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Section 6-1 β Confidence Intervals for the Mean (Large Samples) β’ Estimating Population Parameters VOCABULARY: Point Estimate β 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. FORMULAS: Margin of Error = π¬ = ππ ππ = ππ Ο π To use the formula, it is assumed that the population standard deviation is known. This is rarely the case, but when π β₯ 30, the sample standard deviation s can be used in place of Ο. s The formula effectively becomes π¬ = ππ π ππ = InvNorm of page 311) c-confidence interval sample standard deviation: 1βπ , 2 or 1 (1 2 β π) (explained why on bottom of πβπ¬<π<π+π¬ π= (πβπ₯)π πβπ GUIDELINES: 1)Find the sample statistics π and π₯. π is the sample size. π₯ is the sample mean. 2)Specify Ο, if known. Otherwise, if π β₯ 30, find the sample standard deviation s and use it as a point estimate for Ο. 3)Find the critical value π§π that corresponds to the given level of confidence. The three most commonly used confidence levels are 90%, 95%, and 99%. The corresponding π§π values are: 90% -- 1.645 95% -- 1.96 99% -- 2.575 It would be beneficial to memorize these. You will be using them a lot. s 4)Find the margin of error E. (π¬ = ππ ) π 5)Find the left and right endpoints and form the confidence interval. πβπ¬<π<π+π¬ Now the good news!! The TI-84 can help with this, too. We are going to walk through Example 4 on page 314, using the data points from Example 1 on page 310. 1)Enter the 50 data points into L1 on your calculator (STAT Edit). 2)STAT Calc 1 to find the sample standard deviation (we can use this because we have 50 data points; π β₯ 30). s = 5.01 2)STAT TESTS 7 (Z-Interval) 3)Select Data, since you have the data entered into the calculator. 4)Enter 5.01 as the standard deviation, and .99 as the C-Level (level of confidence). 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 β 10.575)/2 = 1.825. Look at Example 5 on page 315. n = 20, π₯ = 22.9, Ο = 1.5, and c = 90% STAT TESTS 7, select Stats (since you are going to provide the stats instead of the actual data points). Enter 1.5, 22.9, 20, and .9 and then calculate. We can be 90% certain that the actual mean age of all students currently enrolled in college is between 22.3 and 23.5. As a general rule, we round our interval endpoints to the same number of decimal points as the mean that is given to us. We were given 22.9, which is one decimal, so we rounded our interval to one decimal place. CALCULATING MINIMUM SAMPLE SIZE How do you know how many experiments or trials are needed in order to achieve the desired level of confidence for a given margin of error? We take the formula for finding E and solve it for n. s π Ο πΈ = ππ becomes π = ( π )2 . π πΈ Remember, if you donβt know what Ο is, you can use s, as long as π β₯ 30. Example 6 on page 316c = 95%, π§π = 1.96, π β π β 5.01 (from Example 1), E = 1 (given). ππ Ο 2 1.96(5.0) 2 9.8 ) =( ) = ( )2 = πΈ 1 1 π=( 9.82 = 96.04. 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. ASSIGNMENTS Classwork: Pages 317-318; #2-34 Evens Homework: Pages 317-323; #35-40 All, #45-67 Odd Section 6-2 β Confidence Intervals for the Mean (Small Samples) Estimating Population Parameters The t-distribution: What do we do if we donβt know the population standard deviation, and canβt find a sample size of 30 or more? If the random variable is normally distributed (or approximately normally distributed), you can use a t-distribution. t-distribution formula: π= πβπ π π critical values of t are denoted as π‘π , just as critical z values are called π§π . Properties of the t-distribution (Page 325) 1) 2) 3) 4) 5) The t-distribution is bell-shaped and symmetrical about the mean. Degrees of freedom are equal to n-1. The total area under the curve is 1, or 100%. The mean, median, and mode of the t-distribution are equal to zero. 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 zdistribution. Close enough, in fact, that we use the standard normal distribution for d.f. β₯ 30. We just did that in Section 6-1. Guidelines for constructing a Confidence Interval for the Mean: t-distribution (Page 327) 1) Identify the sample statistics n, π₯, and s. 2) Identify the degrees of freedom, the level of confidence c, and the critical value π‘π . 3) Find the margin of error E. s π¬ = ππ ; π find π‘π by using the invT menu under 2nd Vars. 1 nd 2 Vars 4, enter the area ( (1 β π)), and the degrees of freedom. 