Inferences about process quality

Statistical inference is a process of making generalizations about unmeasured populations using data calculated on measured samples. Statistical inference has the advantage of quantifying the degree of certainty for a particular inference. People sometimes get confused about the difference between descriptive statistics and inferential statistics, partly because in many cases the statistical procedures used are identical while the interpretation differs.

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1. INFERENCES ABOUT PROCESS QUALITY — Based on D. C. Montgomery’s Statistical Quality Control — H. Kemal İlter, PhD F @hkilter January 2020
2. WELCOME About Me H. Kemal İlter, PhD Associate Professor of Operations Management Visiting Researcher ENCS Concordia University www.hkilter.com hkilter@encs.concordia.ca kilter@gmail.com Contents 1. Introduction 2. Statistics and Sampling Distributions 3. Point Estimation of Process Parameters 4. Statistical Inference for a Single Sample 5. Statistical Inference for Two Samples
3. PARAMETERS, KNOWN OR UNKNOWN? Unrealistic assumption In the use of probability distributions in modeling or describing the output of a process, we assumed that the parameters of the probability distribution, and, hence, the parameters of the process, are known. For example, in using the binomial distribution to model the number of defective items found in sampling from a production process we assumed that the parameter p of the binomial distribution is known. The physical interpretation of p is that it is the true fraction of defective items produced by the process. It is impossible to know this exactly in a real production process. Furthermore, if we know the true value of p and it is relatively constant over time, we can argue that formal process monitoring and control procedures are unnecessary, provided p is acceptably small.
4. PARAMETERS, KNOWN OR UNKNOWN? In general, the parameters of a process are unknown; furthermore, they can usually change over time. Therefore, we need to develop procedures to; ▶ estimate the parameters of probability distributions, and ▶ solve inference or decision-oriented problems These techniques are the underlying basis for much of the methodology of statistical quality control.
5. INFERENCE The name inferential statistics derives from the term inference, given definition by the Merriam-Webster online dictionary*: the act of passing from statistical sample data to generalizations (as of the value of population parameters) usually with calculated degrees of certainty Inference is a method of making suppositions about an unknown, drawing on what is known to be true. * https://www.merriam-webster.com/dictionary/inference
6. STATISTICAL INFERENCE Statistical inference is a process of making generalizations about unmeasured populations using data calculated on measured samples. Statistical inference has the advantage of quantifying the degree of certainty for a particular inference. People sometimes get confused about the difference between descriptive statistics and inferential statistics, partly because in many cases the statistical procedures used are identical while the interpretation differs. Basic rule Any time you want to generalize your results beyond the specific cases that provided your data, you should be doing inferential statistics.
7. STATISTICS AND SAMPLING DISTRIBUTIONS Basics Observations in a sample are used to draw conclusions about the population. Figure: Relationship between a population and a sample.
8. STATISTICS AND SAMPLING DISTRIBUTIONS Sampling Sampling The process of defining the population and selecting an appropriate sampling method can be quite complex. Nonprobability Sampling (highly subject to sampling bias) ▶ Volunteer sampling ▶ Convenience sampling ▶ Quota sampling Probability Sampling ▶ Simple random sampling ▶ Systematic sampling ▶ Complex random sampling (stratified sampling, cluster sampling) Random sampling Random samples drawn from; ▶ infinite populations ▶ finite populations with replacement ▶ finite populations without replacement Randomness Random behavior a.k.a randomness comes from; ▶ the environment (eg. Brownian motion) ▶ the initial conditions (eg. Chaos theory) ▶ generated by the system, pseudorandomness (Pseudo-random number generators)
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10. STATISTICS AND SAMPLING DISTRIBUTIONS Statistic Statistic is any function of the sample data that does not contain unknown parameters. Let x1 , x2 , . . . , xn represent the observations in a sample. The sample mean ¯ x = ∑ n i=1 xi n the sample variance s2 = ∑ n i=1 (xi − ¯ x)2 n − 1 and, the sample standard deviation s = √∑ n i=1 (xi − ¯ x)2 n − 1 are statistics. The statistics ¯ x and s (or s2 ) describe the central tendency and variability, respectively, of the sample. A statistic is a random variable, because a different sample will produce a different observed value of the statistic. Every statistic has a probability distribution. If we know the probability distribution of the population from which the sample was taken, we can often determine the probability distribution of various statistics computed from the sample data. The probability distribution of a statistic is called a sampling distribution. When a statistic is used to estimate a population parameter, is called an estimator. It can be proved that the mean of a sample is an unbiased estimator of the population mean. This means that the average of multiple sample means will tend to converge to the true mean of the population.
