Chapter 6 Confidence Intervals

Chapter 6 Confidence Intervals Section 6-1 Confidence Intervals for the Mean (Large Samples) VOCABULARY: Point Estimate A value for a parameter. The most point estimate of the population parameter is the sample. Interval Estimate - An interval, or of, used to estimate a population. Level of Confidence - Denoted as, it is the that the interval estimate the. Margin of Error - Sometimes also called the error of estimate, or error. It is denoted as, and is the possible between the estimate and the value of the it is estimating. c-confidence interval - Is found by and from the sample. FORMULAS: Margin of Error = z c = The that the confidence interval contains is. σ E = z c σ x = z c n To use the formula, it is assumed that the standard deviation is known. This is the case, but when n, the standard deviation can be used in place of. The formula effectively becomes E = z c s n of 1 c, or 1 (1 c) (explained why on bottom of page 311). c-confidence interval - x E < μ < x + E sample standard deviation: GUIDELINES (creating interval by hand): 1) Find the sample statistics n and x. n is the. x is the. s = (x x ) n 1 ; It is MUCH easier to use the STAT-Edit function of the calculator to find s. ) Specify, if known. Otherwise, if n 30, find the sample standard deviation s and use it as a point estimate for σ. 3) Find the critical value z c that corresponds to the given. The three most commonly used confidence levels are %, %, and %. The corresponding z c values are: 90% -- 95% -- 99% -- It would help to memorize these. You will be using them a lot. 4) Find the margin of error E. (E = z c s ) n 5) Find the left and right and form the confidence interval. x E < μ < x + E

GUIDELINES (creating interval with calculator): The TI-84 will actually create the interval for you!! 1) If you are given the actual data points (a list of numbers), enter them into L1 using the menu. ) gives you the sample standard deviation, which you will need. Ignore everything else; the calculator will give it to you again. 3) STAT Test (when you have EITHER n > OR you know what is. 4) If you have the list of numbers in L1, choose and enter the standard deviation where it asks for σ. Do NOT use the σ from the calculator!! Use the standard deviation. If you do NOT have a list of numbers, you will be given the mean, standard deviation, and sample size Select and enter the values asked for as given to you. 5) Enter the you are looking for at C-Level and Calculate. The calculator will give you the interval. From here, you can work to find, if necessary. Either take of the between your, or find the difference between the and either. We are going to walk through Example 4 on page 314, using the data points from Example 1 on page 310. TO SOLVE BY HAND: The question asks for a 99% confidence interval for the population mean. To find E using the formula, E =, plug in what we know. z c is found by nd VARS 3 ((1-.99)/, 0,1) =.576 (this is also one of the ones we should be memorizing!!) E = (. 576) 5.01 50 = To build the interval by hand, x E < μ < x + E - < μ < + < μ < We can be certain that the actual population mean is somewhere between and. TO SOLVE ON THE CALCULATOR 1) Enter the 50 data points into L1 on your calculator (STAT Edit). ) STAT Calc 1 to find the sample standard deviation (we can use this because we have 50 data points; n 30). s = ) STAT TESTS ( -Interval) 3) Select, since you have the entered into the calculator. 4) Enter as the standard deviation, and as the C-Level (level of confidence). 5) Select Calculate to get the interval. We can be certain that the actual population mean is somewhere between and. Notice that the calculator also tells us that the mean of the data we entered is data is and that n is., that the standard deviation of the If we need to know what E is, simply find the between the interval endpoints and divide by. ( )/ =. We could also simply find the distance between the and the of the interval to find E. = ; = Again, the same answer we got doing it by hand.