2 The calculator will give you the critical t-value. 4) Find the left and right endpoints and form the confidence interval. πβπ¬<π<π+π¬ The TI-84 can also do a t-distribution interval for you!! STAT TESTS 8 (TInterval) Same as with the z-interval, if you have the data points, select data. If you have the statistics, select stats. Enter the appropriate numbers and select Calculate. EXAMPLE 1 (Page 326) Find the critical value π‘π for a 95% confidence level when the sample size is 15. If n = 15, then the degrees of freedom are (n β 1) = 14 1 If c = .95, the area is (1 β .95) = .025 (this is the same formula we used to find 2 the area for π§π in Section 6-1. 2nd VARS 4, 0.025, 14, Enter gives us a π‘π of 2.145 (use the positive of the number you get). EXAMPLE 2 (Page 327) You randomly select 16 coffee shops and measure the temperature of coffee sold at each. The sample mean temperature is 162.00 F with a sample standard deviation of 10.00 F. Find the 95% confidence interval for the mean temperature. Assume the temperatures are approximately normally distributed. We MUST use the t-distribution for this; the sample size is less than 30 and we donβt know what Ο is, but we do know that the distribution is approximately normal. STAT TESTS 8, select Stats, and enter the values for π₯, s, n, and c. The 95% confidence interval is from 156.7 to 167.3. We round to one decimal because we were given π₯ rounded to one decimal. SO, we are 95% confident that the actual population mean of coffee temperature in ALL coffee shops is between 156.70F and 167.30F. EXAMPLE 3 (Page 328) You randomly select 20 mortgage institutions and determine the current mortgage interest rate at each. The sample mean rate is 6.22%, with a sample standard deviation of 0.42%. Find the 99% confidence interval for the population mean mortgage interest rate. Assume the interest rates are approximately normally distributed. We MUST use the t-distribution for this; the sample size is less than 30 and we donβt know what Ο is, but we do know that the distribution is approximately normal. STAT TESTS 8, select Stats, and enter the values for π₯, s, n, and c. The 99% confidence interval is from 5.95 to 6.49. We round to two decimals because we were given π₯ rounded to two decimals. SO, we are 99% confident that the actual population mean mortgage interest rate for ALL mortgage institutions is between 5.95% and 6.49%. 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!! Is n β₯ 30? YES Use the normal distribution s with πΈ = ππ ; if Ο is π unknown, use s instead. NO Is the population normally, or approximately normally, distributed? NO You can NOT use the normal or the t-distribution. YES Is Ο known? NO Use the t-distribution with s πΈ = ππ and n β 1 degrees π of freedom. YES Use the normal distribution s with πΈ = ππ π 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 Ο is. 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 Ο is, we should use the t-distribution on this one. ASSIGNMENTS Classwork: Page 330; #1-16 All Homework: Pages 331-332; #17-28 Section 6-3 β Confidence Intervals for Population Proportions Sometimes we are dealing with probabilities of success in a single trial (Section 42). This is called a probability proportion. In this section, you will learn how to estimate a population proportion p using a confidence interval. As with confidence intervals for µ, you will start with a point estimate. The point estimate for p, the population proportion of successes, is given by the π₯ proportion of successes in a sample and is denoted by π = , where x is the π number of successes in the sample and n is the number in the sample. The point estimate for the proportion of failures is π = 1 β π. The symbols π and π are read as βp hatβ and βq hatβ Section 6-3 β Confidence Intervals for Population Proportions A c-confidence interval for the population proportion p is π β πΈ < π < π + πΈ, where πΈ = π§π ππ . π The probability that the confidence interval contains p is c. In Section 5-5, you learned that a binomial distribution can be approximated by the normal distribution if np β₯ 5, and nq β₯ 5. When nπ β₯ 5 and nπ β₯ 5, the sampling distribution is approximately normal with a mean of ππ = π and a standard deviation of ππ = ππ . π Section 6-3 β Confidence Intervals for Population Proportions Guidelines (Page 335) Constructing a Confidence Interval for a Population Proportion. Identify the sample statistics n and x. Find the point estimate