Estimation of M Lyman Ott Michael Longnecker

Any time falls in the interval , the interval will con- tain the parameter m see Figure 5.2. The probability of falling in the interval is .95, so we state that is an interval estimate of m with level of confidence .95. We evaluate the goodness of an interval estimation procedure by examining the fraction of times in repeated sampling that interval estimates would encompass the parameter to be estimated. This fraction, called the confidence coefficient, is .95 when using the formula ; that is, 95 of the time in repeated sam- pling, intervals calculated using the formula will contain the mean m. This idea is illustrated in Figure 5.3. Suppose we want to study a commercial process that produces shrimp for sale to restaurants. The shrimp are monitored for size by randomly selecting 40 shrimp from the tanks and measuring their length. We will consider a simulation of the shrimp monitoring. Suppose that the distribution of shrimp length in the tank had a normal distribution with a mean m ⫽ 27 cm and a standard deviation s ⫽ 10 cm. One hundred samples of size are drawn from the shrimp population. From each of these samples we compute the interval estimate See Table 5.1. Note that although the intervals vary in location, only 6 of the 100 intervals failed to capture the population mean m. The fact that six samples produced in- tervals that did not contain m is not an indication that the procedure for pro- ducing intervals is faulty. Because our level of confidence is 95, we would y ⫾ 1.96s 兾1n ⫽ y ⫾ 1.9610兾140. n ⫽ 40 y ⫾ 1.96s 兾1n y ⫾ 1.96s 兾1n y ⫾ 1.96s 兾1n m ⫾ 1.96s 兾1n y y ⫾ 1.96s 兾1n m ⫾ 1.96s 兾1n y FIGURE 5.2 When the observed value of lies in the interval , the interval contains the parameter m. y ⫾ 1.96s 兾1n m ⫾ 1.96s 兾1n y 1.96 + 1.96 f y y y 1.96 Observed y y + 1.96 n n n n FIGURE 5.3 Fifty interval estimates of the population mean 27 33 31 29 27 Limits 25 23 21 19 10

20 30

40 50 Sample 60 70 80 90 100 interval estimate level of confidence confidence coefficient TABLE 5.1 One hundred interval estimates of the population mean 27 Interval Interval Contains Contains Sample Lower Upper Population Sample Lower Upper Population Sample Mean Limit Limit Mean Sample Mean Limit Limit Mean 1 27.6609 24.5619 30.7599 Yes 51 26.9387 23.8397 30.0377 Yes 2 27.8315 24.7325 30.9305 Yes 52 26.4229 23.3239 29.5219 Yes 3 25.9366 22.8376 29.0356 Yes 53 24.2275 21.1285 27.3265 Yes 4 26.6584 23.5594 29.7574 Yes 54 26.4426 23.3436 29.5416 Yes 5 26.5366 23.4376 29.6356 Yes 55 26.3718 23.2728 29.4708 Yes 6 25.9903 22.8913 29.0893 Yes 56 29.3690 26.2700 32.4680 Yes 7 29.2381 26.1391 32.3371 Yes 57 25.9233 22.8243 29.0223 Yes 8 26.7698 23.6708 29.8688 Yes 58 29.6878 26.5888 32.7868 Yes 9 25.7277 22.6287 28.8267 Yes 59 24.8782 21.7792 27.9772 Yes 10 26.3698 23.2708 29.4688 Yes 60 29.2868 26.1878 32.3858 Yes 11 29.4980 26.3990 32.5970 Yes 61 25.8719 22.7729 28.9709 Yes 12 25.1405 22.0415 28.2395 Yes 62 25.6650 22.5660 28.7640 Yes 13 26.9266 23.8276 30.0256 Yes 63 26.4958 23.3968 29.5948 Yes 14 27.7210 24.6220 30.8200 Yes 64 28.6329 25.5339 31.7319 Yes 15 30.1959 27.0969 33.2949 No 65 28.2699 25.1709 31.3689 Yes 16 26.5623 23.4633 29.6613 Yes 66 25.6491 22.5501 28.7481 Yes 17 26.0859 22.9869 29.1849 Yes 67 27.8394 24.7404 30.9384 Yes 18 26.3585 23.2595 29.4575 Yes 68 29.5261 26.4271 32.6251 Yes 19 27.4504 24.3514 30.5494 Yes 69 24.6784 21.5794 27.7774 Yes 20 28.6304 25.5314 31.7294 Yes 70 24.6646 21.5656 27.7636 Yes 21 26.6415 23.5425 29.7405 Yes 71 26.4696 23.3706 29.5686 Yes 22 25.6783 22.5793 28.7773 Yes 72 26.0308 22.9318 29.1298 Yes 23 22.0290 18.9300 25.1280 No 73 27.5731 24.4741 30.6721 Yes 24 24.4749 21.3759 27.5739 Yes 74 26.5938 23.4948 29.6928 Yes 25 25.7687 22.6697 28.8677 Yes 75 25.4701 22.3711 28.5691 Yes 26 29.1375 26.0385 32.2365 Yes 76 28.3079 25.2089 31.4069 Yes 27 26.4457 23.3467 29.5447 Yes 77 26.4159 23.3169 29.5149 Yes 28 27.4909 24.3919 30.5899 Yes 78 26.7439 23.6449 29.8429 Yes 29 27.8137 24.7147 30.9127 Yes 79 27.0831 23.9841 30.1821 Yes 30 29.3100 