Understanding Sample Slope & Confidence Intervals in Statistical Inference & Regression - , Study notes of Data Analysis & Statistical Methods

Download Understanding Sample Slope & Confidence Intervals in Statistical Inference & Regression - and more Study notes Data Analysis & Statistical Methods in PDF only on Docsity! Lecture 29 Nancy Pfenning Stats 1000 Reviewing Confidence Intervals and Tests for Ordinary One-Sample, Matched- Pairs, and Two-Sample Studies About Means Example Blood pressure X was measured for a sample of 10 black men. It was found that x̄ = 114.9, s = 10.84. Give a 90% confidence interval for mean blood pressure µ of all black men. [Note: we can assume that blood pressure tends to differ for different races or genders, and that is why a separate study is made of black men—the confounding variables of race and gender are being controlled.] This is an ordinary one-sample t procedure. A level .90 confidence interval for µ is x̄ ± t∗ s√ n , where t∗ has 10 − 1 = 9 df. Consulting the df = 9 row and .90 confidence column of Table A.2, we find t∗ = 1.83. Our confidence interval is 114.9± 1.83 10.84√ 10 = (108.6, 121.2). Here is what the MINITAB output looks like: N MEAN STDEV SE MEAN 90.0 PERCENT C.I. calcbeg 10 114.90 10.84 3.43 ( 108.62, 121.18) Example Blood pressure for a sample of 10 black men was measured at the beginning and end of a period of treatment with calcium supplements. To test at the 5% level if calcium was effective in lowering blood pressure, let the R.V. X denote decrease in blood pressure, beginning minus end, and µD would be the population mean decrease. This is a matched pairs procedure. To test H0 : µD = 0 vs. Ha : µD > 0, we find differences X to have sample mean d̄ = 5.0, sample standard deviation s = 8.74. The t statistic is t = d̄−µ0s √ n = 5−08.74 √ 10 = 1.81, and the P-value is P (T ≥ 1.81). We refer to Table A.2 for the t(9) distribution, and see that 1.81 is just under 1.83, which puts our P-value just over .05. Our test has not quite succeeded in finding the difference to be significantly greater than zero, in a statistical sense. Populations of black men treated with calcium may experience no decrease in blood pressure. MINITAB output appears below. TEST OF MU = 0.00 VS MU G.T. 0.00 N MEAN STDEV SE MEAN T P VALUE calcdiff 10 5.00 8.74 2.76 1.81 0.052 It is possible that our sample size was too small to generate statistically significant results. An- other concern is the possibility of confounding variables influencing their blood pressure change. The placebo effect may tend to bias results towards a larger decrease. Or, time may play a role: if the beginning or end measurement date happened to be in the middle of a harsh winter or a politically stressful time, results could be affected. Example Data for a control group (taking placebos) of 11 black men at the beginning and end of the same time period produced control sample mean difference d̄2 = −.64, and s2 = 5.87. Now we test H0 : µ1 − µ2 = 0 [same as H0 : µ1 = µ2, or mean difference for calcium-takers same as mean difference for placebo-takers] vs. 114 Ha : µ1 −µ2 > 0 [same as H0 : µ1 > µ2, or mean difference for calcium-takers greater than mean difference for placebo-takers]. The t statistic is t = (d̄1 − d̄2) − 0 √ s2 1 n1 + s2 2 n2 = 5 − (−.64) √ 8.742 10 + 5.901 2 11 = 5.64 3.282 = 1.72. Since 10 − 1 < 11 − 1, use 9 df. In this row of Table A.2, we see that 1.72 is smaller than 1.83, so the P-value is larger than .05. Once again there is not quite enough evidence to reject the null hypothesis. [Note that in the MINITAB output below, degrees of freedom were calculated (in a very complicated way) to be 15, whereas we simply took the smaller sample size minus one, which was 9.] TWOSAMPLE T FOR calcdiff VS contdiff N MEAN STDEV SE MEAN calcdiff 10 5.00 8.74 2.76 contdiff 11 -0.64 5.87 1.77 TTEST MU calcdiff = MU contdiff (VS GT): T= 1.72 P=0.053 DF= 15 Robustness: the sample sizes are quite small, and so MINITAB plots of the above distributions should be consulted to verify that they have no pronounced outliers or skewness. Data values are shown in the following table: Calcium Control Beginning End Difference Beginning End Difference 107 100 7 123 124 -1 110 114 -4 109 97 12 123 105 18 112 113 -1 129 112 17 102 105 -3 112 115 -4 98 95 3 111 116 -5 114 119 -5 112 102 10 112 114 -2 136 125 11 110 121 -11 102 104 -2 117 118 -1 107 106 1 119 114 5 130 133 -3 N MEAN MEDIAN TRMEAN STDEV SEMEAN calcbeg 10 114.90 111.50 113.88 10.84 3.43 calcend 10 109.90 109.00 109.25 7.80 2.47 calcdiff 10 5.00 4.00 4.62 8.74 2.76 contbeg 11 113.27 112.00 113.11 9.02 2.72 contend 11 113.91 114.00 113.89 11.33 3.42 contdiff 11 -0.64 -1.00 -0.89 5.87 1.77 Example A biologist suspects that the antiseptic Benzamil actually impairs the healing process. To test her suspicions with a matched-pairs design, 9 salamanders are randomly selected for treatment. Each has one wounded hind leg treated with Benzamil, the other wounded leg treated with saline (as a control). The healing X (area in square millimeters covered with new skin) is measured 115 

 
 
 
 
 


