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1 Chapter 8 of Devore TESTING A STATISTICAL HYPOTHESIS Maghsoodloo A statistical hypothesis is an assumption about the frequency function(s) (i.e., PDF or pdf) of one or more random variables. Stated in simpler terms, a statistical hypothesis is an assumption about the parameter(s) of one or more population(s) such as µ, σ, µ x µ y, p, and 1 σ / σ, etc. Examples. (a) H 0 : µ = 100, H 1 : µ 100; (b) H 0 : µ = 5000 psi, H 1 : µ > 5000 psi; (c) H 0 : σ = 0.05, H 1 : σ < 0.05; (d) H 0 : = µ x µ y = 0, H 1 : µ x µ y 0; (e) H 0 : σ x = σ y, H 1 : σ x / σ y 1. In the above examples, H 0 is called the null hypothesis while H 1 is called the alternative hypothesis (note that Devore uses the notation H a for the alternative hypothesis but his notation is not as prevalent in Statistics literature as is H 1 ). Further, the above examples indicate that, just like confidence intervals (CIs), there are two types of hypotheses: two-sided and one-sided. The alternative H 1 (or H a ) always determines whether a hypothesis is one or -sided. If the statement under H 1 involves, then the hypothesis is -sided; otherwise, the statement under H 1 will either include < for a left-tailed test, or will include > for a right-tailed test. Therefore, Examples (a), (d) and (e) above constitute - sided hypotheses, while Example (b) formulates a right-tailed test, and (c) is a left-tailed test. Finally, bear in mind that the equality sign "=" generally will be used only in the statement under H 0 (almost never under H 1 ). In the remaining of this chapter, we will study hypothesis testing about a single normal (or nearly normal) population parameter such as µ, σ, and p. Hypothesis testing about parameters of two populations will be discussed in Chapter 9 of Devore. Just like in the case of CIs, the sampling distribution 119

2 (SMD) of point estimator of the parameter under H 0 must be used to conduct a parametric test of hypothesis. That is to say, the null hypothesis H 0 : µ = µ 0 must be tested using the SMD of the point estimator x when σ is known. The null hypothesis H 0 : σ = σ 0 must be tested using the SMD of (n 1) S /σ, which is χ n 1. The null hypothesis H 0 : µ = µ 0, when σ is unknown, must be tested using the SMD of ( x µ ) n / S which is that of the Student s t with ν = n 1 df. The critical (or rejection) region of a null hypothesis is that part of range space of the test statistic that corresponds to the rejection of H 0. In testing a statistical hypothesis, any one of the following four circumstances may occur. Reject H 0 Do not reject H 0 (or accept H 0 ) H 0 is true Event Type I Error Occurrence pr = α Correct Decision Occurrence pr = 1 α H 0 is false Correct Decision Occurrence pr = 1 β Event Type II Error Occurrence pr = β Note that once a decision about H 0 is made based on sample data, then the above 4 circumstances reduce to only two possibilities. For example, if a data set provides sufficient evidence to reject H 0, then the experimenter has either committed a type I error or has made a correct decision [cells (1, 1) and (1, ), respectively]. And vice a versa when sample data does not provide sufficient evidence to reject H 0. Further, the pr of rejecting H 0 given that H 0 is false is called the power of the statistical test, which, from cell (1, ) of the above table, is clearly equal to 1 β. Finally, the question arises as to when an experimenter should conduct a -sided test, and when a one-sided test is warranted? This author recommends 10

