1. Section 6-1 – Confidence Intervals for the Mean (Large Samples) Estimating Population Parameters

2. VOCABULARY: Point Estimate – a single value estimate for a population parameter. The most unbiased point estimate of the population parameter is the sample mean. Interval Estimate -An interval, or range of values, used to estimate a population parameter. Level of Confidence -Denoted as c, it is the probability that the interval estimate contains the population parameter. Margin of Error -Sometimes also called the maximum error of estimate, or error tolerance. It is denoted as E, and is the greatest possible distance between the point estimate and the value of the parameter it is estimating. c-confidence interval -Is found by adding and subtracting E from the sample mean. The probability that the confidence interval contains µ is c.

6. 5)Select Calculate to get the interval. We can be 99% sure that the actual population mean is between 10.575 and 14.225. Notice that the calculator also tells us that the mean of the data we entered is 12.4, that the standard deviation of the data was 5.01 and that n was 50. If we need to know what E is, simply find the distance between the interval endpoints and divide by 2. (14.225 – 10.575)/2 = 1.825.

9. CALCULATING MINIMUM SAMPLE SIZE If you want to be 95% certain that the true population mean lies within the interval created with an E of 1, you need AT LEAST 97 magazine advertisements in your sample. We round up, since 96 advertisements are not quite enough.

10. ASSIGNMENTS Classwork:Pages 317-318; #2-34 Evens Homework:Pages 317-323; #35-40 All, #45-67 Odd

12. Properties of the t-distribution (Page 325) 1)The t-distribution is bell-shaped and symmetrical about the mean. 2)Degrees of freedom are equal to n-1. 3)The total area under the curve is 1, or 100%. 4)The mean, median, and mode of the t-distribution are equal to zero. 5)As the degrees of freedom increase, the t-distribution approaches the normal distribution. After 30 d.f., the t-distribution is very close to the standard normal z- distribution. Close enough, in fact, that we use the standard normal distribution for d.f. ≥ 30. We just did that in Section 6-1.

17. Now, find the 90% and 95% confidence intervals for the population mean mortgage interest rate. What happens to the widths of the intervals as the confidence levels change? There is a flow chart on how to decide which distribution to use (t or normal) on page 329. Study this!!

18. Is n ≥ 30? YES NO Is the population normally, or approximately normally, distributed? NO You can NOT use the normal or the t-distribution. YES Is σ known? NO YES

19. EXAMPLE 4 (Page 329) You randomly select 25 newly constructed houses. The sample mean construction cost is $181,000 and the population standard deviation is $28,000. Assuming construction costs are normally distributed, should you use the normal distribution, the t-distribution, or neither to construct a 95% confidence interval for the population mean construction costs? Explain your reasoning. Although n is less than 30, we can still use the normal distribution because we know that the distribution is normal and we know what σ is.

20. You randomly select 18 adult male athletes and measure the resting heart rate of each. The sample mean heart rate is 64 beats per minute with a sample standard deviation of 2.5 beats per minute. Assuming the heart rates are normally distributed, should you use the normal distribution, the t-distribution, or neither to construct a 90% confidence interval for the mean heart rate? Explain your reasoning. Because n < 30, the distribution is normal, and we do not know what σ is, we should use the t-distribution on this one.

21. ASSIGNMENTS Classwork:Page 330; #1-16 All Homework:Pages 331-332; #17-28

26. .29, or 29%.1799, or 17.99%

31. ASSIGNMENTS Classwork:Page 339-340; #1-20 All Homework:Pages 340-342; #21-28 All

39. Section 6-4 – Confidence Intervals for Variance and Standard Deviation EXAMPLE 2 (Page 347) You randomly select and weigh 30 samples of an allergy medicine. The sample standard deviation is 1.2 milligrams. Assuming the weights are normally distributed, construct 99% confidence intervals for the population variance and standard deviation. d.f. = 29. Look in Table 6 in Appendix B (or on the hand-out I gave you) Find the row for the d.f. = 29 and the columns for.005 and.995.

40. Section 6-4 – Confidence Intervals for Variance and Standard Deviation EXAMPLE 2 (Page 347) Now that you have the interval for the variance, take the square roots of the endpoints to find the interval for the standard deviation. We are 99% confident that the actual population variance of weights of allergy medicines is between 0.8 and 3.2, and that the actual population standard deviation of weights of allergy medicines is between 0.9 and 1.8.

41. Section 6-4 – Confidence Intervals for Variance and Standard Deviation EXAMPLE 2 (Page 347) Find the 90% and 95% confidence intervals for the population variance and standard deviation of the medicine weights. d.f. = 29

42. Section 6-4 – Confidence Intervals for Variance and Standard Deviation EXAMPLE 2 (Page 347) Find the 90% and 95% confidence intervals for the population variance and standard deviation of the medicine weights. d.f. = 29

43. ASSIGNMENTS Classwork:Page 348; #1-6 All Homework:Pages 348-350; #7-20 All