Chapter 9. Hypothesis Testing. 9.1 Introduction

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1 Chapter 9 Hypothesis Testing 9.1 Introduction Many practical problems require us to make decisions about populations on the basis of limited information contained in a sample. For instances, A boss want to decide whether or not to invest money for a new production process. A computer center manager may have to decide whether or not to upgrade the capacity of his installation. Apparently, such a decision is binary yes or not) in nature. In order to arrive at a decision, we often make an assumption or guess about the nature of the underlying population, which may or may not be valid, is called a statistical hypothesis. Example 1 Suppose that 10% of all circuit boards produced by a certain manufacturer. An engineer has suggested a alternative process in the production process in the belief that it will result in a reduced defective rate. Let p denote the true proportion of defective boards, which may never be known, resulting from the alternative process. Consider the following two hypotheses: H 0 : p 0.1 H 1 : p < 0.1 Discuss the relation between hypotheses H 0 and H 1. ans) H 0 and H 1 are two competing hypotheses regarding the population parameters). H 0 and H 1 are mutually complementary, i.e., H 0 = H c 1 and H 1 = H c 0. So, only one of them is true. If H 1 is really true, the investment will be made. Furthermore, the truth of H 1 is really the one we want to prove. Because the investment may be quite huge, we accept H 1 reject H 0 ) only when there is sufficient evidence to support it. 1

2 H 0 is an artificially made hypothesis. i.e., H 0 is simply set to be the complement of H 1. H 0 is called a null hypothesis. The word null means of no value, effect, or consequence, which suggests that H 0 should be identified with the hypothesis of no change, no difference, no improvement, and so on. H 1 is called an alternative hypothesis, which is the one the researcher really like to validate. Statistical Tests Procedures that enable us to decide whether to reject or accept hypotheses are called statistical tests. Let θ be a parameter of a population. We may be given the following different hypotheses for testing. Remark 1. Simple hypothses θ is assigned to a paritular value both in H 0 and H 1. H 0 : θ = θ 0 H 1 : θ = θ 1 2. Composite hypotheses θ may have multiple values in H 0 or H 1. We consider three tests involving composite hypotheses below. Upper one-tailed test Lower one-tailed test Two-tailed test H 0 : θ θ 0 H 1 : θ > θ 0 H 0 : θ θ 0 H 1 : θ < θ 0 H 0 : θ = θ 0 H 1 : θ θ 0 In setting up an experiment, it is common to choose the null hypothesis so that its acceptance represents a weak decision, whereas it rejection represents a strong decision. Two Types of Error Because the information provided from a sample is always not sufficient, a decision that we made may have error such as 2

3 The two types of error: True State of Nature Decision H 0 H 1 Reject H 0 Type-I error Correct decision Accept H 0 Correct decision Type-II error Type-I Error rejecting H 0 when it is true. Type-II Error accepting H 0 when it is false. Let Θ be set all possible values of parameter θ such that H 0 becomes true. Hence, Θ c be set all possible values of parameter θ such that H 0 becomes false. Let V = V X 1,..., X n ) be the random variable corresponding to the statistic obtained from a sample. Let CR be the so-called critical region such that H 0 will be rejected when v CR. Let θ be the true value of parameter θ. Then the probabilities of the type-i error αθ ) and type-ii error βθ ) are: αθ ) = P v CR θ Θ) βθ ) = P v CR θ Θ c ) Example 2 In Example 1, suppose that 100 circuit boards produced by the new process are selected for testing. Let the criterion to reject H 0 be when the number of defectives x 7. Discuss the probabilities of Type-I error and Type-II error by assuming the true proportion to be p = 0.05, 0.06, 0.07, 0.08, 0.09, 0.10, 0.11, 0.12, 0.13, 0.14, and sol) Let X be the random variable represent the number of defectives. Then, X B100, p). Because of large sample, we approximately have X N100p, 100pq). Hence, using ˆp = X/100 to estimate p, we have ˆp Np, pq/100) or, equivalently, Z = ˆp p pq/100 N0, 1) 3

4 Case 1. H 0 is true, i.e., p = p αp 1 ) = P X CR p = p 1 ) Case 2. H 0 is false, i.e., p = p 1 < 0.1 Accordingly, And, = P X 7 p = p 1 ) = P ˆp 0.07 p = p 1 ) = P ˆp p pq/100 pq/ p p = p 1 = P ˆp p 1 p 1 q 1 / p 1 p 1 q 1 /100 = P Z 0.07 p 1 p 1 q 1 /100 βp 1 ) = P X CR p = p 1 ) = P X > 7 p = p 1 ) = P ˆp > 0.07 p = p 1 ) = P ˆp p pq/100 > pq/ p p = p 1 = P ˆp p 1 p 1 q 1 /100 > 0.07 p 1 p 1 q 1 /100 = P Z > 0.07 p 1 p 1 q 1 /100 α0.10) = P Z 1) = Φ 1) = α0.11) = P Z ) = Φ ) = α0.12) = P Z ) = Φ ) = α0.13) = P Z ) = Φ ) = α0.14) = P Z ) = Φ ) = α0.15) = P Z ) = Φ ) = β0.05) = P Z > ) = 1 Φ0.9177) = β0.06) = P Z > ) = 1 Φ0.4211) = β0.07) = P Z > 0) = 1 Φ0) = 0.5 β0.08) = P Z > ) = 1 Φ ) = β0.09) = P Z > ) = 1 Φ ) =

5 Level of significance The probability of committing a type-i error, also called the level of significance, is denoted by Greek letter α. In hypothesis testing, a decision will be made based on the given α, i.e., the critical region for rejecting H 0 is made based on the given alpha. In our illustration, the maximum of α risk appearing at p = 0.10 θ = θ 0 ). Hence, to achieve certain level of significance in hypothesis testing, the critical region will be made based on θ = θ 0. So, we can also write the three composite hypotheses as Upper one-tailed test Lower one-tailed test Two-tailed test H 0 : θ = θ 0 H 1 : θ > θ 0 H 0 : θ = θ 0 H 1 : θ < θ 0 H 0 : θ = θ 0 H 1 : θ θ 0 Example 3 Suppose 6 out 100 circuit boards are found to be defective. Test the hypotheses depicted Example 1 with a 0.05 level of significance. sol) 1. State H 0 and H 1 The null hypothesis H 0 and alternative hypothesis H 1 are given by: H 0 : p = p 0 = 0.10) H 1 : p < p 0 = 0.10) 2. Compute test statistic from the sample data Since ˆp = X/n Np, pq/n), we have Z = ˆp p pq/n N0, 1) Assuming that H 0 is true, i.e., the true proportion p = p 0 = 0.10, then x = 6 ˆp = 6/100 = 0.06 z = )0.90)/100 =