3 a -tailed test for all nominal type parameters, the left-tailed test H 1 : θ < θ 0 for an STB type parameter, and the right-tailed test H 1 : θ > θ 0 if θ is an LTB type parameter. However, exceptions to the above recommendations do exist. Further, it will generally be best to set up the alternative hypothesis H 1 first by always placing the statement with a question mark (or a manufacturer s claim) under H 1, and the null of the question mark statement will then be placed under H 0. Thus far, we have laid down the foundation for testing a statistical hypothesis so that in subsequent sections we will illustrate the specific procedures through many examples. TEST OF HYPOTHESIS ABOUT THE MEAN OF A NORMAL POPULATION WITH KNOWN σ (CHAPTER 8) Example 38. A process manufactures ropes for climbing purposes whose breaking strength, X, is normally distributed with variance σ = 65 psi. A consumer group would like to determine if the manufacturer's process mean strength exceeds 150 psi? (Note that this is the question mark Statement and thus will have to be placed under H 1 ). This question leads to the following null and alternative hypotheses: H 0 : µ 150, versus H 1 ; µ > 150. The null hypothesis in this case can also be stated as H 0 : µ = 150. A random sample of size n = 5 has given the value of x = 160, and the question is "does this sample provide sufficient evidence to reject H 0 in favor of H 1 at a preassigned type I error probability (pr) α = 0.05? Recall that we have to make use of the sampling distribution of the point estimator x as depicted in Figure 13. Since larger values of x (well beyond 150) lead to the rejection of H 0 : µ = 150 in favor of H 1 : µ > 150, we deduce that the rejection region for testing H 0 is on the right-tail of the x sampling distribution. Therefore, the upper limit of acceptance interval (AI) is A u = / (5) 1/ = 158.5, and the 11

4 5 / n = se(x) µ psi 150 A u α = 0.05 x H 0 Figure 13. The Sampling Distribution of x assuming that H 0 is true rejection (or critical) region consists of the interval AI = [158.50, ). Since x = 160 falls inside the rejection region AI, then the sample provides sufficient evidence, at the 5% level, to conclude that µ > 150 psi. The pr of committing a type I error, α, is also called the level of significance (LOS) of the test, or the size of the critical region. Exercise 64. Determine if the sample mean x = 160 provides sufficient evidence to reject H 0 at the 1% level of significance. ANS: No. The obvious question is: "Given the result of a sample, what is the smallest LOS at which H 0 can be barely rejected"? For the Example 38, we could reject H 0 at α = 0.05 but could not reject H 0 at α = Further, we can barely reject H 0 at the.5% level but not at the % LOS. After the test statistic is computed, the smallest LOS at which H 0 can be rejected, denoted by αˆ, is also referred to as the probability level (or the P-value) of the test. That is, the P-value of a statistical 1

5 test is the smallest LOS at which H 0 can be barely rejected after a random sample has been drawn and the corresponding sample statistic computed. Therefore, αˆ = P( x > 160) = P(Z > ) = Note that if H 0 is rejected, then αˆ is always less than the corresponding pre-assigned value of α; further, the smaller αˆ is, the stronger the evidence is against the validity of H 0. Before leaving this section, we must state the fact that all CIs are tests of hypotheses in disguise for all possible values of the parameter under H 0. To illustrate this fact, recall the results of Example 36 (pp of STAT 3600) where we obtained the 95% lower one-sided CI for the mean breaking strength to be < µ <. The rejection of our null hypothesis H 0 : µ = 150 vs H 1 : µ > 150 at α = 0.05 is consistent with the 95% CI < µ < because the hypothesized mean value µ is outside the 95% CI [ , ). Then it must be clear by now that the sample mean x = 160 does not provide sufficient evidence at α = 0.05 to reject H 0 : µ = 15 vs H 1 : µ > 15 because µ 0 15 is inside the 95% CI [ , ). Exercise 65. Determine if the 99% CI of Exercise 55(a) on page 109 is consistent with the test result of Exercise 64. COMPUTING TYPE II ERROR PROBABILITY β Since β is the pr of accepting H 0 if H 0 is false, then when testing H 0 : µ = µ 0, the pr statement for β of the Example 38 is β = P( x falls within the acceptance interval H 0 is false, i.e., µ > 150). To illustrate the computation of β, again consider the Example 38, where H 0 : µ = 150 and H 1 : µ > 150 psi. Suppose the value of true mean µ = 155, i.e., H 0 is now false. Then, what is the pr of accepting H 0 : µ = 150 with a random sample of size 5 before the sample is drawn given that the population mean µ =