26.2110 32.4090 Yes 80 24.4346 21.3356 27.5336 Yes 31 26.6455 23.5465 29.7445 Yes 81 24.7468 21.6478 27.8458 Yes 32 27.9707 24.8717 31.0697 Yes 82 27.1649 24.0659 30.2639 Yes 33 26.7505 23.6515 29.8495 Yes 83 28.0252 24.9262 31.1242 Yes 34 24.9366 21.8376 28.0356 Yes 84 27.1953 24.0963 30.2943 Yes 35 27.9943 24.8953 31.0933 Yes 85 29.7399 26.6409 32.8389 Yes 36 27.3375 24.2385 30.4365 Yes 86 24.2036 21.1046 27.3026 Yes 37 29.4787 26.3797 32.5777 Yes 87 27.0769 23.9779 30.1759 Yes 38 26.9669 23.8679 30.0659 Yes 88 23.6720 20.5730 26.7710 No 39 26.9031 23.8041 30.0021 Yes 89 25.4356 22.3366 28.5346 Yes 40 27.2275 24.1285 30.3265 Yes 90 23.6151 20.5161 26.7141 No 41 30.1865 27.0875 33.2855 No 91 24.0929 20.9939 27.1919 Yes 42 26.4936 23.3946 29.5926 Yes 92 27.7310 24.6320 30.8300 Yes 43 25.8962 22.7972 28.9952 Yes 93 27.3537 24.2547 30.4527 Yes 44 24.5377 21.4387 27.6367 Yes 94 26.3139 23.2149 29.4129 Yes 45 26.1798 23.0808 29.2788 Yes 95 24.8383 21.7393 27.9373 Yes 46 26.7470 23.6480 29.8460 Yes 96 28.4564 25.3574 31.5554 Yes 47 28.0406 24.9416 31.1396 Yes 97 28.2395 25.1405 31.3385 Yes 48 26.0824 22.9834 29.1814 Yes 98 25.5058 22.4068 28.6048 Yes 49 25.6270 22.5280 28.7260 Yes 99 25.6857 22.5867 28.7847 Yes 50 23.7449 20.6459 26.8439 No 100 27.1540 24.0550 30.2530 Yes expect that, in a large collection of 95 confidence intervals, approximately 5 of the intervals would fail to include m. Thus, in 100 intervals we would expect four to six intervals 5 of 100 to not contain m. It is crucial to understand that even when experiments are properly conducted, a number of the experiments will yield results that in some sense are in error. This occurs when we run only a small number of experiments or select only a small subset of the population. In our example, we randomly selected 40 observations from the population and then constructed a 95 confidence interval for the population mean m. If this process were repeated a very large number of times—for example, 10,000 times instead of the 100 in our example—the proportion of intervals not containing m would be very nearly 5. In most situations when the population mean is unknown, the population standard deviation s will also be unknown. Hence, it will be necessary to estimate both m and s from the data. However, for all practical purposes, if the sample size is relatively large 30 or more is the standard rule of thumb, we can estimate the population standard deviation s with the sample standard deviation s in the con- fidence interval formula. Because s is estimated by the sample standard deviation s, the actual standard error of the mean , is naturally estimated by . This estimation introduces another source of random error s will vary randomly, from sample to sample, about s and, strictly speaking, invalidates the level of confidence for our interval estimate of m. Fortunately, the formula is still a very good approximation for large sample sizes. When the population has a normal distribution, a better method for constructing the confidence interval will be presented in Section 5.7. Also, based on the results from the Central Limit Theorem, if the population distribution is not too nonnormal and the sample size is relatively large again, using 30 or more is the standard rule of thumb, the level of the confidence of the interval will be approximately the same as if we were sampling from a normal distribution with s known and using the interval . EXAMPLE 5.1 A courier company in New York City claims that its mean delivery time to any place in the city is less than 3 hours. The consumer protection agency decides to conduct a study to see if this claim is true. The agency randomly selects 50 deliver- ies and determines the mean delivery time to be 2.8 hours with a standard deviation of hours. The agency wants to estimate the mean delivery time m using a 95 confidence interval. Obtain this interval and then decide if the courier com- pany’s claim appears to be reasonable. Solution The random sample of deliveries yields and . Be- cause the sample size is relatively large, , the appropriate 95 confidence interval is then computed using the following formula: With s used as an estimate of s, our 95 confidence interval is The interval from 2.634 to 2.966 forms a 95 confidence interval for m. In other words, we are 95 confident that the average delivery time lies between 2.634 and 2.966 hours. Because the interval has all its values less than 3 hours, we can conclude that there is strong evidence in the data that the courier company’s claim is correct. 