11. EFFECT OF THE SAMPLE SIZE Figure: Histogram of a uniformly distributed population (N = 100) with range 0 − 100. Figure: Distribution of the means of 100 samples of size n = 2, drawn from a uniform distribution. Figure: Distribution of means of 100 samples of size n = 25, drawn from a uniform distribution.
12. STATISTICS AND SAMPLING DISTRIBUTIONS Sampling Distributions Sampling distributions are important in statistics because they provide a major simplification en route to statistical inference. More specifically, they allow analytical considerations to be based on the probability distribution of a statistic, rather than on the joint probability distribution of all the individual sample values. Figure: Many sample observations (black) are shown from a joint probability distribution.
13. STATISTICS AND SAMPLING DISTRIBUTIONS Sampling from a Normal Distribution Suppose that x is a normally distributed random variable with mean µ and variance σ2. If x1 , x2 , . . . , xn is a random sample of size n from this process, then the distribution of the sample mean ¯ x = N(µ, σ2/n) From the central limit theorem we know that, regardless of the distribution of the population, the distribution of ∑ u i=1 xi is approximately normal with mean nµ and variance nσ2. Therefore, regardless of the distribution of the population, the sampling distribution of the sample mean is approximately ¯ x ∼ N(µ, σ2 n ) So, the mean of x is approximately normally distributed with mean µ and variance σ2/n.
14. STATISTICS AND SAMPLING DISTRIBUTIONS Sampling from a Normal Distribution Chi-square or χ2 distribution Suppose that a random sample of n observations — say, x1 , x2 , . . . , xn — is taken from a chi-square or χ2 distribution with mean zero and variance one. Then the random variable y = x2 1 + x2 2 + · · · + x2 n has a chi-square or χ2 distribution with n degrees of freedom. Figure: Chi-square distribution for selected values of n (number of degrees of freedom). t distribution If x is a standard normal random variable and if y is a chi-square random variable with k degrees of freedom, and if x and y are independent, then the random variable t = x √ y k is distributed as t with k degrees of freedom. Figure: The t distribution for selected values of k (number of degrees of freedom). F distribution If w and y are two independent chi-square random variables with u and v degrees of freedom, respectively, then the ratio Fu,v = w/u y/v is distributed as F with u numerator degrees of freedom and v denominator degrees of freedom. Figure: The F distribution for selected values of u (numerator degrees of freedom).
15. STATISTICS AND SAMPLING DISTRIBUTIONS Sampling from a Normal Distribution Chi-square or χ2 distribution t distribution F distribution
16. STATISTICS AND SAMPLING DISTRIBUTIONS Sampling from a Bernoulli Distribution Suppose that a random sample of n observations — say, x1 , x2 , . . . , xn — is taken from a Bernoulli process with constant probability of success p. Then the sum of the sample observations x = x1 + x2 + · · · + xn has a binomial distribution with parameters n and p. Furthermore, since each xi is either 0 or 1, the sample mean ¯ x = 1 n n ∑ i=1 xi is a discrete random variable with range space {0, 1/n, 2/n, . . . , (n − 1)/n, 1}. The distribution of ¯ x can be obtained from the binomial since P{¯ x ≤ a} = P{¯ x ≤ an} = [an] ∑ k=0 ( n k ) pk(1 − p)(n − k) where [an] is the largest integer less than or equal to an. The mean and variance of ¯ x are µ¯ x = p and σ2 ¯ x = p(1 − p) n respectively.
