243 Math Variations on the Inference Theme
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Recall that statistical inference is inferring information about a population from information about a sample. Were generally talking about one of two things: 1. Estimating parameters (condence intervals) 2. Hypothesis testing (signicance tests) The same basic ideas that we used when computing condence intervals and evaluating hypothesis tests for means of a quantitative variable can be applied in a number of related situations. The best way to think of these different situations is as variations on an inference theme. To make this easier, we will use a systematic notation scheme throughout: parameters (population) , proportion (of a categorical variable) , mean (of quantitative variable) , standard deviation statistics (sample) n, sample size X, count (of a categorical variable) p=
X n,
proportion (of a categorical variable)
x, mean (of quantitative variable) s, standard deviation sampling distribution SE, standard deviation of the sampling distribution ( x or p are also used for this) eSE, estimated standard error of the sampling distribution (an estimate for SE) p , x , mean of sampling procedure (for when determining p and x, respectively) Subscripts will be used to indicate The procedures involving the z (normal) and t distributions are all very similar. To do a hypothesis test, compute t or z = data value hypothesis value , SE or eSE
and compare with the appropriate distribution (using tables or computer). To compute a condence interval, rst determine the critical value for the desired level of condence (z or t ), then the condence interval is data value (critical value)( SE or eSE) .
Math 243 Variations on the Inference Theme
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1 Matched Pairs
Inference for matched pairs really is hardly new because generally we combine the values of two quantitative variables to form one new quantitative variable, and then we apply our inference procedures exactly as before. The most common way to combine the variables is simply take the difference between the two variables. Example. Two varieties of oats were compared in an experiment to determine which variety had the higher yield. Since soil type also affects yield, the experimenter blocked out its effect by planting each variety of oats in seven different types of soil. With the data paired by soil types as given below, does it appear that variety A has the higher mean yield?
Yield A B 71.2 65.2 72.6 60.7 47.8 42.8 76.9 73.0 42.5 41.7 49.6 56.6 62.8 57.3
H0 :
x = A-B 6.0 11.9 5.0 3.9 0.8 -7.0 5.5
Soil type 1 2 3 4 5 6 7
Ha : eSE =
p-value =
mean(x) = 3.7286
sd(x) = 5.7792
95% CI for difference:
Example. To test the effect of continuous music on factory workers output, each of 7 workers was observed for one month with music and one month without. Given the following results, does music help? Find a 95% condence interval for the mean difference in output. 1 2 3 8.4 5.0 7.2 7.4 6.1 8.0 1.0 -1.1 -0.8 sd(Diff) = .709 4 6.6 6.4 0.2 5 6 7 6.5 8.7 5.9 6.8 8.8 6.6 -.3 -.1 -.7
Average output with music Average output w/o music Diff mean(Diff) = -.257,
2 Comparing Two Means
A two-sample problem is one in which:
1. 2. 3. The difference between two-sample problems and matched pairs problems is
Math 243 Variations on the Inference Theme
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So in a two-sample problem we want to compare 1 with 2 . We do this by drawing a sample from each population and calculating x1 and x2 . In order to know what x1 and x2 tell us about the difference between 1 and 2 we need to know about the sampling distribution for x1 x2 . Assuming each population has a normal distribution with means i and standard deviations i , we already know that and x2 N (2 , 2 / n2 ) x1 N (1 , 1 / n1 ) Using our rules for combining means and variances, we see that x1 x2 N (1 2 , SE) , where SE =
2 1 2 + 2 n1 n2
Of course, we wont usually know 1 and 2 , so we need to estimate SE using two ways to do this:
and
There are
1. 2.
2.1 Unequal variances
In this case we simple replace each with eSE = , so s2 s2 1 + 2 n1 n2
Unfortunately the t statistic computed from this does does not have a t-distribution with n1 + n2 2 degrees of freedom, as we might have hoped. Why? A t distribution replaces a N(0,1) distribution only when a single population standard deviation is replaced by a single sample standard deviation s. In this case, we replaced two standard deviations (1 and 2 ) with their estimates (s1 and s2 ). The resulting distribution is still approximately a t-distribution, but it is not so easy to specify the correct degrees of freedom. There are two ways to approximate the degrees of freedom we should use:
1. messy formula:
( s2 /n1 + s2 /n2 ) 2 1 2
(s2 /n1 )2 1 n1 1
+
(s2 /n2 )2 2 n2 1
2. easier, but less accurate estimate: Method 1 is typically used by computer software, but is a bit messy to be done by hand. Furthermore, the simpler method is conservative: the value it gives for d f is always on the low side, which means that it will give CIs that are bit and p-values that are a bit Now that we know the distribution involved and a value for eSE, we are all set to do hypothesis testing or to compute CIs.
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Example. An agronomist has developed a new plant food. She hopes it will improve yield. To nd out, she treated 48 plants with the new food and obtained a mean yield of 24.4 lbs (s=4.8 lbs). 45 identical plants were untreated and had a mean yield of 22.3 lbs (s =2.3 lbs). Do the data provide sufcient evidence to determine that the new plant food is better than no treatment?
Example. The weather bureau measured the ozone level at 5 random locations in Orange City before a cool front moved through and at 5 different random locations afterward. Test whether there is a signicant drop in the ozone level after the front has moved through. Time Before front After front n 5 5 mean 0.122 .094 variance .00067 .00016
How could the design of this study need to be changed to make it a matched pairs situation?