             !  " # "  $ # $  % # %   # " $ %  &'() * '+ , -. / 0. 12 3 Example Producers of gasoline want to test which is better, Gas A or Gas B. Miles per gallon are measured for 6 cars using Gas A and for another set of 6 cars using Gas B. Gas A Gas B 15 13 20 17 25 23 25 24 30 28 35 34 x̄1 = 25.00 x̄2 = 23.17 s1 = 7.07 s2 = 7.52 n1 = 6 n2 = 6 The two-sample t statistic is 25.00−23.17√ 7.072 6 + 7.52 2 6 = .43. The P-value, according to MINITAB, is ex- tremely large: .674. There is no evidence at all that µ1 6= µ2, because the P-value is large, because t is small, because s1 and s2 are large. High variation among mileages for various cars prevented us from pinning down the effects of using a different gas. A matched pairs design would be better: Example Producers of gasoline measure mpg for 6 cars using Gas A and for the same 6 cars using Gas B. Which gas is used first is determined by a coinflip. Gas A Gas B Difference 15 13 2 20 17 3 25 23 2 25 24 1 30 28 2 35 34 1 d̄ = 1.833 sd = .753 Now the t statistic is 1.833−0 .753/ √ 6 = 5.97 and, according to MINITAB, the P-value is extremely small: .002. We have strong evidence against H0 : µd = 0 because the P-value is very small, because 118 t is large, because s is small. Concentrating on the difference between mileages, Gas A minus Gas B, wipes out the differences among mileages for various cars, and helped control this outside variable. Lecture 30 Chapter 14: More About Regression In Chapter 5, we displayed the relationship between two quantitative variables with a scatterplot, and summarized it by reporting direction, form, and strength. If the form appeared linear, then we made a much more specific summary by describing the relationship with the equation of a straight line, called the least squares regression line. We also used the correlation r to specify the direction and strength of the relationship. All of this was done for the sample of explanatory and response pairs only. By construction our line was the one that best fitted the sample data points, but we did not attempt to draw conclusions about how the explanatory and response variables were related in the entire population from which the sample was taken. Now that we are familiar with the principles of statistical inference, our goal in this chapter is to use sample explanatory and response values to draw conclusions about how the variables are related for the population. It should go without saying that such conclusions will only be meaningful if the sample is truly representative of the larger population. As usual, all results are based on probability distributions, which tell us what we can expect from random behavior. The first step in this inference process is an important one: examine the scatterplot to decide if the form of the relationship really does appear linear. The methods of inference that we will develop cannot help us in producing evidence that a straight-line relationship holds for the larger population—this is something that we must decide for ourselves, based on the appearance of the scatterplot. If the points seem to cluster around a curve rather than a straight line, then other, more advanced options must be explored. In more advanced treatments of relationships between two quantitative variables, methods are presented for transforming variables so that the resulting relationship is linear. In this book, we will proceed no further if the relationship is non-linear. If linearity seems to be a reasonable assumption, then we can use inference to draw conclusions about what the line should be like for the entire population, and also about how much spread there is around the line. Whereas in Chapter 5 we only went so far as to predict a single value for the response to a given explanatory value, we will now have the tools to make interval estimates. Our first example concerns two variables which common sense suggests should have a positive linear relationship: ages of students’ mothers and fathers. Example Are ages of all students’ mothers and fathers related? If so, what can we say about this relation- ship for a population of age pairs? Because couples tend to have ages that are reasonably close to one another, we would expect the mother to be on the young side if the father is young, and on the old side if the father is old. There is reason to expect a rather steady increase in the variable MotherAge as values of the variable FatherAge increase. Therefore, we do expect the relationship to be positive and linear, not just for a sample of age pairs but also for the larger population. Using the methods of Chapter 5, we can look at a scatterplot of father and mother ages and decide that it appears linear. The least squares method can be used to fit a line that comes “closest” to the data points, in the sense that it minimizes the sum of squared residuals, which are the sample prediction errors. The typical size of sample prediction error is S in the output; it calculates the square root of the “average” squared distance of observed response values minus predicted response values. (This average is calculated dividing by n−2, which will be our regression degrees 119 of freedom.) 