2.8 ⫾ 1.96 .6 150 or 2.8 ⫾ .166 y ⫾ 1.96s 兾 1n n ⫽ 50 s ⫽ .6 y ⫽ 2.8 n ⫽ 50 s ⫽ .6 y ⫾ 1.96s 兾1n y ⫾ 1.96s 兾1n s 兾1n s 兾1n TABLE 5.2 Common values of the confidence coefficient 1 ⫺ a and the corresponding z-value, z a 兾2 Confidence Coefficient Value of Area in Table 1 Corresponding z-Value, 1 ⴚ A A 兾2 1 ⴚ A 兾2 z A 兾2 .90 .05 .95 1.645 .95 .025 .975 1.96 .98 .01 .99 2.33 .99 .005 .995 2.58 z A 兾2 There are many different confidence intervals for m, depending on the confidence coefficient we choose. For example, the interval includes 99 of the values of in repeated sampling, and the interval forms a 99 confidence interval for m. We can state a general formula for a confidence interval for m with a confidence coefficient of 1 ⴚ a, where a Greek letter alpha is between 0 and 1. For a specified value of 1 ⫺ a, a 1001 ⫺ a confidence interval for m is given by the following formula. Here we assume that s is known or that the sample size is large enough to replace s with s. y ⫾ 2.58s 兾1n y m ⫾ 2.58s 兾1n FIGURE 5.4 Interpretation of z a 兾2 in the confidence interval formula f y y z α 2 Area = 2 Area = 1 2 n Confidence Interval for M, S Known y ⫾ z a 兾2 s 兾 1n The quantity z A 兾2 is a value of z having a tail area of a 兾2 to its right. In other words, at a distance of z a 兾2 standard deviations to the right of m, there is an area of a 兾2 under the normal curve. Values of z a 兾2 can be obtained from Table 1 in the Appendix by looking up the z-value corresponding to an area of 1 ⫺ a 兾2 see Figure 5.4. Common values of the confidence coefficient 1 ⫺ a and z a 兾2 are given in Table 5.2. 99 confidence interval 1 ⴚ A ⴝ confidence coefficient EXAMPLE 5.2 A forester wishes to estimate the average number of ‘‘count trees’’ per acre trees larger than a specified size on a 2,000-acre plantation. She can then use this infor- mation to determine the total timber volume for trees in the plantation. A random sample of n ⫽ 50 one-acre plots is selected and examined. The average mean number of count trees per acre is found to be 27.3, with a standard deviation of 12.1. Use this information to construct a 99 confidence interval for m, the mean number of count trees per acre for the entire plantation. Solution We use the general confidence interval with confidence coefficient equal to .99 and a z a 2 -value equal to 2.58 see Table 5.2. Substituting into the formula and replacing s with s, we have This corresponds to the confidence interval 27.3 ⫾ 4.41—that is, the interval from 22.89 to 31.71. Thus, we are 99 sure that the average number of count trees per acre is between 22.89 and 31.71. Statistical inference-making procedures differ from ordinary procedures in that we not only make an inference but also provide a measure of how good that inference is. For interval estimation, the width of the confidence interval and the confidence coefficient measure the goodness of the inference. For a given value of the confidence coefficient, the smaller the width of the interval, the more precise the inference. The confidence coefficient, on the other hand, is set by the experi- menter to express how much assurance he or she places in whether the interval es- timate encompasses the parameter of interest. For a fixed sample size, increasing the level of confidence will result in an interval of greater width. Thus, the experi- menter will generally express a desired level of confidence and specify the desired width of the interval. Next we will discuss a procedure to determine the appropri- ate sample size to meet these specifications.