6 3. Find a critical region to achieve the given level of significance Assume H 0 is true. We are finding an x CR such that P Type-I error) = P X x CR p = p 0 ) = α= 0.05) = P ˆp x CR /n p = p 0 ) = α = P ˆp p 0 p 0 q 0 /n x CR/n) p 0 = α p 0 q 0 /n = P Z x CR/n) p 0 p 0 q 0 /n = α Hence, x CR /n) p 0 p 0 q 0 /n = z α = z 0.05 = x CR /n = p 0 z α p 0 q 0 /n = )0.9)/100 = x CR = np 0 + z α p 0 q 0 /n) = Accordingly, CR = {x x 5.065}. 4. Decision Since x = 6 CR, H 0 cannot be rejected. 5. Conclusion H 0 cannot be rejected at significance level The data does not give strong evidence to the claim that the proportion of defective resulting from the new process is lower than Remark In the above example, the following critical regions are equivalent. CR 1 = {x x 5.065} CR 2 = {ˆp ˆp } CR 3 = {z z 1.645} In general, z-test will be used if a test-statistic possesses normal distribution. In the above example, we have z = 1.33 CR 3. P -Value Hence, H 0 does not be reject. Instead of only giving the decision of reject or do not reject the null hypothesis H 0, The P -value gives richer information to the tester. 6

7 Def 1 The P -value corresponding to an observed value of a test statistic is the lowest level of significance for which the test-statistic value results in rejection of the null hypothesis. Some properties of the P -value is as follows: 1. Given a P -value, someone not directly involved in the experiment can discern at which levels the test-statistic value would be significant. 2. P -value α = reject H 0 P -value > α = accept H 0 Example 4 Finding the P -value of the above example to make decision regarding the truth of H 0 at the levels of significance 0.05 and sol) The z test-statistic is z = Hence, the critical region to reject H 0 is given by z z α. Therefore, the least value of α to reject H 0 P -value) is P -value = Φ 1.33) = For α = 0.05, P -value > α. So, H 0 can not be rejected. For α = 0.10, -value < α. So, H 0 will be rejected. Procedure for hypothesis testing 1. Correctly identify the competing hypotheses. 2. Compute a suitable test-statistic from the sample. You must understand the distribution of the test-statistic. 3. Determined the critical region according to the significance level α. 4. Decide whether or not to reject the null hypothesis. 5. Give a conclusion from your decision. Power Function Def 2 The function P F θ) which denotes the probability of rejecting H 0 concluding H 1 ) when the population parameter is θ is called the power function of the decision rule. A graph of P F θ) is called the power curve of the decision rule. 7

8 Let Θ denote the set of possible values of θ such that H 0 is ture. Then, P F θ) = { αθ) θ Θ 1 βθ) θ Θ c Example 5 Determine the P F p) is Example 2 and plot P F p). sol) P F p) = ) Φ 0.07 p [ p1 p)/100 )] 1 1 Φ 0.07 p p1 p)/ p = Φ p1 p)/100 p 0.10 p < 0.10 The power curve is as follows: Remark 1. The ideal power curve of a decision rule satisfies P F θ) = { 0 θ Θ 1 θ Θ c 2. A decision rule with larger 1 βθ) for θ Θ c is better. Operating Characteristic Function Def 3 The function OCθ) which denotes the probability of accepting H 0 rejecting H 1 ) when the population parameter is θ is called the operating characteristic function of the decision rule. A graph of OCθ) is called the operating characteristic curve of the decision rule. 8

9 Let Θ denote the set of possible values of θ such that H 0 is ture. Then, { 1 αθ) θ Θ P F θ) = βθ) θ Θ c Clearly, OCθ) = 1 P F θ) Example 6 Determine the OCp) is Example 2 and plot OCp). sol) 0.07 p OCp) = 1 Φ p1 p)/100 = Φ 0.07 p p1 p)/100 The operating characteristic curve is as follows: 9.2 Test For A Population Mean A Normal Population with Known σ 2 Example 7 Upper one-tailed test A company that produces bias-ply tires is considering a certain modification in the tread design. An economic study indicates that the modification can be justified only if true average tire life under standard test conditions exceeds 20,000 miles. A random sample of n = 16 prototype tires is manufactured and tested, resulting in a sample average life of x = 20, 758. Suppose tire life is normally distributed with σ = 1, 500. Does this data suggest that the modification meets the condition required for changeover at significance level 0.01? 9

10 sol) In the following we use µ 0 = 20, Define Hypotheses 2. Test-Statistics Fact: H 0 : µ µ 0 H 1 : µ > µ 0 Nµ, σ 2 /n) = Z = X µ σ/ n N0, 1) Suppose that H 0 is true i.e., mu = µ 0 ), we have the z test-statistic as z = 20, , 000 1, 500/ Finding Critical Region Assume that x CR satifies Then, = P Type-I error) = P X x CR µ = µ 0 ) = α = 0.01) P X x CR µ = µ 0 ) = α X µ P σ/ n x ) CR µ σ/ n µ = µ 0 = α X µ0 P σ/ n x ) CR µ 0 σ/ = α n P Z x ) CR µ 0 σ/ = α n Since P Z z α ) = α, we have Hence, x CR µ 0 σ/ n = z α. x CR = µ 0 + z α σ/ n = 20, , 500)/ 16 = 20, This implies that if x-statistic is used, then the corresponding critical rigion is CR x = [x CR, ) = [20, 873, ). On the other hand if z-statistic is used, then the corresponding critical rigion is CR z = [z α, ) = [2.33, ). 4. Decision or, equivalently, x = 20, 758 CR x, H 0 is true, z = CR z, H 0 is true. 10

