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Estimation Estimators & Estimates Estimators are the random variables used to estimate population parameters, while the specific values of these variables are the estimates. Example: the estimator of m is often n X X i 1 i n but if the observed values of X are 1, 2, 3, and 6, the estimate is 3. So the estimator is a formula; the estimate is a number. Properties of a Good Estimator 1. 2. 3. 4. Unbiasedness Efficiency Sufficiency Consistency Unbiasedness An estimator ˆ (“theta hat”) is unbiased if its expected value equals the value of the parameter (theta) being estimated. That is, ˆ E( ) In other words, on average the estimator is right on target. Examples SinceE(X) m, X is an unbiasedestimatorof m. Since E(X/n) , X/n is an unbiased estimator of . Since E(s2 ) 2 , s 2 is an unbiasedestimator of 2 . n Recall that s 2 2 (X X ) i 1 n 1 . If we divided by n instead of by n-1, we would not have an unbiased estimator of 2. That is why s2 is defined the way it is. Bias ˆ bias E( ) The bias of an unbiased estimator is zero. Mean Squared Error (MSE) 2 ˆ MSE E[( ) ] which happens to equal 2 bias2 Efficiency The most efficient estimator is the one with the smallest MSE. Efficiency Since MSE 2 bias2 , for unbiased estimators (where the bias is zero), MSE = 2. So if you are comparing unbiased estimators, the most efficient one is the one with the smallest variance. If you have two estimators, one of which has a small bias & a small variance and the other has no bias but a large variance, the more efficient one may be the one that is just slightly off on average, but that is more frequently in the right vicinity. Example: sample mean & median As we have found, the sample mean is an unbiased estimator of m. It turns out that the sample median is also an unbiased estimator of m. We know the variance of the sample mean is 2/n. The variance of the sample median is (/2)(2/n). Since is about 3.14, /2 >1. So the variance of the sample median is greater than 2/n, the variance of the sample mean. Since both estimators are unbiased, the one with the smaller variance (the sample mean) is the more efficient one. In fact, among all unbiased estimators of m, the sample mean is the one with the smallest variance. Sufficiency An estimator is said to be sufficient if it uses all the information about the population parameter that the sample can provide. Examples Example 1: The sample median is not a sufficient estimator because it uses only the ranking of the observations, and not their numerical values [with the exception of the middle one(s)]. Example 2: The sample mean, however, uses all the information, and therefore is a sufficient estimator. Consistency An estimator is said to be consistent if it yields estimates that converge in probability to the population parameter being estimated as n approaches infinity. In other words, as the sample size increases, The estimator spends more and more of its time closer and closer to the parameter value. One way that an estimator can be consistent is for its bias and its variance to approach zero as the sample size approaches infinity. Example of a consistent estimator distribution of estimator when n = 500 As the sample size increases, the bias & the variance are both shrinking. distribution of estimator when n = 50 distribution of estimator when n=5 m Example: Sample Mean _ We know that the mean of X is m. So its bias not only goes to zero as n approaches infinity, its bias is always zero. The variance of the sample mean is 2/n. As n approaches infinity, that variance approaches zero. So, since both the bias and the variance go to zero, as n approaches infinity, the sample mean is a consistent estimator. A great estimator: the sample mean X We have found that the sample mean is a great estimator of the population mean m. It is unbiased, efficient, sufficient, & consistent. Point Estimators versus Interval Estimators Up