5.3 Choosing the Sample Size for Estimating M

How can we determine the number of observations to include in the sample? The implications of such a question are clear. Data collection costs money. If the sample is too large, time and talent are wasted. Conversely, it is wasteful if the sample is too small, because inadequate information has been purchased for the time and effort expended. Also, it may be impossible to increase the sample size at a later time. Hence, the number of observations to be included in the sample will be a compro- mise between the desired accuracy of the sample statistic as an estimate of the pop- ulation parameter and the required time and cost to achieve this degree of accuracy. The researchers in the dietary study described in Section 5.1 had to deter- mine how many nurses to survey for their study to yield viable conclusions. To de- termine how many nurses must be sampled, we would have to determine how accurately the researchers want to estimate the mean percentage of calories from fat PCF. The researchers specified that they wanted the sample estimator to be within 1.5 of the population mean m. Then we would want the confidence interval for m to be Alternatively, the researchers could specify that the tolerable error in estimation is 3, which would yield the same specification , because the tolerable error is simply the width of the confidence interval. There are two considerations in determining the appropriate sample size for estimating m using a confidence interval. First, the tolerable error establishes the desired width of the interval. The second consideration is the level of confidence. In selecting our specifications, we need to consider that if the confidence interval of m is too wide, then our estimation of m will be imprecise and not very informa- tive. Similarly, a very low level of confidence say 50 will yield a confidence interval that very likely will be in error—that is, fail to contain m. However, to obtain a confidence interval having a narrow width and a high level of confidence may require a large value for the sample size and hence be unreasonable in terms of cost andor time. y ⫾ 1.5 y ⫾ 1.5. 27.3 ⫾ 2.58 12.1 150 y ⫾ 2.58 s 兾 1n What constitutes reasonable certainty? In most situations, the confidence level is set at 95 or 90, partly because of tradition and partly because these lev- els represent to some people a reasonable level of certainty. The 95 or 90 level translates into a long-run chance of 1 in 20 or 1 in 10 of not covering the pop- ulation parameter. This seems reasonable and is comprehensible, whereas 1 chance in 1,000 or 1 in 10,000 is too small. The tolerable error depends heavily on the context of the problem, and only someone who is familiar with the situation can make a reasonable judgment about its magnitude. When considering a confidence interval for a population mean m, the plus-or- minus term of the confidence interval is . Three quantities determine the value of the plus-or-minus term: the desired confidence level which determines the z-value used, the standard deviation s, and the sample size. Usually, a guess must be made about the size of the population standard deviation. Sometimes an initial sample is taken to estimate the standard deviation; this estimate provides a basis for determining the additional sample size that is needed. For a given toler- able error, once the confidence level is specified and an estimate of s supplied, the required sample size can be calculated using the formula shown here. Suppose we want to estimate m using a 1001 ⫺ a confidence interval hav- ing tolerable error W. Our interval will be of the form where E ⫽ W 兾2. Note that W is the width of the confidence interval. To determine the sample size n, we solve the equation for n. This formula for n is shown here: E ⫽ z a 兾2 s 兾1n y ⫾ E, z a 兾2 s 兾1n Sample Size Required for a 1001 ⴚ A Confidence Interval for M of the Form y – ⴞ E n ⫽ z a 兾2 2 s 2 E 2 Note that determining a sample size to estimate m requires knowledge of the population variance s 2 or standard deviation s. We can obtain an approximate sample size by estimating s 2 , using one of these two methods: 1. Employ information from a prior experiment to calculate a sample vari- ance s 2 . This value is used to approximate s 2 . 2. Use information on the range of the observations to obtain an estimate of s. We would then substitute the estimated value of s 2 in the sample-size equation to determine an approximate sample size n. We illustrate the procedure for choosing a sample size with two examples. EXAMPLE 5.3 The relative cost of textbooks to other academic expenses has risen greatly over the past few years, university officials have started to include the average amount ex- pended on textbooks into their estimated yearly expenses for students. In order for these estimates to be useful, the estimated cost should be within 25 of the mean expenditure for all undergraduate students at the university. How many students should the university sample in order to be 95 confident that their estimated cost of textbooks will satisfy the stated level of accuracy?