11 5. Conclusion H 0 cannot be rejected at significance level The data does not give strong support to the claim that true average tire life for the new tread design exceeds 20,000 miles. Production of the current version should be continued. β and Sample Size Determination Example 8 Suppose that the true mean µ = 21, 000 in Example 7. Find β21, 000). Determine the sample size n required if we want to have β21, 000) = 0.1. sol) 1. With α = 0.01, we obtain CR z = [z α, ) = [2.33, ). Hence, with µ 0 = 20, 000, and µ 1 = 21, 000, we have βµ 1 ) = P Z CR z µ = µ 1 ) ) X µ0 = P σ/ n < z α µ = µ 1 = P X < µ 0 + z α σ/ ) n µ = µ 1 X µ = P σ/ n < µ ) 0 µ σ/ n + z α µ = µ 1 X µ1 = P σ/ n < µ ) 0 µ 1 σ/ n + z α = P Z < µ ) 0 µ 1 σ/ n + z α ) µ0 µ 1 = Φ σ/ n + z α ) 20, , 000 = Φ 1, 500/ = Φ 0.34) = From the above, one see βµ 1 ) = P Z < µ ) 0 µ 1 σ/ n + z α = 1 P Z µ ) 0 µ 1 σ/ n + z α We hope β = βµ 1 ) = 0.1. Hence, 1 P Z µ ) 0 µ 1 σ/ n + z α = P Z µ ) 0 µ 1 σ/ n + z α = β = 1 β = µ 0 µ 1 σ/ n + z α = z β 11

12 This leads to n = = [ σzα + z β ) ] 2 µ 0 µ 1 [ ] 2 1, ) = 5.42) 2 = , , 000 To achieve α = 0.01 and β = 0.1, we hence choose n = 30. Example 9 Lower one-tailed test Suppose a paricular application has to use logic gates with propagation delays less than 22 ns. A manufacturer claims that, owing to the use of new semiconductor material, the logic gates produced by his or her company meet the requirement. To check the claim, a random sample of n = 16 is selected and tested, resulting in a sample average propagation delay of x = ns. Assuming the measurement of propagation delay to be normally distrubued with standard deviation σ = 0.4 ns. Does this seem to indicate that the average propagation delay less than 22 ns for the gates producted using the new technology? Use a 0.05 level of significance. sol) In the following we use µ 0 = 22 ns. 1. Define Hypotheses H 0 : µ µ 0 H 1 : µ < µ 0 2. Test-Statistics Fact: X Nµ, σ 2 /n) = Z = X µ σ/ N0, 1) n Suppose that H 0 is true i.e., µ = µ 0 ), we have the z test-statistic as z = / 16 = Finding Critical Region Assume that x CR satifies Then, P Type-I error) = P X x CR µ = µ 0 ) = α = 0.01) P X x CR µ = µ 0 ) = α X µ P σ/ n x ) CR µ σ/ n µ = µ 0 = α X µ0 P σ/ n x ) CR µ 0 σ/ = α n P Z x ) CR µ 0 σ/ = α n 12

13 Since P Z z α ) = α, we have x CR µ 0 σ/ n = z α. Hence, x CR = µ 0 z α σ/ n = )/ 16 = This implies that if x-statistic is used, then the corresponding critical rigion is CR x =, x CR ] =, ]. On the other hand if z-statistic is used, then the corresponding critical rigion is CR z =, z α ] =, 1.645]. 4. Decision or, equivalently, x = CR x, H 0 is true, z = 1.5 CR z, H 0 is true. 5. Conclusion H 0 cannot be rejected at significance level The data does not give strong support to the claim that true average propagation delay is less than 22 ns. Example 10 Consider Example Find β21.7). 2. If we require that β21.7) = 0.05, find the necessary size n. 3. Using sample size obtained above to redo the test x = 21.85). sol) 1. With α = 0.05, we have CR z =, z α ] =, 1.645]. Hence, with µ 0 = 22, and µ 1 = 21.7, we have βµ 1 ) = P Z CR z µ = µ 1 ) ) X µ0 = P σ/ n > z α µ = µ 1 = P X < µ 0 z α σ/ ) n µ = µ 1 X µ = P σ/ n > µ ) 0 µ σ/ n z α µ = µ 1 X µ1 = P σ/ n > µ ) 0 µ 1 σ/ n z α 13

14 2. From the above, one see = P Z > µ ) 0 µ 1 σ/ n z α ) µ0 µ 1 = 1 Φ σ/ n z α ) = 1 Φ 0.4/ = 1 Φ1.355) = Φ 1.355) = βµ 1 ) = P We hope β = βµ 1 ) = Hence, This leads to Z > µ ) 0 µ 1 σ/ n z α P Z > µ ) 0 µ 1 σ/ n z α = P Z µ ) 0 µ 1 σ/ n z α = β = β = µ 0 µ 1 σ/ n z α = z β n = = [ σzα + z β ) ] 2 µ 0 µ 1 [ ] ) = To achieve α = 0.05 and β21.7) = 0.05, we hence choose n = Define Hypotheses H 0 : µ 22 H 1 : µ < 22 Test-Statistics Finding Critical Region z = / 20 = CR =, z α ] =, 1.645]. Decision z = CR, H 0 will be rejected. Conclusion At the significance level of 0.05, the data strongly support that true average propagation delay is less than 22 ns. 14

15 Example 11 Two-tailed test Consider a chemical production process in which it is necessary to maintain a proper ph level. Suppose a ph level of 8.8 is desirable and a chemist plans to take several measurements to detect whether there has been significant deviation from that level. Because a conclusion that ph level varies too greatly from 8.8 can cause a production interruption, the chemist wishes to avoid a false detection. It is decided that a suitable level is α = Suppose 3 measurements are taken and the empirical mean of the observations is x = 9.1. From experience it is known that the measurements tend to be normally distributed with standard deviation σ = 0.3. Test the folowing hypothesis: H 0 : µ = 8.8 H 1 : µ 8.8 sol) In the following we use µ 0 = Hypotheses 2. Test-Statistics Fact: H 0 : µ = µ 0 H 1 : µ µ 0 X Nµ, σ 2 /n) = Z = X µ σ/ n N0, 1) Suppose that H 0 is true i.e., mu = µ 0 ), we have the z test-statistic as z = / 3 = Finding Critical Region Assume that x CR satifies Then, P Type-I error) = P X µ x CR µ = µ 0 ) = α = 0.01) P X µ x CR µ = µ 0 ) = α X µ P σ/ n x ) CR σ/ n µ = µ 0 X µ 0 P σ/ n x ) CR σ/ n X µ0 P σ/ n x CR σ/ n or X µ 0 σ/ n x ) CR σ/ n = α = α = α 15