17. STATISTICS AND SAMPLING DISTRIBUTIONS Sampling from a Poisson Distribution Suppose that a random sample of n observations — say, x1 , x2 , . . . , xn — is taken from a Poisson distribution with parameter λ. Then the sum of the sample observations x = x1 + x2 + · · · + xn has a Poisson distribution with parameter nλ. ¯ x = 1 n n ∑ i=1 xi is a discrete random variable with range space {0, 1/n, 2/n, . . . }. The distribution of ¯ x can be obtained from P{¯ x ≤ a} = P{¯ x ≤ an} = [an] ∑ k=0 e−nλ(nλ)k k! where [an] is the largest integer less than or equal to an. The mean and variance of ¯ x are µ¯ x = λ and σ2 ¯ x = λ n
18. POINT ESTIMATION OF PROCESS PARAMETERS The techniques of statistical inference can be classified into two broad categories: parameter estimation and hypothesis testing. Distributions are described by their parameters. Parameters are generally unknown and must be estimated. Point estimator is a statistic that a single numerical value that is the estimate of the parameter. Properties 1. The point estimator should be unbiased. That is, the expected value of the point estimator should be the parameter being estimated. 2. The point estimator should have minimum variance. Any point estimator is a random variable. Thus, a minimum variance point estimator should have a variance that is smaller than the variance of any other point estimator of that parameter. In many applications of statistics to quality-engineering problems, it is convenient to estimate the standard deviation by the range method. The range of the sample is R = max(xi ) − min(xi ) = xmax − xmin The random variable W = R/σ is called the relative range. The mean of W is a constant d2 that depends on the size of the sample. That is, E(W) = d2 . Therefore, an unbiased estimator of the standard deviation σ of a normal distribution is ˆ σ = R d2 Values of d2 for sample sizes 2 ≤ n ≤ 25 are given in the Appendix Table VI.
19. STATISTICAL HYPOTHESIS Basics A statistical hypothesis is a statement about the values of the parameters of a probability distribution. For example, suppose we think that the mean inside diameter of a bearing is 1.500 in. We may express this statement in a formal manner as H0 : µ = 1.500 H1 : µ ̸= 1.500 The statement H0 : µ = 1.500 is called the null hypothesis, and H1 : µ ̸= 1.500 is called the alternative hypothesis. In the example, H1 specifies values of the mean diameter that are either greater than 1.500 or less than 1.500, and is called a two-sided alternative hypothesis. Depending on the problem, various one-sided alternative hypotheses may be appropriate. An important part of any hypothesis testing problem is determining the parameter values specified in the null and alternative hypotheses. 1. The values may result from past evidence or knowledge. This happens frequently in statistical quality control, where we use past information to specify values for a parameter corresponding to a state of control, and then periodically test the hypothesis that the parameter value has not changed. 2. The values may result from some theory or model of the process. 3. The values chosen for the parameter may be the result of contractual or design specifications, a situation that occurs frequently.
20. STATISTICAL INFERENCE Hypothesis Testing To test a hypothesis, we take a random sample from the population under study, compute an appropriate test statistic, and then either reject or fail to reject the null hypothesis H0 . The set of values of the test statistic leading to rejection of H0 is called the critical region or rejection region for the test. Two kinds of errors may be committed when testing hypotheses. ▶ If the null hypothesis is rejected when it is true, then a type I error has occurred. ▶ If the null hypothesis is not rejected when it is false, then a type II error has been made. The probabilities of these two types of errors are denoted as α = P{ type I error } = P{ reject H0 |H0 is true } β = P{ type II error } = P{ fail to reject H0 |H0 is false } Sometimes it is more convenient to work with the power of a statistical test, where Power = 1 − β = P{ reject H0 |H0 is false } Thus, the power is the probability of correctly rejecting H0 . In quality control work, α is sometimes called the producer’s risk, because it denotes the probability that a good lot will be rejected, or the probability that a process producing acceptable values of a particular quality characteristic will be rejected as performing unsatisfactorily. In addition, β is sometimes called the consumer’s risk, because it denotes the probability of accepting a lot of poor quality, or allowing a process that is operating in an unsatisfactory manner relative to some quality characteristic to continue in operation.