π. Verify that the sampling distribution of π can be approximated by the normal distribution. Find the critical value π§π that corresponds to the given level of confidence c. Find the margin of error, E. Find the left and right endpoints and form the confidence interval. As with the other intervals weβve discussed, the calculator will also create a proportion interval for you. STAT - TESTS β A (1-PropZInt) Enter x, n, and the confidence level to get the interval. Section 6-3 β Confidence Intervals for Population Proportions Finding a Minimum Sample Size to Estimate p. Given a c-confidence level and a margin of error E, the minimum sample size n needed to estimate p is: π§ π = ππ( π )2 . πΈ This formula assumes that you have a preliminary estimate for π and π. If not, use 0.5 for both. Section 6-3 β Confidence Intervals for Population Proportions EXAMPLE 1 (Page 334) In a survey of 1219 U.S. adults, 354 said that their favorite sport to watch is football. Find a point estimate for the population proportion of U.S. adults who say their favorite sport to watch is football. π₯ 354 If n = 1219 and x = 354, then π = = = .29, or 29% π 1219 In a survey of 1006 adults from the U.S., 181 said that Abraham Lincoln was the greatest president. Find a point estimate for the population proportion of adults who say that Abraham Lincoln was the greatest president. π₯ 181 If n = 1006 and x = 181, then π = = = .1799, or 17.99% π 1006 Section 6-3 β Confidence Intervals for Population Proportions EXAMPLE 2 (Page 336) Construct a 95% confidence interval for the proportion of adults in the United States who say that their favorite sport to watch is football. First, check to be certain that nπ β₯ 5 and nπ β₯ 5. (1219)(.29) = 353; (1219)(.71) = 865. Find the margin of error; πΈ = π§π ππ , π πΈ = 1.96 .29 (.71) 1219 = 0.025 Now that we know that the margin of error is .025, we subtract that from .29 to get the lower end of the interval, and add it to .29 to get the upper end of the interval. .29 - .025 < p < .29 + .025; .265 < p < .315. Section 6-3 β Confidence Intervals for Population Proportions EXAMPLE 3 (Page 337) According to a survey of 900 U.S. adults, 63% said that teenagers are the most dangerous drivers, 33% said that people over 75 are the most dangerous drivers, and 4% said that they had no opinion on the matter. Construct a 99% confidence interval for the proportion of adults who think that teenagers are the most dangerous drivers. To find x, you need to multiply the percentage (.63) by the sample size (900) to get 567. n is given to us, at 900. π is also given to us, at .63. This makes π = 1 - .63, or .37. We memorized the critical z score for a 99% confidence interval (2.575). Section 6-3 β Confidence Intervals for Population Proportions EXAMPLE 3 (Page 337) Putting all of this together, πΈ = π§π ππ , π πΈ = 2.575 .63 (.37) 900 = 0.041 Now that we know that the margin of error is .041, we subtract that from .63 to get the lower end of the interval, and add it to .63 to get the upper end of the interval. .63 - .041 < p < .63 + .041; .589 < p < .671. Section 6-3 β Confidence Intervals for Population Proportions EXAMPLE 4 (Page 338) You are running a political campaign and wish to estimate, with 95% confidence, the proportion of registered voters who will vote for your candidate. Your estimate must be accurate within 3% of the true population. Find the minimum sample size needed if (1) no preliminary estimate is available and (2) a preliminary estimate gives π = 0.31. Compare your results. π§π 2 π = ππ( ) . πΈ Remember to use .5 for both π and π when you have no preliminary data. 1.96 2 1) π = .5 (.5)( ) β 1068. 0.03 1.96 2 (.69)( ) β 0.03 2) π = .31 914. You need a larger sample size if you donβt have any preliminary data. ASSIGNMENTS Classwork: Page 339-340; #1-20 All Homework: Pages 340-342; #21-28 All Section 6-4 β Confidence Intervals for Variance and Standard Deviation In manufacturing, it is necessary to control the amount that a process varies. For instance, an automobile part manufacturer must produce thousands of parts to be used in the manufacturing process. It is important that the parts vary little or not at all. How can you measure, and thus control, the amount of variation in the parts? You can start with a point estimate. The point estimate for π 2 is π 2 and the point estimate for π is s. π 2 is the most unbiased estimate for π 2 . You can use a chi-square distribution to construct a confidence interval for the variance and standard deviation. If the random variable x has a normal distribution, then the distribution of π 2 = (πβ1)π 2 π2 as long as n > 1. Section 6-4 β Confidence Intervals for Variance and Standard Deviation There are 4 properties of the chi-square distribution: 1) All chi-square values π 2 are greater than or equal to zero. 2) The chi-square distribution is a family of curves, each determined by the degrees of freedom. To form a confidence interval for π 2 , use the π 2 -distribution with degrees of freedom equal to one less than the sample size. d.f. = n β 1. 