16 P Z x CR σ/ n or Z x ) CR σ/ n P Z x ) CR σ/ n = α = α/2 Since P Z z α/2 ) = α/2, we have x CR σ/ n = z α/2. Hence, x CR = z α/2 σ/ n = )/ 3 = This implies that if x-statistic is used, then the corresponding critical rigion is CR x =, µ 0 x CR ] [µ 0 + x CR, ) =, ] [9.1395, ). On the other hand if z-statistic is used, then the corresponding critical rigion is CR z =, z α/2 ] [z α/2, ) =, 1.96] [1.96, ). 4. Decision or, equivalently, x = 9.1 CR x, H 0 is true, z = CR z, H 0 is true. 5. Conclusion H 0 cannot be rejected at significance level Example 12 Consider Example 11. It is recognized that a deviation of 0.4 ph can portend eventual problems, the conclusion being that given a deviation of 0.4, there only be 0.10 probability of not rejecting the null hypothesis. a) Find the sample size n required. b) Taking a sample with size decided above, the empirical mean of the observations is x = 9.1. Redo the test. sol) a) From the problem statement, one see that the chemist wants to have β9.2) = β8.4) = β =

17 Let µ 1 = 9.2. Then, βµ 1 ) = P Z CR µ = µ 1 ) = P z α/2 < Z < z α/2 µ = µ 1 ) = P z α/2 < X µ ) 0 σ/ n < z α/2 µ = µ 1 = P µ 0 z α/2 σ/ n < X < µ 0 + z α/2 σ/ ) n µ = µ 1 µ0 µ = P σ/ n z α/2 < X µ σ/ n < µ ) 0 µ σ/ n + z α/2 µ = µ 1 µ0 µ 1 = P σ/ n z α/2 < X µ 1 σ/ n < µ ) 0 µ 1 σ/ n + z α/2 µ0 µ 1 = P σ/ n z α/2 < Z < µ ) 0 µ 1 σ/ n + z α/2 = β For µ 1 > µ 0 µ0 µ 1 β = P σ/ n z α/2 < Z < µ ) 0 µ 1 σ/ n + z α/2 P Z < µ ) 0 µ 1 σ/ n + z α/2 Hence, So, that n µ 0 µ 1 σ/ n + z α/2 z β = = 5.9 [ σzα /2 + z β ) ] 2 µ 0 µ 1 [ ) ] 2 Thus, we choose n = 6. For µ 1 < µ 0, we will get the same result. b) 1) Hypotheses H 0 : µ = 8.8 H 1 : µ 8.8 2) Test-Statistics z = / 6 =

18 3) Critical Region CR =, z ] [z 0.025, ) =, 1.96] [1.96, ). 4) Decision z = 2.45 CR x, H 0 is false. 5) Conclusion H 0 is rejected at significance level Summary The important items for z-test with σ 2 being known are summarized as follows: Test Lower one-tailed Two-tailed Upper one-tailed H 0 µ µ 0 µ = µ 0 µ µ 0 H 1 µ < µ 0 µ µ 0 µ > µ 0 Remark Statistic Z = X µ 0 σ/ n CR, z α ], z α/2 ] [z α/2, ) [z α/2, ) ) µ0 µ ) 1 Φ µ0 µ 1 βµ 1 ) 1 Φ σ/ n z σ/ n + z α/2 α ) Φ Φ n µ0 µ 1 σ/ n z α/2 ) µ0 µ 1 σ/ n + z α [ ] 2 [ ] 2 [ σzα + z β ) σzα/2 + z β ) σzα + z β ) µ 0 µ 1 µ 0 µ 1 µ 0 µ 1 P -value Φz) 2[1 Φz)] 1 Φz) Using a large sample n > 30) to test the population-mean not necessary be a normal population) with unknown σ 2, we can still perform the z-test as described above, but, by simply replacing σ with s. Example 13 A manufacture of sports equipment has developed a news synthetic fishing line that he claims has a mean breaking strength of 8 kilograms. Test the hypothesis that µ = 8 kilograms against the alternative that µ 8 kilograms if a random sample of 50 lines is tested and found to have a mean breaking strength of 7.8 kilograms and sample standard deviation 0.5 kilogram. Use a 0.01 level of significance. ] 2 18

19 sol) 1. Hypotheses H 0 : µ = 8 H 1 : µ 8 2. Statistic By n = 50> 30), x = 7.8 and s = 0.5, we have 3. Critical Region By α = 0.01, we have z = / 50 = CR =, z ] [z 0.005, ) =, 2.575] [2.575, ) 4. Decision z = CR, H 0 is flase. 5. Conclusion H 0 will be rejected at a 0.01 level of significance. Example 14 Find the P -value for the test in Example 13. Give your conclusion. sol) According to the table, we have P -value = 21 Φ z )) = 21 Φ2.828)) = ) = P -value = < α = This allows us to reject the null hypothesis that µ = 8 kilograms at a 0.01 level of significance A Normal Population with Unknown σ 2 Small Sample Disribution Regarding X For a normal population, we have the following properties: X Nµ, σ 2 /n). Z = X µ σ/ n N0, 1) n 1)S2 σ 2 χ 2 n 1 19

20 X and S 2 are independent. T = X µ S/ n t n 1 Method for the tests T -test) According to the distribution of T -statistic described above, the method to test the populationmean can be developed as follows: Test Lower one-tailed Two-tailed Upper one-tailed H 0 µ µ 0 µ = µ 0 µ µ 0 H 1 µ < µ 0 µ µ 0 µ > µ 0 Statistic T = X µ 0 S/ n CR, t n 1,α ], t n 1,α/2 ] [t n 1,α/2, ) [t n 1,α/2, ) βµ 1 ) n P -value P T < t) 21 P T < t )) P T > t) Example 15 The Edison Electirc Institude has published figures on the annual number of kilowatt-hours expended by various home appliances. It is claimed that a vacuum cleaner expends an average of 46 kilowatt-hours per year. If a random sample of 12 homes included in a planned study indicates that vacuum cleaners expend an average of x = 42 kilowatt-hours per year with a standard deviation of s = 11.9 kilowatt-hours, does this suggest at the 0.05 level of significance that vacuum cleaners expend, on the average, less that 46 kilowatt-hour annually? Assume the population of kilowatt-hours to be normal. sol) 1. Hypotheses H 0 : µ = 46 kwh s H 1 : µ < 46 kwh s 20