807060504030 80 70 60 50 40 30 FatherAge M ot he rA ge S = 3.28794 MotherAge = 14.5418 + 0.665761 FatherAge Regression Plot R-Sq = 61.0 % The regression equation is MotherAge = 14.5418 + 0.665761 FatherAge S = 3.28794 R-Sq = 61.0 % R-Sq(adj) = 60.9 % 431 cases used 15 cases contain missing values Predictor Coef SE Coef T P Constant 14.542 1.317 11.05 0.000 FatherAge 0.66576 0.02571 25.89 0.000 Pearson correlation of FatherAge and MotherAge = 0.781 P-Value = 0.000 You may have noticed that a p-value is always reported along with the correlation r, and that t statistics and p-values are included with the regression output. We will pay more attention to these once we have established a meaningful hypothesis test procedure for the regression context. First, we consider relationships for populations as opposed to for samples of quantitative explanatory and response values. When we introduced the process of performing statistical inference about a single parameter (such as population mean) based on a statistic (such as sample mean), we acknowledged that although sample mean may be our best estimate for population mean, it is almost surely “off” by some amount. Similarly, although the least squares regression line is our best guess for the line that describes the relationship for the entire population, it is also probably “off” to some extent. Unlike inference about a single parameter like unknown population mean, when we perform inference about a relationship between two quantitative variables, there are actually three unknown parameters. • First of all, we only know the slope b1 = .666 of the line that best fits our sample. The line that best fits the entire population may have more or less slope. We use β1 to denote the unknown slope of the population regression line. The graph below shows that other slopes are plausible candidates for β1. 120 70605040 65 55 45 35 FatherAge M ot he rA ge Each distribution of MotherAges is centered at the mean response to all such FatherAges (on the population regression line) The standard deviation of each distribution is sigma The shape of each distribution is normal Population relationship expressed as µy = β0 + β1x Notice that a new symbol “µy” has been used to model the relationship in the larger population. Because statistics concerns itself with drawing conclusions about populations, based on samples, we must always be sure to distinguish between parameters (describing populations) and statistics (describing samples). In the case of a regression line, we have already used the notation ŷ to refer to the response predicted for the sample: ŷ = b0 + b1x, where b0 and b1 are calculated from the sample data. The corresponding parameter is µy = β0 + β1x, the unknown population mean response to a given explanatory value x, which responds linearly with an unknown intercept β0 and slope β1. [Note: the mean response µy is the same thing as expected response E(y).] Distribution of sample slope b1 As always, we report the long-run behavior of a sample statistic by describing its distribution, specifically by telling its center, spread, and shape. • (Center:) If the previously mentioned requirements are met—linear scatterplot, normally distributed residuals, and apparently constant spread about the line—then slope b1 of the least squares line for a random sample of explanatory/response pairs has mean equal to the unknown slope β1 of the least squares line for the population. • (Spread:) The standard deviation of sample slope b1 is σb1 = σ √ (x1 − x̄)2 + · · · + (xn − x̄)2 which we estimate with the standard error of b1, SEb1 = s √ (x1 − x̄)2 + · · · + (xn − x̄)2 where s, the estimate for spread σ about the population regression line, measures typical residual size. Although the above formula need not be used for calculations as long as software is available, it is worth examining SEb1 to see how the residuals contribute to the spread of the distribution of sample slope. The appearance of s in the numerator of SEb1 should make perfect intuitive sense: if the residuals as a group are small, then there is very little spread about the line and we should be able to pinpoint its slope fairly precisely. Conversely, if the residuals are large, then there is much spread about the line and there is a much wider range of plausible slopes. Note that the quantity √ (x1 − x̄)2 + · · · + (xn − x̄)2, which appears in the denominator, measures combined distances of explanatory values from their mean. This will be larger for larger sample sizes, and so b1 has less spread for larger samples. Again, our intuition tells us that we should be better able to pinpoint the unknown population slope β1 if we obtain a sample slope from a larger sample. 123 • (Shape:) Finally, b1 itself has a normal shape if the residuals are normal, or if the sample size is large enough to offset non-normality of the residuals. The graph below depicts what we have established about the distribution of sample slope b1 for large enough sample sizes: it is centered at population slope β1, has approximate standard deviation SEb1 , and follows a normal distribution. SEb1 beta1 b1 (sample slope) Distribution of standardized sample slope Recall that in Chapters 12 and 13, when we standardized sample mean using sample standard deviation s instead of unknown population standard deviation σ, the resulting random variable x̄−µ s/ √ n followed a t distribution instead of z. This could only be asserted if the sample size was large enough to offset any non-normality in the population distribution, so that the Central Limit Theorem could guarantee sample mean to be approximately normally distributed. In this chapter, we standardize sample slope b1 using SEb1 , calculated from s because σ is unknown. The resulting standardized slope t = b1 − β1 SEb1 follows a t distribution, and its degrees of freedom are n − 2, the same as for s. Again, this can only be asserted if b1 follows a normal distribution, which is the case if the sample is large enough to offset non- normality in the residuals. Remember also that for large samples, the t distribution is virtually