until now we have considered point estimators that provide us with a single value as an estimate of a desired parameter. It is unlikely, however, that our estimate will precisely equal our parameter. We, therefore, may prefer to report something like this: We are 95% certain that the parameter is between “a” and “b.” This statement is a confidence interval. Building a Confidence Interval 0.9750 0.0250 -1.96 We know that Then 0 1.96 Pr(Z < 1.96) = 0.9750 Pr(-1.96 < Z < 1.96) = 0.95 X-m We also know that is distributed as a standard normal (Z). n So there is a 95% probability that - 1.96 X-m n 1.96 Z Continuing from: with 95% probability, - 1.96 X-m 1.96 n Multiplying through by Subtracting off n , X , Multiplying by -1 and flipping the inequalities appropriately, Flipping the entire expression, X - m 1.96 - 1.96 n - X - 1.96 X 1.96 n - m -X 1.96 n m X - 1.96 n n n X - 1.96 m X 1.96 n n So we have a 95% Confidence Interval for the Population Mean m m X 1.96 X - 1.96 n n Example: Suppose a sample of 25 students at a university has a sample mean IQ of 127. If the population standard deviation is 5.4, calculate the 95% confidence interval for the population mean. m X 1.96 X - 1.96 n n 5.4 127 - 1.96 25 5.4 m 127 1.96 25 127 - 2.12 m 127 2.12 124.88 m 129.12 We are 95% certain that the population mean is between 124.88 & 129.12 . When we say we are 95% certain that the population mean m is between 124.88 & 129.12, it means this: The population mean m is a fixed number, but we don’t know what it is. Our confidence intervals, however, vary with the random sample that we take. Sometimes we get a more typical sample, sometimes a less typical one. If we took 100 random samples and from them calculated 100 confidence intervals, 95 of the intervals should contain the population mean that we are trying to estimate. What if we want a confidence level other than 95%? In our formula, the 1.96 came from our the fact that the Z distribution will be between -1.96 and 1.96 95% of the time. X - 1.96 m X 1.96 n n To get a different confidence level, all we need to do is find the Z values such that we are between them the desired percent of the time. Using that Z value, we have the general formula for the confidence interval for the population mean m : m X Z X -Z n n Notice: In our confidence interval formula, we used “less than” symbols: X - Z n m X Z n Your textbook uses “less than or equal to” symbols: X - Z n m X Z n Either of these is acceptable. Recall that the formula is built upon the concept of the normal probability distribution. The probability that a continuous variable is exactly equal to any particular number is zero. So it doesn’t matter whether you include the endpoints of the interval or not. Determining Z values for confidence intervals 0.9800 0.01 -k -2.33 0.01 0 k 2.33 Z Suppose we want a 98% confidence interval. We need to find 2 values, call them –k and k, such that Z is between them 98% of the time. Then Z will be less than k with probability 0.99. Look in the body of the Z table for the value closest to 0.99, which is 0.9901 . The number on the border of the table corresponding to 0.9901 is 2.33. So that is your value of k, and the number you use for Z in your confidence interval. Sometimes 2 numbers in the Z table are equally close to the value you want. For example, suppose you want a 90% confidence interval. Remember the Z table gives cumulative values. So to get a value you can look up in the table, you add the 0.90 from the middle area of your Z graph plus 0.05 from the left tail for a total of 0.95. So you look for 0.95 in the body of the Z table. You find 0.9495 and 0.9505. Both are off by 0.0005. The number on the border of the table corresponding to 0.9495 is 1.64. The number corresponding to 0.9505 is 1.65. Usually in these cases, we use the average of 1.64 and 1.65, which is 1.645. Similarly for the 99% confidence interval, we usually use 2.575. (Draw your graph & work through