21 2. Statistic By n = 12< 30), x = 42, and s = 11.9, we have t = 3. Critical Region By α = 0.05, we have / 12 = 1.16 CR =, t 0.05,11 ] =, 1.796]. 4. Decision t = 1.16 CR, H 0 is ture. 5. Conclusion The average number of kilowatt-hours expended annually by home vacuum cleaners is not significantly less than Test For A Population Proportion Disribution Regarding ˆp Large Samples) For large samples, ˆp has the following properties: ˆp Np, pq/n). Z = ˆp p pq/n N0, 1) Method for the tests According to the distribution of Z-statistic described above, the method to test the population proportion large sample) can be developed as follows: 21

22 Test Lower one-tailed Two-tailed Upper one-tailed H 0 p p 0 p = p 0 p p 0 H 1 p < p 0 p p 0 p > p 0 Statistic Z = ˆp p 0 p 0 q 0 /n CR, z α ], z α/2 ] [z α/2, ) [z α, ) βp 1 ) n ) p0 p ) 1 + z α/2 σ 0 p0 p 1 z α σ Φ ) 0 1 Φ σ p0 1 p 1 + z α σ 0 ) Φ σ 1 p0 p 1 z α/2 σ 0 σ 1 Φ [ zα σ 0 + z β σ 1 ] 2 [ zα/2 σ 0 + z β σ ] 2 1 [ zα σ 0 + z β σ 1 p 1 p 0 p 1 p 0 p 1 p 0 P -value Φz) 2[1 Φ z )] 1 Φz) σ 0 = σ 0 = σ 1 p 0 1 p 0 )/n, σ 1 = p 1 1 p 1 )/n p 0 1 p 0 ), σ 1 = p 1 1 p 1 ) ] 2 Example 16 A plan for an executive traveler s club has been developed by an airline on the premise that 5% of its current customers would qualify for membership. A random sample of 500 customers yielded 40 who would qualify. 1. Using this data, test at level 0.01 the null hypothesis that the company s premise is correct against the alternative that it is not correct. 2. What is the probability that when the test of 1) is used, the company s the company s premise will be judged correct when in fact 10% of all current customers qualify? 3. What sample size would be sufficient to ensure that both α and the probability of 2) are 0.01? 1. Hypothesis H 0 : p = 0.05 H 1 : p

23 Test statistic ˆp = = z = 0.05)0.95)/500 = Critical Region CR =, z ] [z 0.005, ) =, 2.575] [2.575, ) Decision z = CR, H 0 is flase. Conclusion At a level of 0.01, we reject the claim that 5% of its current customers would qualify for membership. 2. We want to find β0.1). From the table, we have β0.10) = Φ )0.95)/ )0.90)/500 Φ )0.95)/ )0.90)/500 = Φ 1.856) P hi 5.597) = We want to have α = β0.10) = From the table, we have Hence, we choose n = 637. Small Sample n = )0.95) )0.90) = For a small sample, we can perform the test using P -value. The steps are briefly described as follows: 1. List H 0 and H 1. Suppose that H 0 is p = p Compute P -value Since X Bn, p), the corresponding P -value is given by Lower one-tailed test P -value = P X x when p = p 0 ) = k=x k=0 n k ) p k 01 p 0 ) n k 23

24 Upper one-tailed test Two-tailed test P -value = P X x when p = p 0 ) = P -value = k=n k=x n k ) p k 01 p 0 ) n k { 2P X x when p = p0 ) if x np 0 2P X x when p = p 0 ) if x > np 0 3. Decision If P -value α, reject H 0 ; otherwise, accept H Makes a conclusion. Example 17 A builder claims that heat pumps are installed in 70% of all homes being constructed today in the city of Richmond. Would you agree with this claim if a random survey of new homes in this city shows 8 out of 15 had heat pumps installed? Use a 0.10 level of significance. sol) 1. Hypotheses H 0 : p = 0.7 H 1 : p Compute P -value n = 15, x = 8, and p 0 = 0.7. Since x < np 0 = 10.5, we have P -value = 2P X 8 when p = 0.7) = 8 2 bx; 15, 0.7) = Decision Since P -value > α = 0.10, H 0 is not be rejected. 4. Conclusion There is infufficient reason to doubt the builder s claim. x=0 9.4 Tests For The Sum Or Difference Of Two Means σ 2 1 and σ 2 2 are known or large samples) Assume that the two populations satisfy X Nµ 1, σ 2 1) and Y Nµ 2, σ 2 2). 24

25 Then X ± Y N Z = X ± Y ) µ 1 ± µ 2 ) N0, 1) σ1 2 n 1 + σ2 2 n 2 µ 1 ± µ 2, σ2 1 n 1 + σ2 2 n 2 ) In the following table, we summarize the test method. Test Lower one-tailed Two-tailed Upper one-tailed H 0 µ 1 ± µ 2 0 µ 1 ± µ 2 = 0 µ 1 ± µ 2 0 H 1 µ 1 ± µ 2 < 0 µ 1 ± µ 2 0 µ 1 ± µ 2 > 0 Statistic Z = X ± Y ) 0 σ 2 1 n 1 + σ2 2 n 2 CR, z α ], z α/2 ] [z α/2, ) [z α, ) ) 0 ) 1 Φ + z 0 α/2 ) 1 σ 0 β 1 ) 1 Φ z X±Y 1 α ) Φ + z α σ X±Y 0 1 σ Φ z X±Y α/2 σ X±Y [ ] 2 [ ] 2 [ σ1 + σ 2 )z α + z β ) σ1 + σ 2 )z α/2 + z β ) σ1 + σ 2 )z α + z β ) n P -value Φz) 2[1 Φz)] 1 Φz) ] 2 In the table entry for computing n, we assume that n = n 1 = n 2. 25