21. STATISTICAL INFERENCE Figure: Type I and Type II errors.
22. CONFIDENCE INTERVALS When we calculate a single statistic, such as the mean, to describe a sample, that is referred to as calculating a point estimate because the number represents a single point on the number line. The sample mean is a point estimate, and is a useful statistic as the best estimate of the population mean. However, we know that the sample mean is only an estimate and that if we drew a different sample, the mean of the sample would probably be different. We don’t expect that every possible sample we could draw will have the same sample mean. It is reasonable to ask how much the point estimate is likely to vary by chance if we had chosen a different sample, and in many professional fields it has become common practice to report both point estimates and interval estimates. A point estimate is a single number, while an interval estimate is a range or interval of numbers. The most common interval estimate is the confidence interval, which is the interval between two values that represent the upper and lower confidence limits or confidence bounds for a statistic. The formula used to calculate the confidence interval depends on the statistic being used. The confidence interval is calculated using a predetermined significance level, often called α, which is most often set at 0.05, as discussed above. The confidence coefficient is calculated as (1 − α) or, as a percentage, 100(1 − α)%. Thus if α = 0.05, the confidence coefficient is 0.95 or 95%. Confidence intervals are based on the notion that if a study was repeated an infinite number of times, each time drawing a different sample from the same population and constructing a confidence interval based on that sample, x% of the time the confidence interval would contain the unknown parameter value that the study seeks to estimate. The confidence interval conveys important information about the precision of the point estimate.
23. CONFIDENCE INTERVALS An interval estimate of a parameter is the interval between two statistics that includes the true value of the parameter with some probability. For example, to construct an interval estimator of the mean m, we must find two statistics L and U such that P{L ≤ µ ≤ U} = 1 − α The resulting interval L ≤ µ ≤ U is called a 100(1 − α)% confidence interval for the unknown mean µ. L and U are called the lower and upper confidence limits, respectively, and 1 − a is called the confidence coefficient. Sometimes the half-interval width U − m or m − L is called the accuracy of the confidence interval. The interpretation of a CI is that if a large number of such intervals are constructed, each resulting from a random sample, then 100(1 − α)% of these intervals will contain the true value of µ. Thus, confidence intervals have a frequency interpretation.
24. P-VALUE The traditional way to report the results of a hypothesis test is to state that the null hypothesis was or was not rejected at a specified α-value or level of significance. The P-value is the smallest level of significance that would lead to rejection of the null hypothesis H0 . Once the P-value is known, the decision maker can determine for himself or herself how significant the data are without the data analyst formally imposing a preselected level of significance.
25. P-VALUE Figure: P-value (Source: https://xkcd.com/1478/).
26. INFERENCE FOR ONE SAMPLE Mean of a Population, Variance Known x is a random variable with unknown mean µ and known variance σ2. Test the hypothesis that the mean is equal to a standard value, µ0 . Hypotheses H0 : µ = µ0 H1 : µ ̸= µ0 Procedure 1. Take a random sample of n observations on the random variable x, 2. Compute the test statistic, Z0 , Z0 = ¯ x − µ0 σ/ √ n 3. Reject H0 if |Z0 | > Zα/2 where Zα/2 is the upper α/2 percentage point of the standard normal distribution.
27. INFERENCE Mean of a Population, Variance Known Example The response time of a distributed computer system is an important quality characteristic. The system manager wants to know whether the mean response time to a specific type of command exceeds 75 millisec. From previous experience, he knows that the standard deviation of response time is 8 mil- lisec. Use a type I error of α = 0.05. Solution H0 : µ = 75 H1 : µ > 75 The command is executed 25 times and the response time for each trial is recorded. We assume that these observations can be considered as a random sample of the response times. The sample average response time is ¯ x = 79.25 millisec. The value of the test statistic is Z0 = ¯ x − µ0 σ/ √ n = 79.25 − 75 8 √ 25 = 2.66 Because we specified a type I error of α = 0.05 and the test is one-sided, then from Appendix Table II we find Zα = Z0.05 = 1.645. Therefore, we reject H0 : µ = 75 and conclude that the mean response time exceeds 75 millisec. Since ¯ x = 79.25 millisec, we know that a reasonable point estimate of the mean response time is ˆ µ = ¯ x = 79.25 millisec. Find a 95% two-sided confidence interval. ¯ x − Zα/2 σ √ n ≤ µ ≤ ¯ x + Zα/2 σ √ n 79.25 − 1.96 8 √ 25 ≤ µ ≤ 79.25 + 1.96 8 √ 25 76.114 ≤ µ ≤ 82.386 Another way to express this result is that our estimate of mean response time is 79.25 millisec ± 3.136 millisec with 95% confidence.