the logic of this number.) Which interval is wider: One with a higher confidence level (such as 99%) or one with a lower confidence level (such as 90%)? Let’s think it through using an unrealistic but slightly entertaining example. You have the misfortune of being stranded on an island, with a cannibal & a bunch of bears. It gets worse… You get captured by the cannibal. The cannibal, who knows the island well, decides to give you a chance to avoid being dinner. He says if you can correctly estimate the number of bears, he’ll let you go. To give you a fighting chance, he’ll let you give him an interval estimate. You think that there are probably about a hundred bears on the island. Would you be more confident of not being dinner if you gave the cannibal a narrow interval like 90 to 110 bears, or a wider one like 75 to 125 bears? You would definitely be more confident with the wider interval. Thus, when the confidence level needs to be very high (such as 99%), the interval needs to be wide. Let’s redo the IQ example with a different confidence level. We had a sample of 25 students with a sample mean IQ of 127. The population standard deviation was 5.4 . Calculate the 99% confidence interval for the population mean. Our general formula is: X -Z m X Z n n We said that the Z value for 99% confidence is 2.575. Putting in our values, 5.4 5.4 127 - 2.575 m 127 2.575 25 25 or 124.22 < m < 129.78 We had for the 95% confidence interval: 124.88 < m < 129.12 We just got for the 99% confidence interval: 124.22 < m < 129.78 The 99% confidence interval starts a little lower & ends a little higher than the 95% interval. So the 99% interval is wider than the 95% interval, as we said it should be. What do we do if we want to compute a confidence interval for m, but we don’t know the population standard deviation ? We use the next best thing, the sample standard deviation s. But with s, instead of a Z distribution, we have a t (with n-1 degrees of freedom). So, X -Z m X Z n n X - t n -1 s m X t n -1 n becomes s n Example: From a large class of normally distributed grades, sample 4 grades: 64, 66, 89, & 77. Calculate the 95% confidence interval for the class mean grade m. X - t n -1 s m X t n -1 n s n is the appropriate formula. So we need to determine the sample mean, sample standard deviation, and the t-value. 4 grades: 64, 66, 89, & 77 95% confidence interval for m X 64 66 89 77 296 Adding our X values, we get 296. 4 grades: 64, 66, 89, & 77 95% confidence interval for m X 64 66 89 77 296 X 74 Dividing by 4, we find our sample mean is 74. 4 grades: 64, 66, 89, & 77 95% confidence interval for m X 64 66 89 77 296 X 74 X X (X X ) -10 -8 15 3 2 Keep in mind that the sample standard deviation is n s (X X ) 2 i 1 n 1 So, next we subtract our sample mean 74 from each of our X values, 4 grades: 64, 66, 89, & 77 95% confidence interval for m X 64 66 89 77 296 X 74 X X (X X ) -10 100 -8 64 15 225 3 9 398 2 square the differences and add them up. n s 2 ( X X ) i 1 n 1 4 grades: 64, 66, 89, & 77 95% confidence interval for m X 64 66 89 77 296 X 74 X X (X X ) -10 100 -8 64 15 225 3 9 398 s2 = 398/3 =132.7 2 Then we divide by n-1 (which is 3) to get the sample variance s2, n s 2 ( X X ) i 1 n 1 4 grades: 64, 66, 89, & 77 95% confidence interval for m X 64 66 89 77 296 X 74 X X (X X ) -10 100 -8 64 15 225 3 9 398 s2 = 398/3 =132.7 s = 11.5 2 and take the square root to get the sample standard deviation s. n s 2 ( X X ) i 1 n 1 So we have X 74 and s = 11.5 Since n = 4, dof = n-1 = 3 Since we want 95% confidence, we want 0.95 as the middle area of our graph, and .025 in each of the 2 tails. 