identical to that of z. The distribution of standardized sample slope is displayed below: centered at zero as is any t distribution, standard deviation subject to degrees of freedom which are determined by sample size (in particular, standard deviation close to 1 if the sample size is large enough to make t roughly the same as z), and bell-shaped like any t distribution. sd as for t distribution with n-2 df 0 (b1-beta1)/SEb1 (standardized sample slope) Now that we know more about the behavior of b1 relative to β1, we will make use of the critical role played by β1 in the relationship between explanatory and response variables, so as to set up a test for evidence of 124 a relationship in the larger population. Because the construction of confidence intervals tends to be more intuitive than carrying out hypothesis tests, we start by setting up a confidence interval for the unknown slope of the linear relationship in the population. After that we will establish a procedure for testing the null hypothesis that slope for the population relationship is zero. In practice, it may make sense to carry out the test first, and then report the confidence interval for slope if there is statistical evidence of a relationship. Inference about β1 If the relationship between sampled values of two quantitative variables appears linear, then methods of Chapter 5 can be used to produce the line that best fits those sample values. For example, ages of students’ fathers and mothers produced the following regression output. Pearson correlation of FatherAge and MotherAge = 0.781 P-Value = 0.000 The regression equation is MotherAge = 14.5 + 0.666 FatherAge 431 cases used 15 cases contain missing values Predictor Coef SE Coef T P Constant 14.542 1.317 11.05 0.000 FatherAge 0.66576 0.02571 25.89 0.000 S = 3.288 R-Sq = 61.0% R-Sq(adj) = 60.9% The fact that r is +.781 tells us there is a fairly strong positive relationship between x and y data values. Based on the fact that b1 = .666, our best guess for how MotherAge responds to FatherAge is to predict that if one student’s father is 1 year older than a second student’s father, his mother would be .666 years older than the second student’s mother. By now we know enough about behavior of samples to realize that there must be some margin of error attached to this slope. For every additional year of FatherAge in the population, does MotherAge tend to be an additional .666 years, give or take about .1 years? Or .666 years, give or take about 1 year? As usual, the size of the margin of error will supply important information. In the former case, having evidence that population slope is in the interval (+.566, +.766) would convince us of a positive relationship, whereas in the latter case, where the range of plausible values (−.334, +1.666) for unknown population slope straddles zero, we could not claim to have statistical evidence of a relationship. Knowing enough about the distribution of sample slope relative to population slope will help us find the answer to our earlier question about the relationship between ages. Thus, we are ready to begin the process of statistical inference to draw conclusions about the relationship between two quantitative variables in a larger population, based on sample data about those variables. Confidence interval for β1 Example For a population of students’ parents, what does the age of the father tell us about the age of a mother? Specifically, if one father is a year older than another, how much older (if at all) do we expect the mother to be? The estimate b1 for the unknown slope β1 of the line that relates the variables MotherAge and FatherAge in the larger population is shown not only in the regression equation, but also as the coefficient of FatherAge in the second row of the output table. Predictor Coef SE Coef T P Constant 14.542 1.317 11.05 0.000 FatherAge 0.66576 0.02571 25.89 0.000 125 Example A website called “ratemyprofessor.com” reports students’ ratings of their professors at universities around the country. These are unofficial in that they are not monitored by the universities themselves. Besides listing average rating of the professors’ teaching on a scale of 1 to 5, where 1 is the worst and 5 is the best, there is also a rating of how easy their courses are, where 1 is the hardest and 5 is the easiest. Is there a relationship between the rating of teaching and the rating of ease? Offhand, we may suspect that students would favor easy teachers, in which case the relationship would be positive. On the other hand, teachers who are more conscientious may maintain higher standards, not just for their students but also for themselves. In this example, because the direction could really go either way, we should keep a more general two-sided alternative, and write H0 : β1 = 0 vs. Ha : β1 6= 0. We have already established that the distribution of sample slope b1, if certain conditions are met, is normal with mean β1 and approximate standard deviation SEb1 . Under the null hypothesis that β1 = 0, the standardized test statistic t = b1 − 0 SEb1 follows a t distribution with n−2 degrees of freedom. If the sample slope b1 is relatively close to zero (taking sample size and spread into account), then the standardized test statistic t is not especially large, and so the p-value is not small and there is no compelling evidence of a non-zero population slope β1. Thus, if b1 isn’t large enough, we cannot produce evidence that the two quantitative variables are related in the larger population. Conversely, if sample slope b1 is relatively far from zero, then t is large, the p-value is small, and we have statistical evidence that the population slope β1 is not zero. In other words, a large t results in a small p-value and a conclusion that the variables are related. If a one-sided alternative has been formulated and the sample slope b1 tends in the direction claimed by that alternative, then the p-value is the one-tailed probability of t being as extreme as the one observed. 