(Z > 0.46) = 0.5 - P(0 < Z < 0.46) = 0.5 – 0.1772 = 0.3228. Now P(X < –1.96) = P(Z < 3.46) = P(Z > 3.46) = 0.5 – P(0 < Z < 3.46) < 0.5-0.4990 = 0.001. So the power of the test is approximately 0.3228 + 0.001 = 0.3238.

9.19 a. n = 12, a = 0.05 one-tailed to the right: t = 1.7939 (df = 11)

b. n = 12, a = 0.01 one-tailed to the right: t = 2.718 (df = 11)

c. n = 19, a = 0.05 one-tailed to the left: t = –1.7341 (df = 18)

d. n = 19, a = 0.05 two-tailed: t = –1.7341 and t = 1.7341 (df = 18)

e. n = 28, a = 0.05 one-tailed to the left: t = –1.7033 (df = 27)

f. n = 41, a = 0.05 two-tailed: t = –1.6839 and t = 1.6839 (df = 40)

g. n = 8, a = 0.10 two-tailed: t = –2.3646 and t = 2.3646 (df = 7)

h. n = 201, a = 0.001 two-tailed: t = –3.3400 and t = 3.3400 (df = 200)

9.22 A meta-analysis is a procedure for drawing statistical inference based on combining information from several independent studies. It is often done because studies are conducted that individually do not have sufficient power to reject a null hypothesis but several such studies could do so if their information could be pooled together. This can be done when the same or similar hypotheses are tested and the subjects are selected and analyzed in similar ways.

9.26 Sensitivity is the probability that the clinical test declares the patient as having the disease (a positive test result), given that he or she does in fact have the disease. If p is the sensitivity, 1 – p is the type II error since the null hypothesis is the hypothesis that the patient does not have the disease and 1 – p is the conditional probability of not declaring the patient to have the disease given that he does have it. Specificity is the probability that a clinical test declares the patient well (a nega-tive test result), given that he or she does not have the disease. If p is the specificity, 1 – p is the type I error since 1 – p is the conditional probability of declaring the pa-tient has the disease when he does not.

Contact Us, Privacy Policy, Terms and Compliant, DMCA Policy and Compliant

TH 2019 - 2023 pharmacy180.com; Developed by Therithal info.