28. INFERENCE Mean of a Population, Variance Unknown x is a random variable with unknown mean µ and unknown variance σ2. Test the hypothesis that the mean is equal to a standard value, µ0 . Hypotheses H0 : µ = µ0 H1 : µ ̸= µ0 Procedure 1. Take a random sample of n observations on the random variable x, 2. Compute the test statistic, t0 . As σ2 is unknown, it may be estimated by s2. If we replace σ by s, we have the test statistic t0 = ¯ x − µ0 s/ √ n 3. Reject H0 if |t0 | > tα/2,n−1 where tα/2,n−1 is the upper α/2 percentage point of the t distribution with n − 1 degrees freedom.
29. INFERENCE Mean of a Population, Variance Unknown Example Rubber can be added to asphalt to reduce road noise when the material is used as pavement. Table shows the stabilized viscosity (cP) of 15 specimens of asphalt paving material. To be suitable for the intended pavement application, the mean stabilized viscosity should be equal to 3200. Test this hypothesis using α = 0.05. Based on experience we are willing to initially assume that stabilized viscosity is normally distributed. Figure: Stabilized viscosity of rubberized asphalt. Solution Hypothesis H0 : µ = 3200 H1 : µ ̸= 3200 Sample mean ¯ x = 1 15 15 ∑ i=1 xi = 48161 15 = 3210.73 Sample standard deviation s = √ ∑ 15 i=1 x2 i − ( ∑ 15 i=1 xi)2 15 15 − 1 = √ 154825783 − (48161)2 15 14 = 117.61 Test statistic t0 = ¯ x − µ0 s/ √ n = 3210.73 − 3200 117.61 √ 15 = 0.35
30. INFERENCE Mean of a Population, Variance Unknown Solution (cont.) Since the calculated value of the test statistic does not exceed t0.025,14 = 2.145 or − t0.025,14 = −2.145, we cannot reject the null hypothesis. Therefore, there is no strong evidence to conclude that the mean stabilized viscosity is different from 3200 cP. The assumption of normality for the t-test can be checked by constructing a normal probability plot of the stabilized viscosity data. Figure shows the normal probability plot. Because the observations lie along the straight line, there is no problem with the normality assumption. Figure: Normal probability plot of the stabilized viscosity data. Find a 95% confidence interval on the mean stabilized viscosity. ¯ x − tα/2,n−1 s √ n ≤ µ ≤ ¯ x + tα/2,n−1 s √ n 3210.73 − 2.145 117.61 √ 15 ≤ µ ≤ 3210.73 + 2.145 117.61 √ 15 3145.59 ≤ µ ≤ 3275.87 Another way to express this result is that our estimate of mean response time is 3210.73 cP ± 65.14 cP with 95% confidence. The manufacturer may only be concerned about stabilized viscosity values that are too low and consequently may be interested in a one-sided confidence bound. The 95% lower confidence bound on mean stabilized viscosity is found using t0.05,14 = 1.761 as 3210.73 − 1.761 117.61 √ 15 ≤ µ 3157.25 ≤ µ
31. INFERENCE Variance of a Normal Distribution Test the hypothesis that the variance of a normal distribution equals a constant, σ2 0 . Hypotheses H0 : σ2 = σ2 0 H1 : σ2 ̸= σ2 0 Procedure 1. Take a random sample of n observations on the random variable x, 2. Compute the test statistic, χ2 0 . As σ2 is unknown, it may be estimated by s2. If we replace σ by s, we have the test statistic χ2 0 = (n − 1)s2 σ2 0 where s2 is the sample variance computed from a random sample of n observations. 3. Reject H0 if χ2 0 > χ2 α/2,n−1 or if χ2 0 > χ2 1−α/2,n−1 where χ2 α/2,n−1 and χ2 1−α/2,n−1 are the upper α/2 and lower 1 − (α/2) percentage points of the chi-square distribution with n − 1 degrees freedom.
32. INFERENCE Variance of a Normal Distribution Solution (cont.) We may use the stabilized viscosity data to demonstrate the computation of a 95% confidence interval on σ2. We have s = 117.61 and s2 = 13832.11. From Appendix Table III, we find that χ2 0.025,14 = 26.12 and χ2 0.975,14 = 5.63. Therefore, we find the 95% two-sided confidence interval on σ2 as (n − 1)s2 χ2 α/2,n−1 ≤ σ2 ≤ (n − 1)s2 χ2 1−α/2,n−1 (14)13832.11 26.12 ≤ σ2 ≤ (14)13832.11 5.63 7413.84 ≤ σ2 ≤ 34396.01 The confidence interval on the standard deviation is 86.10 ≤ σ ≤ 185.46
33. INFERENCE Population Proportion Test the hypothesis that the proportion p of a population equals a standard value, p0 . The test we will describe is based on the normal approximation to the binomial. Hypotheses H0 : p = p0 H1 : p ̸= p0 Procedure 1. Take a random sample of n items is taken from the population and x items in the sample belong to the class associated with p, 2. Compute the test statistic, Z0 . Z0 =          (x + 0.5) − np0 √ np0 (1 − p0 ) if x < np0 (x − 0.5) − np0 √ np0 (1 − p0 ) if x > np0 3. Reject H0 if |Z0 | > Zα/2 where Zα/2 is the upper α/2 percentage point of the standard normal distribution.