0 0 t |t| is small p-value is large sample slope b1 close to zero 0 0 t |t| is large p-value is small sample slope b1 far from zero Example Let’s revisit the output for the regression of MotherAge on FatherAge, carrying out a five-step test of hypotheses. This will require us to focus our attention on the size of the t statistic and p-value. Predictor Coef SE Coef T P Constant 14.542 1.317 11.05 0.000 FatherAge 0.66576 0.02571 25.89 0.000 1. The null hypothesis states that the slope β1 of the line that relates MotherAge to FatherAge in the population is zero; alternatively, it may state that MotherAge and FatherAge are not related. Since common sense would suggest a positive relationship between the two variables, our alternative hypothesis would be that β1 > 0; alternatively, it could make a more general 128 claim that MotherAge and FatherAge are related for the general population of parents’ ages, that is, that β1 6= 0. 2. When we set up a confidence interval for unknown population slope β1, we noted that b1 and its standard error SEb1 are reported in the second row of the regression table. (The first row concerns intercept b0, which is not of particular interest to us, since we tend not to perform inference about intercept β0. Remember that it is the slope that provides key information about if and how the explanatory and response values are related.) Thus, the standardized sample slope t is reported as 25.89. Notice that the t statistic is easily calculated from b1 and SEb1 : t = b1 − 0 SEb1 = .66576− 0 .02571 = 25.89. (Remember that 0 is subtracted from sample slope b1 because for random samples b1 is centered at population slope β1, which is proposed to be 0 in the null hypothesis.) For a large sample like this, our cut-off point for “large” values of t is like that for z, namely 2. Obviously 25.89 is extremely large compared to 2. 3. The p-value corresponding to our t statistic is shown to be 0.000 (in the second row, not the first). 4. Just as t was extremely large, the p-value is extremely small. The fact that the p-value is so small tells us that obtaining a sample slope as far from zero as +.66576 would be extremely unlikely if population slope were zero, and so we conclude population slope is not zero. This p-value actually corresponds to a two-sided alternative; technically, if we suspected all along that the slope would be positive and formulated the alternative as β1 > 0, the p-value should be half of the one shown in the output. This only serves to strengthen our conclusion that ages are related. 5. To summarize, we have strong statistical evidence that MotherAge and FatherAge have a positive relationship, not just in our sample, but also in the larger population of students. Motivated by the example above, we summarize the process of testing for a relationship between two quantitative variables, by testing the null hypothesis that slope of the regression line for the population of explanatory and response values is zero. Hypothesis Test about β1 Just as for any hypothesis procedure, there are five basic steps to test for a relationship between two quantitative variables in the population of interest, based on a random sample of size n. 1. Assuming the relationship (if it exists) between two quantitative variables to be linear rather than curved, we test the null hypothesis that the variables are not related, which is equivalent to the claim H0 : β1 = 0 where β1 is the slope of the population least squares regression line. The alternative may state more generally that the two variables are related, which is equivalent to the claim Ha : β1 6= 0 or a more specific one-sided alternative may be formulated as Ha : β1 < 0 if we suspect in advance that the relationship is negative, or Ha : β1 > 0 if we suspect the relationship is positive. 2. Software should be used to produce the standardized sample slope t = b1−0SEb1 , which, when conditions below are met, follows a t distribution with n− 2 degrees of freedom. This t statistic is a standardized measure for how far sample slope b1 is from zero. 129 3. The p-value to accompany the t test statistic is the probability of a t random variable being as extreme as the one observed. It is reported alongside t as part of the regression output. A small p-value suggests that t is unusually extreme if the null hypothesis were true; that is, that the sample slope could be considered unusually steep if it were coming from a population where the explanatory and response variables were not related. 