Look at Example 5 on page 315. n = 0, x =.9, σ = 1.5, and c = 90% NOTICE Even though n < 30, we were given, so we can still run the z-test. The question asks for a 90% confidence interval for the population mean. To find E using the formula, E = z c s n z c is found by nd VARS 3 ((1-.90)/, 0,1) =, plug in what we know. E = (1.645) 5.01 50 = To build the interval by hand, x E < μ < x + E - < μ < + < μ < (again, one of the ones we should be memorizing) NOTE: as a general rule, round the interval to the same number of digits as the data you are given. Since the mean and standard deviation were given to you to one decimal point, round the interval to one decimal point. We can be 90% certain that the actual population mean is somewhere between and. On the calculator, STAT TESTS, select (since you are going to provide the stats instead of the actual data points). Enter,,, and and then calculate. Again, as a general rule, we round our interval endpoints to the same number of decimal points as the data that is given to us. We were given.9 and 1.5, which have one decimal, so we should round our interval to one decimal place. < μ < SAME ANSWER we got doing it by hand! 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 E = z c becomes n = z n c( σ E ). Remember, if you don t know what σ is, you can use s, as long as n 30. Example 6 on page 316- c = 95%, σ s 5.01 (from Example 1), E = 1 (given). z c is found by going to nd VARS 3 and entering (1-.95)/ for the area. Remember to use the positive of the value given to you. z c = 1.96, (AGAIN, one of the ones we should be memorizing!!) n = ( z cσ E ) = ( 1.96(5.01) ) = ( 9.8196 ) = 9.8196 =. 1 1 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 We round up, since magazine advertisements in your sample. advertisements are not quite enough. REMEMBER!! Sample size is a NUMBER!! You can t do part of an experiment; you either get a data point or you don t. Section 6- Confidence Intervals for the Mean (Small Samples) The t-distribution: What do we do if we don t know the population standard deviation, and can t 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: t = x μ s ; critical values of t are denoted as t c just as critical z values are called z c. n THERE IS A FLOW-CHART ON PAGE 39 IN YOUR BOOK OUTLINING HOW TO KNOW WHEN TO USE WHICH TEST! It would be a good idea to study it and be able to determine which test to use when. Properties of the t-distribution (Page 35) 1) The t-distribution is and about the. ) of are equal to n-1. 3) The total area under the curve is, or. 4) The,, and of the t-distribution are equal to. 5) As the degrees of freedom, the t-distribution approaches the distribution. After 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. The TI-84 can also do a t-distribution interval for you!! In the interest of time, we are not going to cover doing these by hand. We do need to know how to find the critical t-score, though EXAMPLE 1 (Page 36) Find the critical value t c for a 95% confidence level when the sample size is 15. To do this, go to nd VARS ( ) instead of invnorm. t c = The area is (1.95)/ (this is the same formula we used to find the area for z c in Section 6-1). If n = 15, then the degrees of freedom are (n 1) = nd VARS 4, (1-.95)/, 14, Enter gives us a t c of, but we will use the positive of that, just like we did with the z c. EXAMPLE (Page 37) You randomly select 16 coffee shops and measure the temperature of coffee sold at each. The sample mean temperature is 16.0 0 F with a sample standard deviation of 10.0 0 F. Find the 95% confidence interval for the mean temperature. Assume the temperatures are approximately normally distributed. We MUST use the for this; the sample size is than and we don t know what is, but we do know that the distribution is. STAT TESTS, select Stats, and enter the values for x, s, n, and c and calculate. The 95% confidence interval is from 0 F to 0 F. We round to one decimal because we were given x and the standard deviation rounded to one decimal. SO, we are 95% confident that the actual population mean of coffee temperature in ALL coffee shops is between EXAMPLE 3 (Page 38) 0 F and 0 F. You randomly select 0 mortgage institutions and determine the current mortgage interest rate at each. The sample mean rate is 6.%, with a sample standard deviation of 0.4%. Find the 99% confidence interval for the population mean mortgage interest rate. Assume the interest rates are approximately normally distributed. We MUST use the for this; the sample size is than and we don t know what is, but we do know that the distribution is. STAT TESTS, select Stats, and enter the values for x, s, n, and c and calculate. The 99% confidence interval is from % to %. We round to two decimals because we were given x rounded to two decimals.

