6. CONFIDENCE INTERVALS. Training is everything cauliflower is nothing but cabbage with a college education.

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1 CIVL 3103 Approximation and Uncertainty J.W. Hurley, R.W. Meier 6. CONFIDENCE INTERVALS Training is everything cauliflower is nothing but cabbage with a college education. Mark Twain At the beginning of the course, we said that we use data obtained from samples to make inferences about the population being sampled. We called this inferential statistics. By applying the appropriate statistics to the sample data, we can infer what the population parameters look like. In this chapter, we ll examine some of those statistics. THE CENTRAL LIMIT THEOREM Suppose we have a population described by a random variable X with a mean µ and a standard deviation σ. We place no restrictions on the probability distribution of X. It may be normally distributed, uniformly distributed, exponentially distributed, it doesn t matter. Suppose we now take random samples from this population, each with a fixed and large sample size n. Each sample will have a sample mean X, and this X will not, in general, be equal to the population mean µ. After repeated samplings, we will have built a population of X s. The X s are themselves random variables and they have their own probability distribution! The probability distribution of a sample statistic is called a sampling distribution of that statistic. The distribution of the X s is a sampling distribution of the mean. The Central Limit Theorem says that, as long as n is reasonably large, σ X N µ, n Let s put that in words as long as n is reasonably large, the X s are normally distributed with a mean equal to the mean of the population being sampled and a variance equal to the variance of the population divided by the sample size. This is true regardless of the population distribution! So, how large is reasonably large? It depends. If the population is normally distributed, any sample size will do. If the probability distribution of the population is at least symmetrical (e.g., a uniform distribution) then n 15 is considered sufficient. If the probability distribution of the underlying population is asymmetrical (e.g., an exponential distribution) then n 30 is considered sufficiently large. THE STANDARD ERROR If σ n is the variance of the sampling distribution, then the standard deviation is σ n. This is commonly referred to as the standard error of the mean. As we ll see very shortly, you can use the standard error to estimate how closely your sample mean approximates the population mean. In other words, you can use the standard error to determine the amount by which your sample mean may be in error!

2 You are making ball bearings for use in an industrial sewing machine. Because of minor variations in the manufacturing process, the diameters always vary slightly from one bearing to the next. If the diameters stray too far from the target value, the sewing machine will not operate properly, so the ball bearings are sampled at regular intervals. If the machine is running properly, the ball bearings should have a mean value of in. with a standard deviation of in. If I collect a random sample of 50 ball bearings, what is the likelihood that the sample mean will be somewhere between 0.50 and inches? 55

3 CONFIDENCE INTERVALS ON THE MEAN In the preceding example, we used the population mean to deduce what the sample means should look like. When we use inferential statistics, we do exactly the opposite. Knowing (usually) one sample mean, we wish to infer the population mean. Suppose that in the previous example, the population standard deviation was known to be inches (i.e., we know the amount of variability from one ball bearing to the next) but the population mean is unknown (perhaps some machine parts have worn down and the mean diameter has changed). If we obtain a random sample of 100 ball bearings and their mean diameter is 0.50 inches, what can we infer about the mean diameter of the ball bearing population being produced today? If, according to the Central Limit Theorem, σ X N µ, n then we know, for example, that 95% of all sample means should fall within two standard errors of the population mean. Why? Well, 95% of the area under the normal probability distribution is located within two standard deviations of the mean and the standard deviation of this normal distribution is the standard error σ n Now, if we know that 95% of all sample means should fall within two standard errors of the population mean, we can also say that the population mean will fall with two standard errors of the sample mean in 95% of all samples. (You may have to think about this for a minute, but it s a valid assertion.) Thus, σ x ± n can be said to be a 95% confidence interval on the mean. If our 100-bearing sample has a mean of inches and, based on the known population variance, a standard error of σ = = n 100 then we are 95% confident that the interval [0.501, 0.508] contains the population mean. Thus, it is safe to assume that we are producing ball bearings that are out of spec! In some cases, we might desire a higher degree of assurance, while in others, we may be willing to accept less assurance. In general, the level of confidence is expressed as (1 α) 100%, where α represents the area under the curve that falls outside the confidence interval. If the total area outside the interval is α, then the upper tail area is α/ and the lower tail area is α/, as shown on the next page. 56