26 Example 18 To compare the performace in the average turnaround times sec) of a computer using two different scheduling methods A and B. Experimental results are as follows: Method A Method B n 1 = 40 n 2 = 40 x 1 = 3.64 x 2 = 4.25 s 1 = 0.53 s 2 = A claim is made that the mean turnaround time for method A is more than 0.4 sec faster than B. Is this claim correct? Test at a 0.05 level of significance. 2. Find βµ 1 µ 2 = 0.5). sol) Because large sample, we can use Z-test. 1. Hypotheses H 0 : µ 1 µ H 1 : µ 1 µ 2 < 0.4 Statistic Critical Region z = = 1.91 CR =, z 0.05 =, 1.65] Decision z = 1.91 CR, H 0 is false. 2. Conclusion Method A results in a mean turnaround time of more than 0.4 second faster than method B. β 0.5) = 1 Φ ) = 1 Φ 0.74) =

27 Example 19 In a studty carried out to compare the effectiveness of a contract grading method with the traditional method of grading, ninth-grade students were tested for retention of material five weeks after completing a venereal disease unit in a health course. Contract Traditional n 1 = 32 n 2 = 31 x 1 = x 2 = s 1 = 8.00 s 2 = 8.12 A pretest revealed virtually no difference in knowledge level prior to the unit.) Does the data suggest that there is a difference in true average retention level for the two methods? Compute the P -value, and use it to reach a conclusion at level of significance sol) Because of large samples, we use z-statistic. Hypothese H 0 : µ 1 = µ 2 µ 1 µ 2 = 0) H 1 : µ 1 µ 2 µ 1 µ 2 0) Statistic P -value z = = 0.54 P -value = 2[1 Φ z )] = 2[1 Φ0.54)] = Decision P -value = > α = 0.01, H 0 is not rejected Conclusion The data does not suggest that there is a difference in true average retention level for the two methods. σ 2 1 = σ 2 2 = σ 2 but unknown Assume that the two populations satisfy X Nµ 1, σ 2 1) and Y Nµ 2, σ 2 2). Facts: X ± Y N µ 1 ± µ 2, σ2 1 n 1 + σ2 2 n 2 Z = X ± Y ) µ 1 ± µ 2 ) N0, 1) σ1 2 n 1 + σ2 2 n 2 T = X ± Y ) µ 1 ± µ 2 ) S p 1 n n 2 ) t n1 +n

28 where S p = n 1 1)S n 2 1)S 2 2 n 1 + n 2 2 In the following table, we summarize the test method. Test Lower one-tailed Two-tailed Upper one-tailed H 0 µ 1 ± µ 2 0 µ 1 ± µ 2 = 0 µ 1 ± µ 2 0 H 1 µ 1 ± µ 2 < 0 µ 1 ± µ 2 0 µ 1 ± µ 2 > 0 Statistic T = X ± Y ) 0 s p 1 n n 2 CR, t α,n1 +n 2 2], t α/2,n1 +n 2 2] [t α/2,n1 +n 2 2, ) [t α,n1 +n 2 2, ) β 1 ) n P -value Example 20 An experiment was performed to compare the abrasive wear of two different laminated materials. Twelve pieces of material 1 were tested, by exposing each piece to a machine measuring wear. Ten pieces of material 2 were similarly tested. In each case, the depth of wear was observed. The samples of material 1 gave an average coded) wear of 85 units with a sample standard diviation of 4, while the samples of material 2 gave an average of 81 and a sample standard deviation of 5. Can we conclude at the 0.05 level of significance that the abrasive wear of material 1 exceeds that of material 2 by more than 2 units? Assume the populations to be approximately normal with equal variances. sol) 1. Hypotheses H 0 : µ 1 µ 2 = 2 H 1 : µ 1 µ 2 > 2 28

29 2. Statistic n 1 = 12, x 1 = 85, s 1 = 4 and n 2 = 10, x 2 = 81, s 2 = 5. s 2 p = 12 1) )5 2 = s p = t = 85 81) /12) + 1/10) = Critical Region α = CR = [t 0.05,20, ) = [1.725, ) 4. Decision t = 1.04 CR, H 0 is considered true. 5. Conclusion At the level of 0.05, we are unable to conclude that the abrasive wear of material 1 exceeds that of material 2 by more than 2 units. σ 2 1 σ 2 2 and unknown The test method is summarized as follows. Test Lower one-tailed Two-tailed Upper one-tailed H 0 µ 1 ± µ 2 0 µ 1 ± µ 2 = 0 µ 1 ± µ 2 0 H 1 µ 1 ± µ 2 < 0 µ 1 ± µ 2 0 µ 1 ± µ 2 > 0 Statistic T = X ± Y ) 0 s 2 1 n 1 + s2 2 n 2 CR, t α,ν ], t α/2,ν ] [t α/2,ν, ) [t α,ν, ) β 1 ) n P -value Remark Analysis of Paired Data { s 2 ν = round 1/n 1 + s 2 2/n 2 ) 2 }. [s 2 1/n 1 ) 2 /n 1 1)] + [s 2 2/n 2 ) 2 /n 2 1)] 29

30 In certain environment, it is difficult to control the experiments to be performed under the same condition. Therefore, to measure the difference of two population means, we conduct the measurement by taking paired observation. In this paring structure, the conditions of the two populations are assigned randomly with homogeneous units. Let D 1 = X 1 Y 1,..., D n = X n Y n represent the difference of n paired experiments. We will assume D i Nµ D, σ 2 ). Facts: D = 1 n D 1 Nµ d, σ 2 /2) n i=1 SD 2 1 n = D i D) 2 = n 1)S2 D n 1 i=1 σ 2 χ 2 n 1 T = D µ d S D / n t n 1 Accordingly, given paired data, we can test the difference of ture means as follows: Test Lower one-tailed Two-tailed Upper one-tailed H 0 µ D µ 0 or µ 1 µ 2 µ 0 µ D = µ 0 or µ 1 µ 2 = µ 0 µ D µ 0 or µ 1 µ 2 µ 0 H 1 µ D < µ 0 or µ 1 µ 2 < µ 0 µ D µ 0 or µ 1 µ 2 µ 0 µ D > µ 0 or µ 1 µ 2 > µ 0 Statistic T = D µ 0 S D / n CR, t n 1,α ], t n 1,α/2 ] [t n 1,α/2, ) [t n 1,α/2, ) βµ 1 ) n P -value P T < t) 21 P T < t )) P T > t) 30