4. If the p-value is small, we reject the null hypothesis of no relationship (equivalent to rejecting the claim that slope β1 for the population is zero). If the p-value is not small, we conclude that the null hypothesis may be true. 5. Conclusions should be stated in context: if the null hypothesis has been rejected, we conclude that there is statistical evidence of a relationship between the explanatory and response variables. If it has been rejected against a one-sided alternative, we conclude there is evidence of a negative or of a positive relationship, depending on how the alternative has been expressed. If the null hypothesis has not been rejected, we conclude there is not enough statistical evidence to convince us of a relationship between the two quantitative variables. Results of the above test are only valid if the following conditions are met: • the scatterplot appears linear • the sample size is large enough to offset any non-normality in the response values • spread of responses is fairly constant over the range of explanatory values • explanatory/response pairs are independent of one another In the previous example, not only did we have strong evidence of a relationship (by virtue of the p-value being close to zero), but we also could assert that the relationship was strong (by virtue of the correlation r being .78, which is pretty close to one). It is nevertheless possible to produce weak evidence of a strong relationship, or strong evidence of a weak relationship. These possibilities will be explored in the following examples. We will also consider an example where there is no statistical evidence of a relationship. Example While most voters in a presidential election vote for the democratic or republican candidate, other parties do account for a small percentage of the popular vote in each state. The table below looks at the relationship between percentages voting democratic and republican in the year 2000 for just a few states. State Democratic Republican Alabama 48.4 47.9 California 53.4 41.7 Ohio 46.4 50.0 Minnesota 47.9 45.5 The points in the scatterplot below do appear to cluster around some straight line, rather than a curve. The line has a negative slope because when the percentage voting republican is low, then 130 By now we have established that his property’s assessed value ($40,000) is higher than average ($34,624), although its size (4,000 square feet) is smaller than average (5,619 square feet). This in itself is not enough evidence to argue that the assessment is unfair; the homeowner needs to show that in general the relationship between size and value is such that $40,000 would be an unreasonably high value for a lot of size 4,000 square feet. First let’s look at a scatterplot of the 29 size and value pairs: 0 2000 4000 6000 8000 10000 12000 0 10000 20000 30000 40000 50000 60000 70000 Size La nd V al ue Certainly the relationship appears to be positive, linear, and quite strong, suggesting that a smaller-than-average lot should be given a smaller-than-average assessment. This is confirmed by the output below, which shows the correlation to be quite close to one. Furthermore, the p- value is close to zero, providing evidence that this relationship should hold in the larger population from which the sample of land sizes and values was obtained. Pearson correlation of Size and LandValue = 0.927 P-Value = 0.000 An option when using software to perform a regression is to request a “prediction interval for new observation”, with results shown below for an observed size of 4,000 square feet: Predicted Values for New Observations New Obs Fit SE Fit 95.0% CI 95.0% PI 1 25094 1446 ( 22127, 28060) ( 11066, 39121) Values of Predictors for New Observations New Obs Size 1 4000 The output includes two very different intervals: one labeled “95.0% CI” that ranges roughly from $22,000 to $28,000, and one labeled “95.0% PI” that ranges roughly from $11,000 to $39,000. Both intervals are centered at $25,094, the predicted value for a lot of size 4,000 square feet. The first of the intervals is not especially relevant to the homeowner, because it presents a set of plausible values for mean value of all 4,000-square-foot lots in the neighborhood. The second interval reports a 95% prediction interval for the value of one individual lot whose size is 4,000 square feet. Since the assessed value of $40,000 falls above the interval ($11,066, $39,121), the homeowner does have statistical evidence that the assessment is unusually high, given the size of his lot. In order to put new inference skills in perspective, the following example includes a variety of estimates: estimating an individual or mean value of a quantitative variable; estimating an individual or mean response for a given explanatory value; and estimating an individual or mean response for a different explanatory value. We present a series of questions, all alike in that they seek estimates concerning male weight, but all different in terms of whether an estimate is sought for an individual or a mean, and also in terms of what height information, if any, is provided. 133 Example 1. Based on a sample of male weights, how do we estimate weight of an individual male? 2. Based on a sample of male weights, how do we estimate mean weight of all males? 3. Based on a sample of male heights and weights, how do we estimate weight of an individual 71-inch-tall male? 4. Based on a sample of male heights and weights, how do we estimate mean weight of all 71-inch-tall males? 