SO, we are % confident that the actual population mean mortgage interest rate for ALL mortgage institutions is between % and %. 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? 90% interval is to 95% interval is to 99% interval is to The more we need to be, the the interval must become. Put another way, the c is, the the interval will be. EXAMPLE 4 (Page 39) You randomly select 5 newly constructed houses. The sample mean construction cost is $181,000 and the population standard deviation is $8,000. 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 than, we can still use the distribution because we know what 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.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, the distribution is normal, and we do not know what is, we should use the -distribution on this one. Section 6-3 Confidence Intervals for Population Proportions Sometimes we are dealing with probabilities of success in a single trial (Section 4-). This is called a. In this section, you will learn how to a population proportion using a confidence interval. As with confidence intervals for µ, you will start with a. The point estimate for p, the population proportion of successes, is given by the proportion of successes in a sample and is denoted by p = x, where x is the number of in the sample and n is the number in the. n The point estimate for the proportion of The symbols p and q are read as p hat and q hat is q = 1 p. A c-confidence interval for the population proportion p is p E < p < p + E, where E = z c p q n. The probability that the confidence interval contains p is c. In Section5-5, you learned that a binomial distribution can be approximated by the normal distribution if np 5, and nq 5. When np 5 and nq 5, the sampling distribution is approximately normal. Guidelines (Page 335) Constructing a Confidence Interval for a Population Proportion. 1) Identify the sample statistics n and x. ) Find the point estimate p. 3) Verify that the sampling distribution of p can be approximated by the normal distribution. 4) Find the critical value z c that corresponds to the given level of confidence c. 5) Find the margin of error, E. 6) Find the left and right endpoints and form the confidence interval. As with the other intervals we ve discussed, the calculator will also create a proportion interval for you. STAT - TESTS A (1-PropZInt)

Enter x, n, and the confidence level get the interval. 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: n = p q ( z c E ). This formula assumes that you have a preliminary estimate for p and q. If not, use 0.5 for both. EXAMPLE 1 (Page 334) In a survey of 119 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. If n = and x =, then p = x =, or %. n 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. If n = and x =, then p = x =, or %. n EXAMPLE (Page336) 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 np and nq. ( )( ) = ; ( )( ) =. both of which are safely more than. To do this on the calculator, STAT TESTS (1 PropZInterval ) Enter for x, for n, and for C-Level You MUST enter a If you don t, you will get an error message. The interval is from It also tells you that p = to for x!! If this is not the right value for p, you put the numbers into the calculator incorrectly. To do this by hand, find the margin of error; E = z c p q, E = 1.96 (.9)(.71) = n 119 p E < p < p + E - < p < + < p < This is the same interval the calculator gave us, to 3 decimals. 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 ( ) by the ( ) to get. n is given to us, at. p is also given to us, at. This makes q = 1 -, or. We memorized the critical z-score for a 99% confidence interval (.576). Putting all of this together,

E = z c p q, E =.575 (.63)(.37) = n 900 Now that we know that the margin of error is, we that from to get the end of the interval, and it to to get the end of the interval. - < p < + ; < p <. Remember to round to two decimals, since p was given to you to two decimals. On the calculator, STAT-TESTS- Enter ( * ) for x Enter Enter for n for C-Level You get < p <, same as by hand. Remember to round to two decimals, since p was given to you to two decimals. 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 () a preliminary estimate gives p = 0.31. Compare your results. n = p q ( z c E ). Remember to use.5 for both p and q when you have no preliminary data. 1) n = (. 5)(.5)( 1.96 0.03 ). ) n = (. 31)(.69)( 1.96 0.03 ). You need a sample size if you don t have any data. AGAIN, remember that sample sizes MUST be NUMBERS!! Section 6-4 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 is and the point estimate for is. is the most unbiased estimate for. You can use a distribution to construct confidence intervals for the and deviation. If the random variable x has a distribution, then the distribution of X = (n 1)s as long as n >. σ There are 4 properties of the chi-square distribution: 1) All chi-square values X are greater than or equal to. ) The chi-square distribution is a family of curves, each determined by the of. To form a confidence interval for σ, use the X -distribution with degrees of freedom equal to sample size. d.f. =. 3) The area under each curve of the chi-square distribution equals. 4) Chi-square distributions are. There are critical values for each level of confidence. The value represents the critical value and represents the critical value. Area to the right of X R = 1 c and the area to the right of X L = 1+c less than the