0.025 0.95 We find the 3.182 in our t table. s Our formula is m X t X - t n -1 n -1 n 0 0.025 3.182 t3 s n Putting in our numbers we have 11.5 74 - 3.182 4 11.5 m 74 3.182 4 So our 95% confidence interval is 56 < m < 92. The interval is very wide, because we only have 4 observations. If we had more information, we’d be able to get a narrower interval. From our previous confidence intervals, we can see that we have a basic format, which can be used when the point estimator is roughly normal. X -Z m X Z n n X - t n -1 point estimate z or t s m X t n -1 n std . dev. or estimate of the std. dev. of our pt. estimate Desired parameter s n point estimate z or t std . dev. or estimate of the std. dev. of our pt. estimate Calculating confidence intervals for the binomial proportion parameter When the number events of interest (X) and the number of events not of interest (n-X) are each at least five, the binomial distribution can be approximated by the normal and we can develop a confidence interval for the binomial proportion parameter . That is, we can develop a confidence interval for , if X ≥ 5 and n-X ≥ 5 . We need a point estimate for , & the standard deviation of our point estimate. For the point estimate we will use the binomial proportion variable X/n or p . Its standard deviation was (1 ) n . Since we don’t know , we will use our sample proportion p in the standard deviation formula. Use our format to get the confidence interval for the binomial proportion . point estimate z or t std . dev. or estimate of the std. dev. of our pt. estimate Desired parameter point estimate p (1 p) p(1 p) pz pz n n z or t std . dev. or estimate of the std. dev. of our pt. estimate We have our confidence interval for the binomial proportion . p (1 p) p(1 p) pz pz n n Example: Consider a random sample of 144 families; 48 have 2 or more cars. Compute the 95% confidence interval for the population proportion of families with 2 or more cars. pz p (1 p) p(1 p) pz n n n = 144 48 1 p 144 3 2 1 p 3 0.95 0.0250 -1.96 0.0250 0 1.96 Z Looking up the cumulative area 0.9500 + 0.0250 = 0.9750, we find our z value is 1.96 . 1 2 We now have n = 144, z = 1.96, p and 1 p 3 3 pz p (1 p) p(1 p) pz n n 1 2 1 2 3 3 3 3 1 1 1.96 1.96 3 144 3 144 0.333 0.077 0.333 0.077 So our 95% confidence interval for is: 0.256 < < 0.410 . Suppose we want a confidence interval not for a mean but for the difference in two means (m1-m2). For example, we may be interested in the difference in the mean income for two counties, or the difference in the mean exam scores for two classes. We will use the same basic format, but it will be a bit more complicated. point estimate z or t std . dev. or estimate of the std. dev. of our pt. estimate Desired parameter point estimate z or t std . dev. or estimate of the std. dev. of our pt. estimate Our “desired parameter” is m1 – m2 . Our point estimate is X X 1 2 . Initially, we will assume that we have the population standard deviations, so we will use a z. We need the standard deviation of the point estimate, X X 1 To get that we will first determine the variance of X X 1 2 . 2 . Recall: V(aX + bY) = a2V(X) + b2V(Y) + 2ab[C(X,Y)] Letting a = 1, b = -1, X X 1 and Y X 2 V ( X 1 X 2 ) (1) 2 V ( X 1 ) (1) 2 V ( X 2 ) 2(1)(1)C( X 1 , X 2 ) If our samples are independent, the covariance term is zero, and the expression becomes V ( X 1 X 2 ) (1) 2 V ( X 1 ) (1) 2 V ( X 2 ) or V (X 1 X 2 ) V (X 1) V (X 2 ) We now have V ( X 1 X 2 ) V ( X 1 ) V ( X 2 ) 2 Recall that V ( X ) . n Applying subscripts for our samples, 12 22 V (X 1 X 2 ) n1 n2 & the standard deviation of X X 1 12 n1 22 n2 2 is Apply our basic format point estimate z or t std . dev. or estimate of the std. dev. of our pt. estimate (X 1 X 2 ) z 12 n1 22 n2 Desired parameter point estimate z or t m1 m 2 ( X 1 X 2 ) z std . dev. or estimate of the std. dev. of our pt. estimate 12 n1 22 n2 Example: From 2 large classes, with normally distributed grades, sample 4 grades (64, 66, 89, & 77) & 3 grades (56, 71, & 53). If the population variances for the 2 classes are both 96, compute the 90% confidence interval for the difference in means of the class grades. We will use the formula we just developed: (X 1 X 2 ) z 12 n1 22 n2 m1 m 2 ( X 1 X 2 ) z 12 n1 22 n2 We need the 2 sample means & the z value. Adding the observations & dividing by the number of observations, our sample means are (64 + 66 + 89 + 77) / 4 = 74 and (56 + 71 + 53) / 3 = 60 The z value for 90% confidence, as we found before, is 1.645 . 