5. Based on a sample of male heights and weights, how do we estimate weight of an individual 76-inch-tall male? 6. Based on a sample of male heights and weights, how do we estimate mean weight of all 76-inch-tall males? Estimating an individual weight with no height information In Chapter 2, we learned that if a distribution is roughly normal, and we know its mean and standard deviation, we can report a range for most of its values using the 68-95-99.7 Rule, which is based on a normal distribution. For example, the output below shows male weights to have mean 170.83 and standard deviation 33.06. Variable N N* Mean Median TrMean StDev WT_male 162 2 170.83 165.00 168.24 33.06 Variable SE Mean Minimum Maximum Q1 Q3 WT_male 2.60 115.00 315.00 150.00 185.00 If the shape of the distribution of weights is approximately normal, then about 95% of the time any one individual weight should fall within 2 standard deviations of the mean, from 170.83−2(33.06) to 170.83 + 2(33.06); that is, in the interval (104.71, 236.95). The accuracy of this interval is not necessarily to be trusted, because the shape of the distribution of weights shown in the histogram below is not entirely normal, but is rather right-skewed. 100 200 300 0 10 20 30 40 WT_male F re qu en cy Histogram of WT_male Although weights are often purported to be normal for specific age and gender groups, the reality is that most populations include individuals with weights that are unusually high to the point where they cannot be balanced out by unusually low weights. In our sample, for instance, the highest weight (315) is (315 − 170.83)/33.06 = 4.4 standard deviations above the mean; a man would have to weigh just 25 pounds to be this many standard deviations below the mean! In fact, the lowest weight (115) has a z-score of (115 − 170.83)/33.06 = −1.7, so it is only 1.7 standard deviations below the mean. 134 Estimating mean weight with no height information In Chapter 12, we learned to perform inference about the mean of a single quantitative variable. These methods can be used to set up a confidence interval for the mean weight of all male college students, based on a sample of weights. One-Sample T: WT_male Variable N Mean StDev SE Mean 95.0% CI WT_male 162 170.83 33.06 2.60 ( 165.70, 175.96) Thus, a 95% confidence interval for the mean weight of all male college students is (165.70, 175.96). Notice how much narrower this interval is than the interval that should contain an individual weight. The interval for individuals ranged all the way from about 105 to 237 pounds, with a width of 132 pounds. In contrast, the interval for mean weight ranged only from about 166 to 176 pounds, with a width of only 10 pounds. It is much harder to pinpoint an individual as opposed to a mean value. Remember that the spread of all values is estimated with s, while the spread of sample mean is estimated with s√ n . Whereas the non-normality of the distribution of weights presented a problem in setting up a range for 95% of individual weights, by virtue of the Central Limit Theorem, the large sample size guarantees sample mean weight to be approximately normal, and so this interval should be quite accurate. Including Height Information In fact, the confidence interval above is of limited usefulness, because instead of asking, “What is a typical weight for any male college student?” we would be more inclined to wonder, “What is a typical weight for a male college student who is x inches tall?” A range of plausible values for the mean weight of all male college students is no longer appropriate if we are specifically interested in what is plausible for the mean of, say, all 71-inch-tall male college students. In order to really do justice to the variable weight, the variable height should be taken into account. Inference for regression can be used to produce a range of plausible values for the mean weight of all male college students of a given height. Along the way, we will also take a look at the range of plausible values for the weight of an individual male college student of a given height, in order to contrast such intervals. We have already examined the distribution of weights alone. Now let’s examine the heights of our sample of 162 male college students, and look at the relationship between height and weight. Then, we’ll produce a 95% prediction interval for the weight of an individual 71-inch-tall male, along with a 95% confidence interval for the mean weight of all 71-inch-tall males. These in turn will be compared to 95% prediction and confidence intervals for a given height of 76 inches. In the end, we will contrast these to the intervals already discussed, which do not take height into account. Variable N N* Mean Median TrMean StDev HT_male 163 1 70.626 71.000 70.626 2.940 Variable SE Mean Minimum Maximum Q1 Q3 HT_male 0.230 63.000 79.000 68.000 73.000 135 confidence interval for the mean weight of all 71-inch-tall men (9.27). In the next section, we will see that this difference is due to the fact that 76 is much further from the mean height than 71 is. For now, we summarize our interval estimates with the display below. 