4 f(x) α/ α/ x Now we can write a generic equation for a (1 α) 100% confidence interval on the mean of a population when the standard deviation of the population is known: CONFIDENCE INTERVAL ON A MEAN (σ KNOWN) x σ ± zα n where zα is the critical point corresponding to a tail area of α a. Compute a 90% confidence interval for µ when σ = 3.0, x = 58.3, and n = 5. b. Compute a 99% confidence interval for µ when σ = 3.0, x = 58.3, and n = 100. c. How large must n be for the width of the 99% confidence interval to be less than 1.0? 57

5 STUDENT S t DISTRIBUTION In the real world, we hardly ever know the true value of the population variance, σ, which makes it hard to calculate a standard error. More than a century ago, a statistician named William Gosset studied this problem and came up with a solution. First, if a random variable X is normally distributed, the Central Limit Theorem tells us that the statistic x µ z = σ follows a standard normal distribution regardless of sample size. Now, if you don t know the population standard deviation, σ, maybe you could use the sample standard deviation, s, in its place. If the sample size is large, this is not a bad assumption, but as the sample shrinks, the sample standard deviation becomes an increasingly poor approximation of the population standard deviation. The end result is that a 95% confidence interval computed using s instead of σ may actually only contain the population mean 90% of the time, or 85% of the time, or even less. So William Gosset developed a new probability distribution, which he called the t distribution, to describe the probabilities associated with the statistic n µ t = x s n At the time, Gosset was an employee of the Guinness brewery in Ireland and Guinness policy prohibited employees from publishing their findings under their own name, so Gosset published his results under the pen name A. Student. As a result, his distribution is known as Student s t distribution. Student s t distribution is actually a family of probability distributions, each corresponding to a different sample size n. The distributions are all bell-shaped and symmetrical but, because s is an imperfect substitute for σ, the t statistic has more variability than the z statistic. In other words, compared to the z (standard normal) distribution the t distribution contains more area in the tails and less in the center. Now we can write a generic equation for a (1 α) 100% confidence interval on the mean of a population when the standard deviation of the population is unknown: CONFIDENCE INTERVAL ON A MEAN (σ UNKNOWN) x ± tα, n 1 s n where t α, n 1 is the critical point corresponding to a tail area of α 58

6 As n increases, the t distribution approaches the standard normal distribution. When n 30, the two distributions are virtually indistinguishable and you might as well use the standard normal distribution. What happens if the population is not normally distributed? As long as the sample size is reasonably large and the probability distribution underlying the population is reasonably symmetrical, the t distribution can be used to estimate the population mean when σ is unknown. An unconfined compression test performed on 15 concrete cylinders produced the following strength results (in psi): Find a 95% confidence interval for the true average strength of the concrete. 59

7 CONFIDENCE INTERVALS ON DIFFERENCES Often, we use confidence intervals to compare one population to another. If we draw random samples from two different populations, it would be pure serendipity if their means match exactly. But how far apart can they be for us to still consider the two populations to be the same? If random samples of size n 1 and n are drawn from two populations with known variances σ 1 and σ, respectively, then a 100(1 α)% confidence interval on the difference between the sample means, µ 1 µ is: CONFIDENCE INTERVAL ON A DIFFERENCE IN MEANS (σ 1 and σ KNOWN) ( ) 1 σ σ x1 x ± zα / + n n 1 where zα is the critical point corresponding to a tail area of α This relationship is exact if the two populations are normally distributed. Otherwise, the confidence interval is approximately valid for large sample sizes (n 1 30 and n 30). Aluminum spars from two different suppliers are used in manufacturing the wing of a commercial aircraft. You have been asked to determine if the latest shipments from each supplier are equally strong. From past experience, the standard deviations of the tensile strengths are known to be 1.5 kg/mm for Supplier 1 and 1.0 kg/mm for Supplier (who has tighter quality control). A sample of 1 spars from Supplier 1 has a mean tensile strength of 87.6 kg/mm and a sample of 10 spars from Supplier has a mean tensile strength of 7.5 kg/mm. If µ 1 and µ denote the true mean tensile strengths for the two shipments of spars, find a 90% confidence interval on the difference in mean strength, µ 1 µ. 60

