Review / Sample Test ST 540

Review / Sample test ST 540

Chapter 6 and chapter 7

Point estimation, Interval estimation and Testing of hypothesis based on a single sample:
Answer all the review questions printed in green at the end of each section.
Write the formal statement of Central Limit Theorem.
What is the difference between normal distribution and t-distribution, when used as sampling distribution?
Compare them with respect to shape, parameters, estimation of those parameters, asymptotic behavior ( as sample size becomes large).
Write a short note on interval estimation based on following points
What width is preferable?
What issues affect width favorable/ unfavorably
How do we get the best interval estimate.
How do we interprete the interval estimate.
How do u connect the interval estimate with testing of hypothesis
What is the meaning of confidence level.
Chi-square is a sampling distribution, write a short note of chi square distribution, its range, shape, parameters asymptotic behavior, its application in different scenario, underlying assumptions.
Know about one sided confidence intervals.
Solve example 6.62 in the textbook.
Solution to the problem 6.5 and 6.6 of text book (page 217)
for normal men and for men with chronic airflow limitation.

It means that the distribution of mean triceps skin-fold thickness from repeated samples of size 40 drawn from the population of normal men can be considered to be normal with mean and variance . A similar statement holds for men with chronic airflow limitation.

Solution to the problems 6.13-6.21

13. We have that days. Therefore, a 95% confidence interval for is given by

14The 95% confidence interval is computed from . We have that , . There-fore, we have the following 95% confidence interval

15A 90% confidence interval is given by

16The 90% confidence interval should be shorter than the 95% confidence interval, since we are requiring less confidence. This is indeed the case.

17We refer to Table 6. The lower 2.5th percentile is 0.0506 and is denoted by The upper 2.5th percentile is 7.38 and is denoted by .

18Our best estimate is given by . Because is unbiased and has minimum variance among all estimators of p.

19The standard error = .

20Since , we can use the normal theory method. Therefore, a 95% confidence interval for the percentage of males discharged from Pennsylvania hospitals is given by

21There are 16 people after excluding women of childbearing age. Of these 16 people, 4 received a bacterial culture while in the hospital. Thus, the best point estimate

Solution to 6.41. .42

41We assume that , where , are unknown, and find that , . Thus, a two-sided 95% confidence interval for the mean is given by

42A two-sided 99% confidence interval for the unknown variance is given by

In chapter 7 you should be able to write a short note on type one and type errors describing their definitions, their interrelationship, effect of different issues in testing on type one an type two error. Their behavior with respect to values in null and alternative hypothesis. How do we control them in our decision making process.
Solutions to 7.39-7.46

.39 We wish to test the hypothesis versus , where true mean daily iron intake for 911-year-old boys below the poverty level and true mean daily iron intake for 911-year-old boys in the general population.

.40We must use a one-sample t test. We reject if , or where , and accept otherwise. We have that , , , , . Therefore,

The critical values are and . Since , it follows that we reject at the 5% level. We conclude that 911-year-old boys below the poverty level have a significantly lower mean iron intake than comparably aged boys in the general population.

41To obtain the p-value, we must compute 2x (t0 >.2917). Since , and , if we had 40 df, then or . Similarly, since , and , if we had 60 df, it would also follow that . Therefore, since we actually have 50 df, and we reach the same conclusion with either 40 or 60 df, it follows that .

42The hypotheses to be tested are versus where  underlying variance in low-income population, underlying variance in the general population.

43We use a one-sample chi-square test. We reject if or .

We have , , , . Thus,

The critical values are and . Since it follows that we accept at the 5% level and conclude that there is no significant difference between the variance of iron intake for the low-income population and the general population.

.44Since , and , it follows that or .

45A 95% confidence interval for the underlying variance is given by

)

Since this confidence interval contains , we conclude that the underlying variance of the low-income population is not significantly different from that of the general population.

46The inferences made with the hypothesis-testing approach in Problems 7.43 and 7.44 are the same as those made with the CI approach in Problem 7.45, viz. there is no significant difference between the variance of iron intake for the low income population and the variance of the general population.