Example. The scores of two groups of prison inmates on a rehabilitation test are summarized below: First offenders Mean 300 Variance 20 Sample size 16 Repeat offenders 305 4 13
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Example. Ten old frisbees and ten new freesbees were tested to see how much water they could hold. The results are summarized below:
> m243.summary(f$water[age=="old"]) Min. 1st Qu. Median Mean 3rd Qu. 1.6900 1.7300 1.8100 1.7800 1.8100 > m243.summary(f$water[age=="new"]) Min. 1st Qu. Median Mean 3rd Qu. 1.9000 1.9100 1.9400 1.9400 1.9500
Max. St Dev. 1.8500 0.0583
Max. St Dev. 1.9800 0.0259
The statistical package R reports the following: t.test(water~age, alternative=two.sided, conf.level=.95, var.equal=FALSE, data=f) Welch Two Sample t-test data: water by age t = 7.835, df = 12.424, p-value = 3.741e-06 alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval: 0.11423 0.20177 sample estimates: mean in group new mean in group old 1.936 1.778
2.2 Equal variances: pooled two-sample procedures
If the two population distributions have the same variance, then we do have a situation that is exactly a t distribution, and we pool data from both samples to estimate the common variance. We will use the pooled estimate if we have reason to believe that the two population variances are equal. One common rule of thumb is that if the variances of the two samples are with a factor of 4 (so s1 and s2 will be within a factor of ), then we can use the pooled estimate. Computer software will usually have clearly marked options for the two types of two sample tests or else print both out results and leave it to you to decide which should be used. When we do this, our pooled estimate for the variance in each population is sp = and eSE = s2 p n1
( d f 1 ) s2 + ( d f 2 ) s2 1 2 d f1 + d f2
s2 p n2 1 1 + n1 n2
+
= sp
The degrees of freedom of the t distribution in this case is the sum of the degrees of freedom from the two samples: d f = d f 1 + d f 2 = (n1 1 ) + (n2 1 ) = n1 + n2 2 Once again, with the distribution in hand, we are all set to do hypothesis testing or to compute CIs.
Math 243 Variations on the Inference Theme
Example. Here are the results of the frisbee data using an assumption of equal variances. t.test(water~age, alternative=two.sided, conf.level=.95, var.equal=TRUE, data=f) Two Sample t-test data: water by age t = 7.835, df = 18, p-value = 3.293e-07 alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval: 0.11563 0.20037 sample estimates: mean in group new mean in group old 1.936 1.778 Compare the output here with the output above. What differences do you notice?
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Is it justiable to use the equal variances assumption in this case? Why or why not? (Note: the computer is more than happy to spit out the results for you regardless of whether you should be asking it to do so.)
In Rcmdr there is a check-box where you can indicate whether or not to use the equal variances assumption.
2.3 Robustness
Robustness of a statistical test or procedure refers to how well it works even when the underlying assumptions upon which it is based are not true. All of the t-based procedures are based on an assumption that the populations involved have a distribution. But we have seen that even when this is not the case, these procedures are often still at least approximately correct, at least in the one-sample case. Here is a summary of robustness for the two-sample t-procedures: In certain situations, the two-sample t procedures are actually somewhat more robust than the one-sample t procedures. When the sizes of the two samples are equal and the distributions of the two populations being compared have similar shapes, p-values based on the t-distribution are quite accurate for a broad range of distributions even when the sample sizes are as small as n1 = n2 = 5. When the two population distributions have different shapes, larger samples are needed. The guidelines we gave before (sample sizes < 15, 15 40, 40) can be adapted to two-sample procedures by replacing the one-sample sample size (n) with the sum of the sample sizes: n1 + n2 .
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3 Categorical Data
3.1 One-proportion procedures
We already know 1. the sampling distribution for p is based on the 2. the sampling distribution for p can be approximated by a 3. SE p = distribution, distribution provided
3.1.1 Hypothesis testing This is all we need to know for hypothesis testing, since the hypothesis provides us with a value for , from which we can get the value of SE. Example. A pollster contacts 225 people in Statville and asks if they plan to vote for or against Referendum A. 99 say they will vote in favor, 126 say they will vote against. Is this sufcient evidence to predict the outcome of the vote?
3.1.2 Condence intervals In order to compute condence intervals we need to estimate SE, since we dont know what is. Our estimate will come by replacing with Once we do that we get eSE =
The distribution is still approximately normal (not a t distribution because the standard deviation is not an independent parameter in this case, it is determined by ). So we use the normal distribution for condence intervals, too. Example. The National Transportation Safety Board conducted a study of truck drivers killed in highway accidents. They found that 24 of 185 drivers tested positive for alcohol. Obtain a 95% condence interval for the true percentage of truck driver deaths in which the truck driver had a positive level of alcohol.
If they were reporting this on the news, they would say: It is estimated that 13% of truck drivers killed in highway accidents tested positive for alcohol (margin of error percentage points).
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Example. A special Newsweek on aging in fall/winter 2001 reported the results of a national survey of U.S. adults. They reported that 20% of women and 6% of men would like a face lift. At the bottom of the percentages, they wrote, For this special Newsweek poll, Princeton Survey Research Associates interviewed a random national sample of 801 adults 45-65 years old by telephone July 13-17. The margin of error is 5 percentage ...