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Hw 1 Prob 1 A sample of 20 cigarettes is tested to determine nicotine content and the average value observed was 1.2 mg. Compute a 99 percent two-sided confidence interval for the mean nicotine content of a cigarette if it is known that the standard deviation of a cigarette’s nicotine content is σ = 0.2 mg. 1 P{ X Z / 2 1 C . I . given by X z / 2 n 1.2 2.575 0.2 20 (1.08 1.32) n X Z / 2 n } Hw 1 Prob 2 • In Problem 1, suppose that the population variance is not known in advance of the experiment. If the sample variance is 0.04, compute a 99% two-sided confidence interval for the mean nicotine content. In problem 1, we knew = 0.2. Now we must estimate with the sample standard deviation s=.2. 1 P{ X t / 2 ,n1 s s X t / 2 ,n1 } n n 1 C . I . given by s X t / 2 ,n1 n 1.2 2.861 0.2 20 (1.07 1.33) Note: this is just slightly larger than C.I. in problem one because we have a fairly large sample. For large n, the difference between the normal and t distributions starts to disappear. Hw 1 Prob 3 • The daily dissolved oxygen concentration for a water stream has been recorded over 30 days. If the sample average of the 30 values is 2.5 mg/liter and the sample standard deviation is 2.12 mg/liter, determine a value which, with 90 percent confidence, exceeds the mean daily concentration. Since n=30, we could choose to use a t-distribution with 29 degrees of freedom or a standard normal distribution. The difference will be small. I will use the standard normal. For a standard normal a confidence interval is given by X z / 2 s n 1.2 1.645 02.12 30 (1.86 3.14) Note that the way the problem, we are more likely interested only in an upper limit. In this case, we could conduct a one sided confidence interval. In this case, we have X z s n 2.5 1.282 ( 2.996) 02.12 30 Hw 1 Prob 3 • The daily dissolved oxygen concentration for a water stream has been recorded over 30 days. If the sample average of the 30 values is 2.5 mg/liter and the sample standard deviation is 2.12 mg/liter, determine a value which, with 90 percent confidence, exceeds the mean daily concentration. Since n=30, we could choose to use a t-distribution with 29 degrees of freedom or a standard normal distribution. The difference will be small. I will use the standard normal. For a standard normal a confidence interval is given by X z / 2 s n 2.5 1.282 02.12 30 ( 2.996) If we do a one sided C.I. using the t distribution with 29 degrees of freedom, we get s X t ,n1 n 2.5 1.311 ( 3.007 ) 02.12 30 Hw 1 Prob 4 • Suppose that when sampling from a normal population having an unknown mean µ and unknown variance σ2, we wish to determine a sample size n so as to guarantee that the resulting 100(1 – α) percent confidence interval for µ will be of size no greater than A, for given values α and A. Explain how we can approximately do this by a double sampling scheme that first takes a subsample of size 30 and then chooses the total sample size by using the results of the first subsample. 1 C . I . given by s X t / 2 ,n1 n which means the precision is E t / 2 ,n1 s t s or n / 2 ,n1 n E 2 Since we do not know s until we collect data, we cannot know the t value to look up. Iterative solution 1. 2. 3. 4. 5. Estimate a sample size n Collect data and calculate s If we have achieved the desired precision, stop. If not, calculate a sample size based on formula above. Collect additional data and calculate new s. Go to step 3.