6 µ 0? The SMD of x given that H 1 is true (or H 0 is false, i.e, µ 150) is depicted in Figure 14 below. From Figure 14 we deduce that the value of β (at µ = 155) = P( < x < µ = 155) = P( x = P(Z 0.645) = Φ (0.645) = ) = Figure 14 and the above computation of β (at µ = 155) show that as the true value of µ changes, so does the value of β, i.e., Type II Error pr is a function of the parameter under H 0 (in this case β is a function of µ). For example, if µ = , then β = 0.50; if µ > 158.5, then β < 0.50; if µ = 170, β = ; if µ = 180, β = , and if µ = 1300, then β is practically 0. Note that the larger the true value of µ is than 150, i.e., the falser H 0 is, then the smaller is the pr of committing a Type II error. For example, if µ = 1300, then the P( < x < µ = 1300) = P(Z 8.355) = , which is less than 1 in a billion. se( x ) = 5 β 1 β 155 A u = x H 1 Figure 14. The Sampling Distribution of x Given that H 0 is false The graph of β as a function of the parameter under H 0 is called an operating 14

7 characteristic (OC) curve. The OC curve for a -sided H 0 : µ = µ 0 is symmetrical about µ 0 with maximum value of β = 1 α which occurs at µ = µ 0. For onesided tests, β = 1 α at µ = µ 0, but for right-tailed tests, β > 1 α if µ < µ 0, and for a left-tailed test β > 1 α if µ > µ 0. Exercise 66. (a) For the null hypothesis H 0 : µ = 150 of the above Example 38, draw the OC curve by computing and tabulating the values of β at µ = 145, 150, 153, 158.5, 160, 163, and 165. (b) Repeat part (a) using a LOS α = It must be clear by now that if H 0 is -sided, then the critical region of the test is divided equally at the left and right tails of the distribution of the test statistic, and for a left-tailed test the entire value of α is placed on the distribution's left tail and vice a versa for a right-tailed test. Further, always bear in mind that β gives the pr over the AI while α = pr( AI ), and almost for all statistical tests 1 β α. A test for which 1 β α is said to be unbiased, and a statistical test for which the value of 1 β 1 as n is said to be consistent Exercise 67. (a) Work Exercises 8-10, 8-1 (pp. 30-1), 8-18, 8-19, 8-0, and 8-5 on pp (b) For the Exercise 8-10(d), derive the minimum value of n if α = 0.05 and β = 0.01 at µ = Compute the P-values of your tests in all cases. 15

8 TEST OF HYPOTHESIS ABOUT THE MEAN OF A NORMAL UNIVERSE WITH UNKNOWN VARIANCE σ Since the population variance σ is unknown, then we may obtain a point unbiased estimator of σ using the sample statistic S. Clearly, ( x µ )/ σx = ( x µ ) n /σ has a standardized normal distribution, but σ is unknown and has to be estimated by S. However, ( x µ ) n /S is not normally distributed but its sampling distribution follows that of the Student s t with (n 1) df. We illustrate the procedure for testing H 0 : µ = µ 0 at a LOS α thru the following example. Example 39. A manufacturer claims that the loudness of its compressors is on the average less than 50 decibels. A random sample of n = 14 compressors has noise levels 53, 48, 49, 57, 45, 46, 54, 47, 49, 50, 48, 44, 46, 51 db. Our objective is to test H 0 : µ = 50, at α = 0.05, versus the alternative H 1 : µ < 50 db putting the burden of proof on the manufacturer. Note that a claim by a manufacturer should generally be placed under H 1. Since this is a left-tailed test, the rejection region of the test statistic ( x µ ) n /S is AI = (, t 0.95; 13 ) = (, 1.771). Note that this test is left-tailed because smaller x values (relative to 50 db), which in turn lead to smaller values of ( x µ ) n /S, will reject H 0 in favor of H 1 : µ < 50. For our sample of 14 observations we have x = , S x = and se( x ) = S x / 14 = so that the value of our test statistic t 0 = ( x 50) / se( x ) = ( ) / = Since t 0.95 (13) = 1.771, we do not quite have sufficient evidence to reject H 0 at the 5% LOS. AS expected, the critical level αˆ = P(T 13 < 0.947) = far exceeds the pre-assigned LOS α = 0.05 because the sample did not provide sufficient evidence to reject H 0. 16