0.90 0.05 0.05 0 1.645 Z Assembling our formula: (X 1 X 2 ) z 12 n1 22 n2 m1 m 2 ( X 1 X 2 ) z 12 n1 22 n2 96 96 96 96 (74 60) 1.645 m1 m 2 (74 60) 1.645 4 3 4 3 14 12.31 m1 m 2 14 12.31 1.69 m1 m2 26.31 Interpreting the results 1.69 m1 m2 26.31 We are 90% certain that the difference in class mean grades is between 1.69 and 26.31 . Notice that this interval does not include zero. If m1 – m2 = 0, then m1 = m2 . That implies that the probability is less than 10% that the class mean grades are equal. What do we do if we want to compare means, but we don’t know the population variances? As before, we use the sample variances & the t distribution. Our formula was (X 1 X 2 ) z 12 n1 22 n2 m1 m 2 ( X 1 X 2 ) z 12 n1 22 n2 Now the formula is s12 s22 s12 s22 (X 1 X 2 ) t m1 m 2 ( X 1 X 2 ) t n1 n2 n1 n2 For the t, the number of degrees of freedom is determined by a very messy formula. The degrees of freedom for the t for the confidence interval for the difference between means with unknown variances dof the integer part of 2 s s n1 n2 2 2 2 2 s1 s2 n1 n2 n1 1 n2 1 2 1 2 2 Let’s do the same example as before, but without knowing the population variances. From 2 large classes, with normally distributed grades, sample 4 grades (64, 66, 89, & 77) & 3 grades (56, 71, & 53). Compute the 90% confidence interval for the difference in means of the class grades. This time we need to calculate the sample variances. n Recall s 2 2 ( X X ) i i 1 n 1 We calculate the sample means as before. Class 1 X1 X 1 X 1 ( X1 X 1 )2 Class 2 X2 64 56 66 71 89 53 77 296 296 4 74 X1 180 180 3 60 X2 2 X2 X 2 (X2 X 2 ) n Recall s 2 2 ( X X ) i i 1 n 1 Then subtract the sample mean from each observation, square that difference, Class 1 X1 Class 2 X 1 X 1 ( X1 X 1 )2 X2 2 X2 X 2 (X2 X 2 ) 64 -10 100 56 -4 16 66 -8 64 71 11 121 89 15 225 53 -7 49 77 3 9 296 296 4 74 X1 180 180 3 60 X2 n Recall s 2 2 ( X X ) i and add up. i 1 n 1 Class 1 X1 Class 2 X 1 X 1 ( X1 X 1 )2 X2 2 X2 X 2 (X2 X 2 ) 64 -10 100 56 -4 16 66 -8 64 71 11 121 89 15 225 53 -7 49 77 3 9 296 296 4 74 X1 398 180 180 3 60 X2 186 n Recall s 2 2 ( X X ) i i 1 n 1 Dividing by n-1, we have our sample variances. Class 1 X1 Class 2 X 1 X 1 ( X1 X 1 )2 X2 2 X2 X 2 (X2 X 2 ) 64 -10 100 56 -4 16 66 -8 64 71 11 121 89 15 225 53 -7 49 77 3 9 296 296 X1 4 74 398 398 3 132.67 s12 180 180 X2 3 60 186 186 2 93.0 s 22 What are the dof & t value? dof the integer part of 2 s12 s22 n1 n2 2 2 s12 s22 n1 n2 n1 1 n2 1 2 132.67 93.0 4 3 = 4.860 2 2 132.67 93.0 4 3 3 2 So the degrees of freedom is the integer part of 4.86 or 4. 0.90 0.05 For 90% confidence & 4 dof, the t value is 2.1318 . 0 2.1318 t4 Assemble our formula s12 s22 s12 s22 (X 1 X 2 ) t m1 m 2 ( X 1 X 2 ) t n1 n2 n1 n2 (74 60) 2.1318 132.67 93.0 132.67 93.0 m1 m2 (74 60) 2.1318 4 3 4 3 14 17.08 m1 m2 14 17.08 3.08 m1 m2 31.08 Notice here that zero is contained in our 90% confidence interval. So we can’t rule out the possibility that the class mean grades are equal. Sometimes we believe the variances of 2 populations are equal, even though we don’t know the actual values. We have another confidence interval for this situation. (X 1 X 2 ) z 12 n1 22 n2 m1 m 2 ( X 1 X 2 ) z 12 n1 22 n2 In our earlier formula above, we can drop the distinguishing subscripts on our variances. (X 1 X 2 ) z 2 n1 2 n2 m1 m 2 ( X 1 X 2 ) z 2 n1 2 n2 Factoring out the variance, we have 1 1 1 1 2 ( X 1 X 2 ) z m1 m 2 ( X 1 X 2 ) z n1 n2 n1 n2 2 Next we replace the variance by a pooled sample variance, based on information from both samples. 1 1 1 1 2 ( X 1 X 2 ) t s p m1 m 2 ( X 1 X 2 ) t s p n1 n2 n1 n2 The dof for the t value is n1 + n2 – 2 . 