31 Example 21 A taxi company is trying to decide whether the use of radial tires instead of regular belted tires improves fuel economy. Twelve cars were equipped with radial tires and driven over a prescribed test course. Without changing drivers, the same cars were equipped with regular belted tires and driven once again over the test course. The gasoline confumption, in kilometers per liter, was recorded as follows: Kilometers per Liter Car Radial Tires Belted Tires At the level of significance, can we conclude that cars equipped with radial tires give better fuel economy than those equipped with belted tires? Assume the populations to be normally distributed. sol) 1. Hypotheses H 0 : µ D 0 or µ 1 µ 2 0 H 1 : µ D > 0 or µ 1 µ 2 > 0 2. Statistic 31

32 Kilometers per Liter Car Radial Tires Belted Tires d i d 2 i Total Critical Region α = d = 1.7/12 = s 2 D = /12) = s D = t = / 12 = CR = [t 0.025,11, ) = [2.201, ) 4. Decision t = CR, H 0 is considered false 5. Conclusion At level 0.025, we conclude that cars equipped with radial tires give better fuel economy than those equipped with belted tires. 9.5 Tests For The Proportions Of Two Populations Large Sample For large samples, we have ˆp 1 Np 1, p 1 q 1 /n 1 ) and ˆp 2 Np 2, p 2 q 2 /n 2 ). Hence, ˆp 1 ˆp 2 N p 1 p 2, p 1q 1 + p ) 2q 2 n 1 n 2 Z = ˆp 1 ˆp 2 ) p 1 p 2 ) p1 q 1 n 1 + p 2q 2 N0, 1) n 2 32

33 Suppose we have H 0 : p 1 p 2 = 0 or p 1 = p 2 ), and assume H 0 is true. Then, we will use ˆp = X 1 + X 2 = n 1 ˆp 1 + n 2 ˆp 2 n 1 + n 2 n 1 + n 2 n 1 + n 2 to approximate p 1 and p 2. Accordingly, the test method is summarized as follows: Test Lower one-tailed Two-tailed Upper one-tailed H 0 p 1 p 2 0 p 1 p 2 = 0 p 1 p 2 0 H 1 p 1 p 2 < 0 p 1 p 2 0 p 1 p 2 > 0 Statistic Z = ˆp 1 ˆp 2 ˆpˆq ) 1 n n 2 CR, z α ], z α/2 ] [z α/2, ) [z α, ) Example 22 A vote is to be taken among the residents of a town and the surrounding county to determine whether a proposed chemical plant should be constructed. The construction site is within the town limits and for this reason many votes in the county feel that the proposal will pass because of the large proportion of town voters who favor the counstruction. To determine if there is a significant difference in the proportion of town voters and county voters favoring the proposal, a poll is taken. If 120 of 200 town voters favor the proposal and 240 of 500 county residents favor it, would you agree that the proportion of town voters favoring the proposal is higher than the proportion of county voters? Use a level of significance. sol) 1. Hypotheses H 0 : p 1 p 2 H 1 : p 1 > p 2 2. Statistic ˆp 1 = 120/200 = 0.60 ˆp 2 = 240/500 = 0.48 ˆp = = 0.51 z = )0.49)[1/200) + 1/500)] =

34 3. Critical Region CR = [z 0.025, ) = Decision z = 2.9 CR, H 0 is considered false 5. Conclusion At level 0.025, we agree that the proportion of town voters favoring the proposal is higher than the proportion of county voters. 9.6 Tests For The Two population Variances Let X 1,..., X m and Y 1,..., Y n be two random samples for two normal populations Nµ 1, σ 2 1) and Nµ 2, σ 2 2), respectively. Define S 2 1 = 1 m 1 m X i X) 2, S2 2 = 1 n Y i Y ) 2 i=1 n 1 i=1 Then, S2 1/σ 2 1 S 2 2/σ 2 2 F m 1,n 1 Accordingly, we summarize the test methods for the two population variance as follows: Test Lower one-tailed Two-tailed Upper one-tailed H 0 σ 2 1 σ2 2 or σ2 1 /σ2 2 1) σ2 1 = σ2 2 or σ2 1 /σ2 2 = 1) σ2 1 σ2 2 or σ2 1 /σ2 2 1) H 1 σ 2 1 < σ2 2 or σ2 1 /σ2 2 1) σ2 1 σ2 2 or σ2 1 /σ2 2 1) σ2 1 > σ2 2 or σ2 1 /σ2 2 > 1) Statistic F = S2 1 S 2 2 CR [0, f 1 α,m 1,n 1 ] 0, f 1 α/2,m 1,n 1 ] [f α/2,m 1,n 1, ) [f α,m 1,n 1, ) 34

35 Example 23 A study is conducted to compare the length of time between men and women to assemble a certain product. Past experience indicates that the distribution of times for both men and women is approximately normal but the variance of the times for women is less than that for men. A random sample of times for 11 men and 14 women produced the following data: Men Women n 1 = 11 n 2 = 14 s 1 = 6.1 s 2 = 5.3 Does the data suggest that the time of variance for men is greater than that of women? Use a 0.01 level of significance. sol) 1. Hypotheses 2. Statistic H 0 : σ 2 1 = σ 2 2 H 1 : σ 2 1 > σ 2 2 f = = Critical Region α = 0.01, n 1 = 11, n 2 = 14. CR = [f 0.01,10,13, ) = [4.10, ) 4. Decision f = CR, H 0 is considered true 5. Conclusion At level α = 0.01, the data does not give strong evidence to suggest that the time of variance for men is greater than that of women. Example 24 Two machines are used to fill juice containers that are normally rated at one quart. The following data are recorded in quarts) by measuring actual juice content in randomly selected containers: Machine A Machine B Assume that the content populations are normally distributed. At α = 0.05, does these data suggest that the two machines give different quantities of juice? sol) 35

36 1. First, test whether or not that the two populations have equal variaces. Hypotheses Statistic H 0 : σ 2 1 = σ 2 2 H 1 : σ 2 1 σ 2 2 s 2 1 = s 2 2 = f = / = Critical Region α = 0.05, n 1 = n 2 = 6. Decision CR = [0, f 0.975,5,5 ] [f 0.025,5,5, ) = [0, ] [9.60, ) f = CR, H 0 is considered true Conclusion At α = 0.05, there is not sufficient evidence to conclude that σ 2 1 σ Based on the above result, we can perform T test as follows: Hypotheses Statisctic n 1 = n 2 = 6. x = H 0 : µ 1 = µ 2 H 1 : µ 1 µ 2 y = 0.99 s ) ) p = = s p = t = = Critical Region α = 0.05, n 1 = n 2 = 6 Decision CR =, t 0.025,10 ] [t 0.025,10, ) =, 2.228] [2.228, ) t = CR, H 0 is considered true Conclusion At level α = 0.05, the data does not give strong evidence to conclude that the machines give different quantities of juice. 36