9 Now suppose the true mean noise level of the manufacturer's compressors is µ = 48.5 db. What is the pr of rejecting H 0 : µ = 50 before the random sample of size n = 14 is drawn, assuming that the true mean µ = 48.5 db. The OC curves in Table A.17, page 744, graph the values of β vs the abscissa d = µ µ ) / σ ( 0 ( µ µ ) /S. Since d = / , then β 0.57 so that the pr of 0 rejecting H 0 : µ = 50 is approximately 1 β = 0.43, i.e., the power of the test at µ = 48.5 is roughly 43%. The exact value of 1 β from Minitab is 0.48 (Open Minitab STAT Power and Sample Size 1-sample t, in the Dialogue Box insert 14 for sample sizes, differences = 1.5, σ = 3.67 Options Alternative hypothesis Less than, and then ok. Exercise 68. (a) Work Exercises 8.4, 8.6 and 8.9 on page 333. (b) For Exercise 8.4, estimate the power of the test if the true population average µ = 900 from the OC curves in Table A.17 and then use Minitab to compute the exact value of 1 β TEST OF HYPOTHESIS ABOUT THE VARIANCE OF A NORMAL POPULATION ( I doubt that Devore covers this test) 0 < σ < Since a process variance σ is an STB type parameter, a CI of the type σ u is generally most meaningful and hence a left-tailed test on σ the most appropriate. However, different authors quite often conduct a -sided test H 0 : σ = σ > σ = σ 0 vs H 1 : σ σ 0, or even a right-tailed test on σ where H 1 : σ 0. For all three alternatives, the statistic (n 1)S / σ 0 is used to test H 0 : σ 0. Recall that the SMD of χ = (n 1)S / 0 is σ 0 follows a χ with (n 1) df, 17

10 and hence for the alternative H 1 : σ < σ 0, the critical region is (0, the -tailed test, the acceptance interval for the test statistic (n 1)S / χ 1 α;n 1 ); for σ 0 is AI = ( χ 1 α / ;n 1, χ α / ;n 1 ), while for the right-tailed test the rejection region for χ 0 is AI = ( χ α; n 1, ). Example 40. The % of titanium in an alloy, X, in aerospace castings is N(µ, σ ), and n = 51 randomly selected parts yielded the standard deviation S = Our objective is to test H 0 : σ = 0.5 vs H 1 : σ 0.5 at the LOS α = The AI for the test statistic (n 1) S / σ 0 = (n 1) S / 0.5 is AI = ( χ, 0.975;50 χ 0.05;50 ) = (3.36, 71.4) as depicted in Figure 15 below. The value of our test statistic is χ = 0 50(0.31) 0.5 = 76.88, which is outside the AI = (3.36, 71.4) and hence we have sufficient evidence to reject H 0 in favor of H 1 : σ 0.5. Since the test is -tailed, the critical level of the test is αˆ = P-value = P( χ > 76.88) = ( ) = , which as expected, is 50 less than the prescribed LOS α = Although, S is not a linear sum of independent rvs, yet its SMD very slowly approaches the N(σ, σ / n) pdf as n as depicted in Figure 16. For practical applications the normal approximation is fairly accurate for n > 60. From Figure 16, αˆ P(S > 0.31) = P(Z >.44) = Φ (.44) = as compared to the exact value Note that even a sample of size n = 51 is not sufficiently large for an adequate normal approximation. We now use the above example to illustrate the computation of β when H 0 18

11 is false. To this end, assume that the true value of σ = Then f( χ 50 ) AI = (3.36, 71.4) (n 1)S / σ 0 Figure 15. The Sampling Distribution of (n 1) S / σ 0 under H / 10 = ˆα / S H 0 Figure 16. The Approximate SMD of S β(at σ = 0.30) = P( χ is inside the AI H 0 is false) 0 19