120 140 160 180 200 220 240 male weight 95% prediction interval for individual 95% prediction interval for individual 95% prediction interval for individual 95% confidence interval for mean 95% confidence interval for mean 95% confidence interval for mean info used no height height 71 inches height 76 inches intervals centered at sample mean weight 170.83 intervals centered at weight 172.84 predicted for height 71 intervals centered at weight 198.21 predicted for height 76 100 260 Especially in the case of prediction intervals, if there is a substantial relationship between two quantitative variables, we can produce a narrower interval if we include information in the form of a given explanatory value. Confidence intervals for means will be considerably narrower than prediction intervals for individuals. In the next section we will see that sample size plays an important role in the width of the confidence interval. Naturally enough, if the relationship is positive, then for higher values of the explanatory variable, both confidence and prediction intervals are centered at a higher response. Role of s in Confidence and Prediction Intervals As is the case for any interval estimates, our confidence and prediction intervals are of the form estimate ± margin of error = estimate ± multiplier ∗ standard error. No matter if we are constructing a prediction interval for an individual response, or a confidence interval for the mean response to a given explanatory value, the estimate at the center of our interval is the regression line’s predicted response for that explanatory value. Example The regression line for estimating male weight from height is WT_male = - 188 + 5.08 HT_male. The confidence and prediction intervals for male weight when height is 71 are both centered at the estimate −188 + 5.08(71) = 172.68. The confidence and prediction intervals for male weight when height is 76 are both centered at the estimate −188 + 5.08(76) = 198.08. Both confidence and prediction intervals are centered at the predicted response ŷ = b0 + b1x∗, but the confidence interval for mean response is narrower. When sample size is large, the prediction interval extends roughly 2s on either side of the predicted response. If there were no relationship between explanatory and response values, this interval would be no different from the interval that extends two ordinary standard deviations in y (sy) on either side of the mean response. If there is a strong relationship between explanatory and response values, this interval is noticeably more precise than the interval obtained without taking explanatory value into account. 138 Example For the regression of male weight on height, based on a large sample of 162 height/weight pairs, the regression output showed s = 29.6. The prediction interval should have a margin of error equal to roughly twice this, and so its entire width should be about four times 30, or 120. New Obs Fit SE Fit 95.0% CI 95.0% PI 1 172.83 2.35 ( 168.20, 177.47) ( 114.20, 231.47) Values of Predictors for New Observations New Obs HT_male 1 71.0 New Obs Fit SE Fit 95.0% CI 95.0% PI 1 198.21 4.88 ( 188.58, 207.84) ( 138.97, 257.45) Values of Predictors for New Observations 1 76.0 In fact, the width of the prediction interval for weight when height equals 71 is 231.47−114.20 = 117.27 and the width of the prediction interval for weight when height equals 76 is 257.45 − 138.97 = 118.48. Both of these are quite close to our ad hoc calculation of 120. When sample size is large and a confidence interval for mean response is desired for an explanatory value that is close to the mean x̄, this interval extends roughly 2 s√ n on either side of the predicted response. Example Since the mean male height is 70.626, as shown in our summary output, a height of 71 is close to the mean, and so for our sample of size 162, the standard error should be roughly s√ n = 29.6√ 162 = 2.3. If our confidence interval for mean weight extends roughly 2 standard errors on either side of the predicted weight 172.83, its width should be about 4(2.3) = 9.2. In fact, the 95% confidence interval has width 177.47− 168.20 = 9.27. Variable N N* Mean Median TrMean StDev HT_male 163 1 70.626 71.000 70.626 2.940 WT_male 162 2 170.83 165.00 168.24 33.06 New Obs Fit SE Fit 95.0% CI 95.0% PI 1 172.83 2.35 ( 168.20, 177.47) ( 114.20, 231.47) Values of Predictors for New Observations New Obs HT_male 1 71.0 On the other hand, when predicting the mean response to an explanatory value far from the mean of all explanatory values, the standard error is considerably larger than s√ n . Example A height of 76 inches is rather far from the mean height of 70.626. The confidence interval for mean weight of all 76-inch-tall men has a width of 207.84− 188.58 = 19.26, which is more than eight times the standard error, 2.3, instead of just four times, as was the case for estimating mean weight when height was 71, close to the mean of all heights. The illustration shows that whereas prediction interval width remains fairly uniform throughout the range of explanatory values, the confidence interval band widens considerably for explanatory values far below or above average. 139 New Obs Fit SE Fit 95.0% CI 95.0% PI 1 198.21 4.88 ( 188.58, 207.84) ( 138.97, 257.45) Values of Predictors for New Observations 1 76.0 80757065 300 200 100 HT_male W T _m al e S = 29.5971 R-Sq = 20.3 % R-Sq(adj) = 19.8 % WT_male = -187.554 + 5.07587 HT_male 95% PI 95% CI Regression Regression Plot margin of error in PI for individual is approximately 2s margin of error in CI for mean is approximately 2s/sqrt(n) for heights near mean margin of error in CI for mean is more than 2s/sqrt(n) for heights far from mean S=29.5971 These rough estimates are presented here merely as a reference point; in practice, the precise prediction interval and confidence interval should be found using software. Sample size plays its usual role, in that smaller samples result in wider intervals. Exercise: Find two quantitative variables from our survey, summarize their relationship as in Chapter 5, and then test H0 : β1 = 0. State your conclusions in terms of the variables of interest. 140