3) The area under each curve of the chi-square distribution equals one. 4) Chi-square distributions are positively skewed. Section 6-4 β Confidence Intervals for Variance and Standard Deviation There are two critical values for each level of confidence. The value ππ 2 represents the right-tail critical value and ππΏ2 represents the left-tail critical value. 1βπ 1+π 2 2 Area to the right of ππ = and the area to the right of ππΏ = 2 2 Table 6 in Appendix B lists critical values of π 2 for various degrees of freedom and areas. Each area in the table represents the region under the chi-square curve to the right of the critical value. Section 6-4 β Confidence Intervals for Variance and Standard Deviation Confidence Intervals for π 2 and π. You can use the critical values ππ 2 and ππΏ2 to construct confidence intervals for a population variance and standard deviation. The formula for the confidence interval for a population variance π 2 is: (πβ1)π 2 ππ 2 < π2 < (πβ1)π 2 . ππΏ2 Remember that the population standard deviation π is simply the square root of the variance. (πβ1)π 2 ππ 2 <π< (πβ1)π 2 ππΏ2 Section 6-4 β Confidence Intervals for Variance and Standard Deviation Constructing a Confidence Interval for a Variance and Standard Deviation. First the bad news; there is no magic calculator button to do this for you. 1) Verify that the population has a normal distribution. 2) Identify the sample statistic n and the degrees of freedom. estimate π 2 . π 2 (π₯βπ₯)2 . πβ1 3) Find the point = Many times s will be given. 4) Find the critical values ππ 2 and ππΏ2 that correspond to the given level of confidence c. Use Table 6 in Appendix B or the chart I gave you. 5) Find the left and right endpoints and form the confidence interval for the population variance. 6) Find the confidence interval for the population standard deviation by taking the square root of each endpoint. Section 6-4 β Confidence Intervals for Variance and Standard Deviation EXAMPLE 1 (Page 345) Find the critical values ππ 2 and ππΏ2 for a 90% confidence interval when the sample size is 20. Because n = 20, the degrees of freedom = 19. 1βπ 1β.9 2 ππ = = = .05 2 1+π 2 2 1+.9 2 ππΏ2 = = = .95 Look in Table 6 in Appendix B (or on the hand-out I gave you) Find the row for the d.f. = 19 and the columns for .05 and .95. ππΏ2 = 10.117 and ππ 2 = 30.144 Section 6-4 β Confidence Intervals for Variance and Standard Deviation EXAMPLE 1 (Page 345) Now find the critical values ππ 2 and ππΏ2 for a 95% confidence interval when the sample size is 25. d.f. = 24. 1β.π 1β.95 2 ππ = = = .025 2 1+π 2 2 1+.95 2 ππΏ2 = = = .975 Look in Table 6 in Appendix B (or on the hand-out I gave you) Find the row for the d.f. = 24 and the columns for .025 and .975. ππΏ2 = 12.401 and ππ 2 = 39.364 Section 6-4 β 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. 1 β π 1 β .99 1 + π 1 + .99 2 2 ππ = = = .005 ππΏ = = = .995 2 2 2 2 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. ππΏ2 = 13.121; ππ 2 = 52.336 2 (π β 1)π 2 (π β 1)π 2 < < π ππ 2 ππΏ2 2 (29)(1.2)2 (29)(1.2) 0.80 < π 2 < 3.2 < π2 < 52.336 13.121 Section 6-4 β 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. 0.80 < π < 3.2 .9 < π < 1.8 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. Section 6-4 β 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 ππ 2 1 β .90 = = .05 2 2 (π β 1)π 2 (π β 1)π < π2 < 2 ππ ππΏ2 ππΏ2 1 + .90 = = .95 2 (29)(1.2)2 (29)(1.2)2 2 <π < 42.557 17.708 ππΏ2 = 17.708 1.0 < π 2 < 2.4 ππ 2 = 42.557 1.0 < π < 1.5 Section 6-4 β 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 ππ 2 1 β .95 1 + .95 2 2 = = .025 ππΏ = = .975 ππΏ = 16.047 2 2 2 2 (π β 1)π 2 (π β 1)π (29)1.22 (29)1.2 2 <π < < π2 < 2 2 ππ ππΏ 45.722 16.047 0.9 < π 2 < 2.60 ππ 2 = 45.722 1.0 < π < 1.6 ASSIGNMENTS Classwork: Page 348; #1-6 All Homework: Pages 348-350; #7-20 All