2 The pooled sample variance (n1 1)s1 (n2 1)s2 sp n1 n2 2 2 2 2 When the 2 samples are the same size, this estimator gives an estimate that is halfway between the two sample variances. When the samples are not the same size, the estimate will be closer to the sample variance from the larger sample. So our confidence interval for the difference in the population means, when we don’t know the population variances but we believe that they are equal is: 1 1 1 1 2 ( X 1 X 2 ) t s p m1 m 2 ( X 1 X 2 ) t s p n1 n2 n1 n2 2 where 2 2 ( n 1 ) s ( n 1 ) s 1 2 2 s p2 1 n1 n2 2 and the number of degrees of freedom is n1 + n2 – 2 . Let’s do the same example as before, assuming that the unknown population variances are believed equal. We had: X 1 74, X 2 60, s1 132.67, s 2 93.0 2 2 2 ( n 1 ) s ( n 1 ) s 1 2 2 s p2 1 n1 n2 2 (3)132.67 (2)93.0 432 584 116.8 5 2 We have: X 1 74, X 2 60, s 2p 116.8 We want 90% confidence. dof = n1 + n2 – 2 = 4 + 3 – 2 = 5 So our t value is 2.015 . 0.90 0.05 0 2.015 t5 We have: X1 74, X 2 60, s 2p 116.8, t 2.015 1 1 1 1 2 ( X 1 X 2 ) t s p m1 m 2 ( X 1 X 2 ) t s p n1 n2 n1 n2 2 1 1 1 1 (74 60) 2.015 116.8 m1 m 2 (74 60) 2.015 116.8 4 3 4 3 14 16.63 m1 m 2 14 16.63 2.63 m1 m 2 30.63 We are 90% certain that the difference in the population means is between -2.63 & 30.63. Again, since zero is in this interval, we can’t rule out the possibility that the class mean grades are equal. We can also develop a confidence interval for the difference in population proportions 1 – 2 The point estimate is the difference in the sample proportions p1 p2 Next we need the standard deviation of our point estimate. Similarly to the case of the difference in population means, V ( p1 p2 ) V ( p1 ) V ( p2 ) Recalling that our previous estimate of V ( p) was we have p (1 p ) , n p1 (1 p1 ) p2 (1 p2 ) V ( p1 p2 ) n1 n2 The estimated standard deviation of our point estimate becomes p1 (1 p1 ) p2 (1 p2 ) n1 n2 Using our basic format, we find the confidence interval for the difference in population proportions. point estimate z or t ( p1 p2 ) z std . dev. or estimate of the std. dev. of our pt. estimate Desired parameter point estimate z or t std . dev. or estimate of the std. dev. of our pt. estimate p1 (1 p1 ) p2 (1 p2 ) p (1 p1 ) p2 (1 p2 ) 1 2 ( p1 p2 ) z 1 n1 n2 n1 n2 Example: Samples from 2 states show proportions of Democrats 1/3 & 1/5 with sample sizes 100 & 225. Calculate the 99% confidence interval for the difference in population proportions. ( p1 p2 ) z p1 (1 p1 ) p2 (1 p2 ) p (1 p1 ) p2 (1 p2 ) 1 2 ( p1 p2 ) z 1 n1 n2 n1 n2 The z value for a 99% confidence interval is 2.575 . 0.99 0.005 0.005 0 2.575 Z We have: p1 0.33, 1-p1 0.67, p2 0.20, 1 p2 0.80, n1 100, n 2 225, z 2.575 Applying the formula ( p1 p2 ) z p1 (1 p1 ) p2 (1 p2 ) p (1 p1 ) p2 (1 p2 ) 1 2 ( p1 p2 ) z 1 n1 n2 n1 n2 yields (0.33)(0.67) (0.20)(0.80) (0.33 0.20) 2.575 1 2 100 225 (0.33 0.20) 2.575 or (0.33)(0.67) (0.20)(0.80) 100 225 0.13 0.14 1 2 0.13 0.14 So the 99% confidence interval for the difference in population proportions is 0.01 0.27 1 2 Given our confidence interval: 0.01 1 2 0.27 Can we conclude that the two population proportions are not equal? No. Since zero is in the interval, 1 may equal 2 . How do you decide the appropriate sample size for a project? 2 Decisions: • Desired confidence level • Maximum difference D between the estimate of the population parameter & the true value of the population parameter (that is, the maximum error you’re willing to accept) For example, if you’re estimating the population mean m using the sample mean X , D is themaximumdifferencebetween X andm . Suppose you have chosen 95% as your desired confidence level. You know that there is a z value (call it z0) such that - z0 < Z < z0 95% of the time. Xm You also know that is distributed as a Z. n So, z 0 Xm n z 0 with 95% probabilit y . With 95% probabilit y , we hav e z 0 Xm z0 . n Multiply ing by n , we hav e z 0 n X m z0 . n We see here that the largest v alue of the dif f erencebetween X and m , which we called D, is z 0 So, D