Table 6 in Appendix B lists critical values of X for various degrees of freedom and areas. Each area in the table represents the region under the chi-square curve to the Confidence Intervals for σ and σ. of the critical value. You can use the critical values and to construct confidence intervals for a population variance and standard deviation. The formula for the confidence interval for a population variance σ is: (n 1)s < σ < (n 1)s. Remember that the population standard deviation σ is simply the (n 1)s < σ < (n 1)s Constructing a Confidence Interval for a Variance and Standard Deviation. There is no magic calculator button to do this for you. 1) Verify that the population has a distribution. ) Identify the sample statistic and the of. 3) Find the point estimate s. s = (x x ). ( is usually given to you) n 1 4) Find the critical values and that correspond to the given level of confidence c. Use Table 6 in Appendix B or the chart I gave you. of the variance. 5) Find the left and right endpoints and form the confidence interval for the population. 6) Find the confidence interval for the population standard deviation by taking the of each endpoint. EXAMPLE 1 (Page 345) Find the critical values and for a 90% confidence interval when the sample size is 0. Because n = 0, the degrees of freedom =. = 1 c = 1.9 = = 1+c = 1+.9 = Look in Table 6 in Appendix B (or on the hand-out I gave you) Find the row for the d.f. = and the columns for and. = and = Now find the critical values and for a 95% confidence interval when the sample size is 5. = 1 c = 1.9 = = 1+c = 1+.9 = Look in Table 6 in Appendix B (or on the hand-out I gave you) EXAMPLE (Page 347) Find the row for the d.f. = and the columns for and. = and = You randomly select and weigh 30 samples of an allergy medicine. The sample standard deviation is 1. milligrams. Assuming the weights are normally distributed, construct 99% confidence intervals for the population variance and standard deviation. = 1 c = 1.9 = = 1+c = 1+.9 = Look in Table 6 in Appendix B (or on the hand-out I gave you) Find the row for the d.f. = and the columns for and.

= and = (n 1)s < σ < (n 1)s, < σ <, < σ < 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 < σ < 3.18, < σ < Remember, round the interval to the same number of decimal points as your data has. You were given the standard deviation (1.) to one decimal, so state your interval to one decimal. We are confident that the actual population variance of weights of allergy medicines is between and, and that the actual population standard deviation of weights of allergy medicines is between and. Find the 90% and 95% confidence intervals for the population variance and standard deviation of the medicine weights. d.f. = 9. d.f. = 9. X R = 1.c = 1.90 = X L = 1+c = 1+.90 = X L = and X R = (n 1)s < σ < (n 1)s, < σ <, < σ < ; < σ < Rounded to one decimal, your 90% intervals should be: < σ < ; < σ < X R = 1.c = 1.95 = X L = 1+c = 1+.95 = X L = and X R = (n 1)s Helpful Hints for Calculator: < σ < (n 1)s, < σ <, < σ < ; < σ < Rounded to one decimal, your 95% intervals should be: < σ < ; < σ < 1) Use Alpha y= to access fraction function. ) Put (n 1)(s ) in the numerator, and your value in the denominator. This gives you the left end of the interval for the variance. 3) Press nd x and then nd and the negative sign to take the square root of the answer. This gives you the left end of the interval for the standard deviation. 4) Press nd Enter twice to get back to the fraction. Use the arrow key to put your cursor in the denominator, press Clear and enter the value. This gives you the right end of the interval for the variance. 5) Repeat step 3 to find the right end of the interval for the standard deviation. 6) Now that you have the exact answers, round to the appropriate number of decimals.