8 What happens if you don t know the standard deviations a priori? If random samples of size n 1 and n are drawn from two normal populations with equal but unknown variances, a 100(1 a)% confidence interval on the difference between the sample means, µ 1 µ is: CONFIDENCE INTERVAL ON A DIFFERENCE IN MEANS (σ 1 = σ UNKNOWN) 1 1 x1 x ± tα /, n + n Sp + n n ( ) 1 1 where S p is a pooled estimator of the unknown standard deviation and is calculated as S p = ( 1) + ( 1) 1 1 n s n s n + n 1 But this can only be used if both populations are normally distributed. The drying time of pavement marking paint is of concern to transportation engineers. Of two such paints from a particular manufacturer, it is suspected that yellow paint dries faster than white paint. Sample measurements of the drying times of both paints (in minutes) are given below. White: 10, 13, 13, 1, 140, 110, 10, 107 Yellow: 16, 14, 116, 15, 109, 130, 15, 117, 19, 10 Find a 95% confidence interval on the difference in mean drying times, assuming that the drying times are normally distributed and the standard deviations of the drying times are equal. 61

9 Experimental situations frequently exist where there is only one set of individuals or objects, and two observations are made on each individual or object. If, for example, you wanted to test the accusation that Auburn s punter got such great distance because Auburn used helium-inflated footballs, you might have 30 punters each punt two footballs one inflated with helium and the other inflated with air and examine the differences in each punter s kicking distance. This testing arrangement eliminates the punters individual capabilities as a source of error in comparing the two footballs. The differences in kicking distance, d, can be treated as though they came from a single population with mean d and standard deviation s d. If the samples are from two normal populations and the data is paired, a 100(1 α)% confidence interval on the difference in means, µ = µ 1 µ is: CONFIDENCE INTERVAL ON A DIFFERENCE IN MEANS (PAIRED SAMPLES) d t d ± α, n 1 s n But this can only be used if both populations are normally distributed. The manager of a fleet of automobiles is testing two brands of radial tires. He assigns one tire of each brand at random to the two front wheels of eight different cars and runs the cars until the tires wear out. The tire lives (in miles) are shown below. Assuming that the tire lives for both brands are normally distributed, find a 99% confidence interval on the difference in mean life. Car Brand 1 36,95 45,300 36,40 3,100 37,10 48,360 38,00 33,500 Brand 34,318 4,80 35,500 31,950 38,015 47,800 37,810 33,15 6

10 CONFIDENCE INTERVALS ON THE VARIANCE If a random sample of size n is taken from a normally distributed population, a 100(1 α)% confidence interval on the variance of the population is CONFIDENCE INTERVAL ON A VARIANCE ( n 1) s ( n ) 1 s σ χ χ α/, n 1 1 α/, n 1 But this can only be used if the population is normally distributed. Here, χ and χ α /, n 1 1 α /,n 1 are the upper and lower critical points of the chi-square distribution with n-1 degrees of freedom. Because the χ distribution is asymmetrical, the upper and lower tails are not the same. The compressive strength of concrete is being tested by a civil engineer. He tests 1 specimens and obtains the following data: Find the 95% confidence interval on the population variance. 63

11 CONFIDENCE INTERVALS ON THE RATIO OF VARIANCES If random samples of size n 1 and n are drawn from two normal populations with unknown variances, a 100(1 a)% confidence interval on the ratio of the population variances is: CONFIDENCE INTERVAL ON RATIO OF VARIANCES 1 σ1 s1 F 1 α, n 1, n1 1 F α, n 1, n1 1 1 σ1 s1 s s But this can only be used if both populations are normally distributed. Here, n 1 and n 1 1 are the degrees of freedom of the F distribution, which has separate degrees of freedom for the numerator and the denominator. The F distribution is not symmetrical, but F 1 α, n 1, n1 1 = F 1 α, n 1, n1 1 The diameter of steel rods manufactured on two different extrusion machines is being investigated. Two random samples of sizes n 1 = 15 and n = 18 were selected from the two machines. The sample means and variances are x 1 = 8.73, s 1 = 0.35, x = 8.68, s = Construct a 95% confidence interval on the ratio of the population variances, σ1 σ. 64