z 0 n n . We hav e now, D z 0 , n and we can solv e f or the sample size n. First, square both sides of the equation: Multiply through by n: Divide through by D 2: Dropping the subscript on z for convenience, we have the formula: D z0 2 2 2 n nD z 0 2 2 n z0 2 nz 2 D 2 2 2 D 2 2 So we have a formula for determining the appropriate sample size n when we want to estimate the population mean. nz 2 D 2 2 Example: Suppose you’re trying to estimate the mean monthly rent of 2-bedroom apartments in towns of 100,000 people or less. The population standard deviation is 20. You want to be 95% sure that your estimate is within $3 of the true mean. How large a sample should you take? nz 2 2 2 D 2 2 (20 ) (1.96) (3) 2 170.3 You need to sample 171 observations. It’s not 170, because sample sizes smaller than 170.3 provide you with less information & therefore less than the desired level of confidence. Our formula for n has the population standard deviation in it. What do we do if we don’t know ? In the past, we used the sample standard deviation s. Why can’t we do that here? s came from the sample. We haven’t taken the sample yet. We’re still trying to figure out how many observations our sample should have. If previous researchers have done related work, you may be able to use their estimate for the standard deviation. Alternatively, you can do a small preliminary sample, & based on that information, estimate the standard deviation. Determining the appropriate sample size n for estimating the population proportion . Again we will use - z0 < Z < z0 with the desired confidence level as our starting point. We know that p- (1- ) is approximately standard normal. n So with the desired level of confidence, -z 0 p (1 ) n z0 . Starting from p -z 0 (1 ) z0 n Multiply through by (1- ) n , -z 0 (1- ) n p z0 We see here that the maximum difference D between our estimator p & our parameter is: z0 (1- ) n (1- ) n We have now, D z0 (1 ) n and we can solve for the sample size n. First, square both sides of the equation: Multiply through by n: Divide through by D 2: Dropping the subscript on z for convenience, we have the formula: D z 2 2 0 (1 ) n nD z (1 ) 2 2 0 nz nz 2 0 2 (1 ) D 2 (1 ) D2 There’s one big problem with this formula. What is it? nz 2 (1 ) D2 We want to collect a sample in order to estimate , but we have the unknown in our equation for determining the sample size! We can’t use the sample proportion as we did before, because we haven’t taken the sample yet. As it happens, we can resolve this problem fairly easily. The largest possible value for (1 ) occurs when is ½, and that largest value is ¼. Play with some values for & 1- , and convince yourself that this is true. For example, (1/3)(2/3) = 2/9 < 1/4 (3/10)(7/10) = 21/100 < 1/4 (1/100)(99/100) = 99/10,000< 1/4 If we know the largest possible value for pq, we can determine the largest sample size we should need for nz 2 (1 ) D2 Plugging in the maximum value of ¼ for (1 ), we have 1 4 2 1 1 2 z 2 nz 2 D 4 D z2 1 2 4 D 2 So our formula for n is: z n 4D2 2 z 4D 2 Sometimes you have a rough idea of what is, but you’re trying to get a more precise value. You can use your rough idea to determine the sample size. If 1 is your rough idea, then the sample size formula becomes nz 2 1 (1 1 ) D2 So we have 2 formulae for determining the appropriate sample size for estimating the population proportion. If you have no idea at all what is, you use: If you have a rough idea of 1 for the value of , you use: 2 z n 2 4D nz 2 1 (1 1 ) D2 Example: We are estimating the proportion of families with 2 or more cars. We want to be 95% certain that the estimate is within 3% (0.03) of the correct percentage. What is the necessary sample size? We’re clueless on the proportion , so we use the formula The z value for 95% confidence is 1.96 . 2 z n 2 4D 0.95 0.0250 0.0250 0 Filling in our values, we get 2 2 1.96 Z (1.96) z n 1067.1 2 2 4(0.03) 4D So the needed sample size is 1068.