12 = P(3.36 (n 1)S / = P(3.36 σ σ = 0.30) σ0 (n 1)S / σ 71.4 σ = P( σ σ χ ) 50 0 σ = 0.30) = P(.47 χ ) 50 β(at σ = 0.30) = The cdf of χ at the cdf (at.47) 50 = = The above exact pr of type II error could also be approximated from the OC curves if Devore had provided them. The abscissa of such OC curves would be λ = σ / σ 0 = 0.30 / 0.5 = 1.0. I used chart (i) on page A-16 of Montgomery and Runger (Applied Statistics and Probability For Engineers) to obtain the approximate ordinate value of β As far as I have been able to ascertain, Minitab does not provide power values for a χ test either. Therefore, Excel has to be used for all χ test computations. Exercise 69. (a) For the above example, use Excel to compute the pr of a type II error for testing H 0 : σ = 0.5 (with n = 51) if the true value of σ were equal to (b) Obtain a 95% CI for σ and compare your CI with test result of Example 40 and draw conclusions. (c) For the Example 40, derive the necessary sample size n such that the pr of type II error at λ = σ / σ 0 = 0.30 / 0.5 = 1.0 is reduced from o.5103 to β =

13 TEST of HYPOTHESIS ABOUT A POPULATION PROPORTION p To illustrate the procedure for testing H 0 : p = p 0 vs one of the 3 alternatives H 1 : p < p 0, H 1 : p p 0, or H 1 : p > p 0, we consider Devore s Example 8.11 on pp of your text. Before proceeding, however, you will need to review your STAT 3600 notes pp As stated at the beginning of this chapter, the? statement must be placed under H 1. In this example, n = 10 doctors were randomly surveyed only 47 of whom knew the generic name for Methadone, i.e., there were 47 successes in n = 10 Bernoulli trials. Therefore, a point unbiased estimate of the population proportion (of all doctors), p, is pˆ = 47/ 10 = The question is does this sample provide sufficient evidence at the LOS α = 0.05 to warrant the rejection of H 0 : p = 0.50 in favor of H 1 : p < 0.50? The approximate large-sample SMD of pˆ is depicted in Figure 17. se(pˆ ) = A L 0.50 pˆ Figure 17. The Approximate SMD of pˆ Figure 17 clearly shows that A L = = so that 131

14 the AI = (0.4186, 1). Since the sample test statistic pˆ = is well inside this AI, the null hypothesis cannot be rejected at the 5% LOS. The critical level of this test is given by P-value P(pˆ ) = P(Z 0.791) =0.141, which as expected, far exceeds the pre-assigned LOS of 5%. The 95% CI for p is given by 0 p 0.540, which clearly agrees with the above test result because the hypothesized value of p, p , is well inside this CI. If the normal approximation to the binomial is not adequate, then the binomial test on a population proportion must be conducted using the Bin(n, p) distribution directly. To illustrate, I will compute the exact critical level of the test for the Example 8.11 of Devore. The procedure consists of using MS Excel to compute αˆ from αˆ = 47 x 10 x 10 Cx (0.50) (0.50) x= 0 = BINOMDIST(47, 10, 0.50, True) = Note that the approximate value of αˆ = from the normal approximation is fairly inadequate because a the correction for continuity of 1/( n) was not applied to αˆ = If we apply the correction for continuity in the normal approximation to the binomial, we will obtain αˆ P(pˆ / 04 ) = P(pˆ ) = P( Z ) = Φ ( ) = 0.441, which is almost identical to the exact value of Therefore, if n < 50, use the exact procedure for the binomial test of proportion by computing the exact αˆ from MS Excel and compare αˆ against 13

15 α = If αˆ < 0.05, then reject H 0 : p = p 0 ; otherwise, the data does not cast sufficient doubt about the validity of H 0. Exercise 70. Work exercises 35, 36, 37, 4 on pp and exercises 69, 7,74 on page 35 of Devore. Test 1 133

1; (f) H 0 : = 55 db, H 1 : < 55.

1; (f) H 0 : = 55 db, H 1 : < 55. Reference: Chapter 8 of J. L. Devore s 8 th Edition By S. Maghsoodloo TESTING a STATISTICAL HYPOTHESIS A statistical hypothesis is an assumption about the frequency function(s) (i.e., pmf or pdf) of one