34. INFERENCE Population Proportion Example A foundry produces steel forgings used in automobile manufacturing. We wish to test the hypothesis that the fraction conforming or fallout from this process is 10%. In a random sample of 250 forgings, 41 were found to be nonconforming. What are your conclusions using α = 0.05? Solution H0 : p = 0.1 H1 : p ̸= 0.1 Test statistic Z0 = (x − 0.5) − np0 √ np0 (1 − p0 ) = (41 − 0.5) − (250)(0.1) √ 250(0.1)(1 − 0.1) = 3.27 Using α = 0.05 we find Z0.025 = 1.96, and therefore H0 : p = 0.1 is rejected (the P-value here is P = 0.00108). That is, the process fraction nonconforming or fallout is not equal to 10%. Example In a random sample of 80 home mortgage applications processed by an automated decision system, 15 of the applications were not approved. The point estimate of the fraction that was not approved is ˆ p = 15 80 = 0.1875 Assuming that the normal approximation to the binomial is appropriate, find a 95% confidence interval on the fraction of nonconforming mortgage applications in the process. Solution ˆ p − Zα/2 √ ˆ p(1 − ˆ p) n ≤ p ≤ ˆ p + Zα/2 √ ˆ p(1 − ˆ p) n The desired confidence interval is 0.1875−1.96 √ 0.1875(0.8125) 80 ≤ p ≤ 0.1875+1.96 √ 0.1875(0.8125) 80 0.1020 ≤ p ≤ 0.2730
35. INFERENCE Power of a Test Example The mean contents of coffee cans filled on a particular production line are being studied. Standards specify that the mean contents must be 16.0 oz, and from past experience it is known that the standard deviation of the can contents is 0.1 oz. The hypotheses are H0 : µ = 16.0 H1 : µ ̸= 16.0 A random sample of nine cans is to be used, and the type I error probability is specified as α = 0.05. Therefore, the test statistic is Z0 = ¯ x − 16.0 0.1 √ 9 and H0 is rejected if Z0 > Z0.025 = 1.96. Find the probability of type II error and the power of the test, if the true mean contents are µ1 = 16.1 oz. Solution Since we are given that δ = µ1 − µ0 = 16.1 − 16.0 = 0.1, we have β = Φ(Zα/2 − δ √ n σ ) − Φ(−Zα/2 − δ √ n σ ) β = Φ(1.96 − 0.1(3) 0.1 ) − Φ(−1.96 − 0.1(3) 0.1 ) β = Φ(−1.04) − Φ(−4.96) = 0.1492 That is, the probability that we will incorrectly fail to reject H0 if the true mean contents are 16.1 oz is 0.1492. Equivalently, we can say that the power of the test is 1 − β = 1 − 0.1492 = 0.8508.
36. INFERENCE Operating-Characteristic Curves Operating-characteristic curves are useful in determining how large a sample is required to detect a specified difference with a particular probability. As an illustration, in the last example we wish to determine how large a sample will be necessary to have a 0.90 probability of rejecting H0 : µ = 16.0 if the true mean is µ = 16.05. Since δ = 16.05 − 16.0 = 0.05, we have d = |δ|/σ = |0.05|/0.1 = 0.5. From Figure with β = 0.10 and d = 0.5, we find n = 45, approximately. That is, 45 observations must be taken to ensure that the test has the desired probability of type II error. Figure: Operating-characteristic curves for the two-sided normal test with α = 0.05.
37. INFERENCE Operating-Characteristic Curves Figure: Operating-characteristic curves for the two-sided normal test with α = 0.05.
38. INFERENCE FOR TWO SAMPLES Difference in Means, Variances Known Example A product developer is interested in reducing the drying time of a primer paint. Two formulations of the paint are tested; formulation 1 is the standard chemistry, and formulation 2 has a new drying ingredient that should reduce the drying time. From experience, it is known that the standard deviation of drying time is eight minutes, and this inherent variability should be unaffected by the addition of the new ingredient. Ten specimens are painted with formulation 1, and another 10 specimens are painted with formulation 2; the 20 specimens are painted in random order. The two sample average drying times are ¯ x1 = 121 min and ¯ x2 = 112 min, respectively. What conclusions can the product developer draw about the effectiveness of the new ingredient, using α = 0.05? Solution H0 : µ1 − µ2 = 0 H1 : µ1 − µ2 ̸= 0 Test statistic Z0 = 121 − 112 √ (8)2 10 + (8)2 10 = 2.52 Because the test statistic Z0 = 2.52 > Z0 .05 = 1.645, we reject H0 : µ1 = µ2 at the α = 0.05 level and conclude that adding the new ingredient to the paint significantly reduces the drying time. Alternatively, we can find the P-value for this test as P-value = 1 − Φ(2.52) = 0.0059 Therefore, H0 : µ1 = µ2 would be rejected at any significance level α ≥ 0.0059.
39. INFERENCE FOR TWO SAMPLES Difference in Means of Two Normal Distributions, Variances Unknown Example Two catalysts are being analyzed to determine how they affect the mean yield of a chemical process. Specifically, catalyst 1 is currently in use, but catalyst 2 is acceptable. Since catalyst 2 is cheaper, it should be adopted, providing it does not change the process yield. An experiment is run in the pilot plant and results in the data shown in the table. Is there any difference between the mean yields? Use α = 0.05 and assume equal variances. Figure: Catalyst yield data. Solution H0 : µ1 − µ2 = 0 H1 : µ1 − µ2 ̸= 0 We have ¯ x1 = 92.255, s1 = 2.39, n1 = 8, ¯ x2 = 92.733, s2 = 2.98, and n2 = 8 s2 p = (n1 − 1)s2 1 + (n2 − 1)s2 2 n1 + n2 − 2 s2 p = (7)2.392 + (7)2.982 8 + 8 − 2 = 7.30 sp = 2.70 t0 = ¯ x1 − ¯ x2 sp √ 1 n1 + 1 n2 = 92.255 − 92.733 2.70 √ 1 8 + 1 8 = −0.35 Because t0.025,14 = 2.145, and −2.145 < −0.35 < 2.145, the null hypothesis cannot be rejected. That is, at the 0.05 level of significance, we do not have strong evidence to conclude that catalyst 2 results in a mean yield that differs from the mean yield when catalyst 1 is used.
40. INFERENCE FOR TWO SAMPLES Variances of Two Normal Distributions See p. 137 of the textbook.
41. INFERENCE FOR TWO SAMPLES Two Population Proportions See p. 139 of the textbook.
42. INFERENCE What If There Are More Than Two Populations? The analysis of variance, ANOVA, can be used for comparing means when there are more than two levels of a single factor.
43. INFERENCE Other Diagnostic Tools ▶ Standardized and Studentized residuals ▶ R-student – an outlier diagnostic ▶ The PRESS statistic ▶ R2 for prediction based on PRESS – a measure of how well the model will predict new data ▶ Measure of leverage – hat diagonals ▶ Cook’s distance – a measure of influence
44. COMMON PROBLEMS
45. CHECK LIST ▶ Research questions should be stated up front. Investigators must have formulated hypotheses (and the corresponding null hypotheses) well before they begin to collect data. ▶ The relationship between the population of interest and the sample obtained must be clearly understood. ▶ Hypotheses must relate to the effect of specific independent variables on dependent variables. ▶ In complex designs, all of the possible combinations of main effects and interactions and their possible interpretations must be noted. ▶ Procedures for random sampling and handling missing data or refusals must be formalized early on, in order to prevent bias from arising. A truly representative sample must be randomly selected. ▶ Always select the simplest test that will allow you to explore the inferences that you need to examine. ▶ Selection of tests must always be balanced against known or expected characteristics of the data. ▶ Don’t be afraid to report deviations, nonsignificant test results, and failure to reject null hypotheses — not every experiment can or should result in a major scientific result!