37 9.7 Exercise 1. Suppose a random sample x 1, x 2,..., x 10 of size 10 comes from a distribution with pdf fx; p) = { p x 1 p) 1 x x = 0, 1 0 otherwise where it is known that 0 < p < 1/2. Find the power function for the testing the null hypothesis H 0 : p = 1/2 against the composite alternative hypothesis H 1 : p < 1/2. Assume X = X X 10 3 is used as the critical region. 2. A random sample of 100 recorded deaths in the United States during the past year showed an average life span of 71.8 years. Assuming a population standard deviation of 8.9 years, does this seem to indicate that the average life span today is greater than 70 years? Use a 0.05 level of significance. 3. A certain machine is said to be in control if the proportion of defective items manufactured by the machine is not greater than 10%. To check whether the machine is in control, 10 finished items are randomly selected from its output. The implicit hypothesis that the machine is in control will be rejected if 3 or more defectives are found. a) What is the probability of type-i error for this test? b) If the machine is really out of control and the probability of a defective is 0.3. What is the probability of type-ii error for the test? 4. A researcher wants to test H 0 : θ = 0 versu H 1 : θ = 1, where θ is the parameter of a population of interest. The static W, based on a random sample from a population, is used to test the hypothesis. Suppose that under H 0, W N0, 1) and under H 1, W N4, 1), and H 0 is rejected when W > 1.5. Find the probability of type-i and type-ii errors, respectively. 5. The length of time necessary to complete a manufacturing part is a normal distribution measured in seconds and it is known that σ = 5.2 sec. The following times were recorded for 6 parts: a) Test H 0 : µ = 119 vs. H 1 : µ < 119, using α = b) For the test in a), what is the probability os a Type-II Error if µ = 117 sec. 6. A soft-drink machine at Longhorn Steak House is regulated so that the amount of drink dispensed is approximately normally distributed with a mean of 200 milliliters and a standard deviation of 15 milliliters. The machine is checked periodically by taking a sample of 9 drinks and computing the average content. If x falls in the interval 191 < x < 209, the machine is thought to be operating satisfactory; otherwise, we conclude that µ 200 milliliters. a) Find the probability of committing a type-i error when µ = 200 milliliters. 37

38 b) Find the probability of committing a type-ii error when µ = 215 milliliters. 7. It is claimed that an automobile is driven on the average more than 20,000 kilometers per year. To test this claim, a random sample of 100 automobile owners are asked to keep a record of the kilometers they travel. Would you agree with this claim if the random sample showed an average of 23,500 kilometers and a standard deviation of 3,900 kilometers? Use a P -value in your conclusion. 8. In order to test gasoline mileage performance for a new version of one of its compact cars, an automobile manufacturer selected six nonprofessional drivers to drive a car from Phoenix to Los Angeles. The miles per gallon values for the six cars at the conclusion of the trip were 27.2, 29.3, 31.5, 28.7, 30.2, and The manufacturer wishes to advertise that cars of this type average at least) 30 mpg on such long trips. Assume the population of mpg to be normal. Does the sample data contradict the validity of this claim? Use a 0.05 level of significance. 9. The relative conductivity of a semiconductor device is determined by the amount of impurity doped into the device during its manufacture. A silicon diode to be used for a specific purpose requires a cut-on voltage of 0.60 volt, and if this is not achieved then the mechanism governing the amount of impurity must be adjusted. A sample of 12 such diodes yielded a sample average voltage of 0.62 volt and sample deviation of At level 0.01, does the data indicate that the true average cut-on voltage is something other than 0.60? 10. A cigarette manufacturer advertises that its new low-tar cigarette contains on average no more than 4 milligrams of tar. You have been asked to test the claim using following sample information: n = 25, x = 4.16 milligrams, s = 0.30 milligrams. Does the sample information disagree with the manufacturer s claim? Test using α = List any assumption you make. 11. At a certain colledge it is estimated that at most 25% of the students ride bicycles to class. Does this seem to be a valid estimate if, in a random sample of 90 students, 28 are found to ride bicycles to class? Use a 0.05 level of significance. 12. A supplier claims that at least 96% of the parts it supplies meet the product specifications. In a sample of 100 parts received over the past 6 months, 7 were defective. Answer the following questions: a) Does the data indicate that the supplier is cheating the costomers? Test at the 0.05 level of significance. b) Using the sample result, develop a 95% confidence interval for the proportion of the population that are defective. Explain your results. c) Use the interval estimate of part b) to test whther the supplier is cheating the costomers. What is your conclusion? Explain. d) Explain Type-I and Type-II errors in terms of this problem. e) What is the probability of Type-II error if the true proportion of the parts which meet the product specification is

39 13. A certain machine is said to be in control if the proportion of defective items manufactured by the machine is not greater than 10%. To chect whether the machine is in control, 10 finished items are randomly selected from its output. The implicit hypothesis that the machine is in control will be rejected if 3 or more defectives are found. a) What is the probability of the Type-I error for this test? b) If the machine is really out of control and the probability of a defective is 0.3. What is the probability of Type-II error for the test? 14. A distributor of cigarettes claims that 20% of the smokers in Miami prefer Kent cigarettes. To test this claim, 20 cigarette smokers are selected at random and asked what brand they prefer. If 6 of the 20 named Kent as the preference, what conclusion do we draw? Use a 0.05 level of significance. 15. To test whether it can be concluded that there is a difference in the hardness of gray cast iron resulting from two production sources, ten independent measurements of Brinell hardness are taken from the iron produced by each of the two sources. Assuming both measurement populations are normally distributed with known standard deviations σ 1 = σ 2 = 0.8, set up the appropriate hypothesis test to check a difference in the means yield the emprical value x = and y = 160.8, what conclusion is drawn? 16. The lifetime of a metal-cutting tool varies according to a number of factors. A turning cutter is tested at feed rates 0.30 and 0.35 mm per revolution, the speed of the tool being held constant and the test metal being a steel of fixed hardness. Let X and Y denote the random variables corresponding to the lifetimes in minutes) of tools tested at the feed rates 0.30 and 0.35, respectively. Given that 30 tests are independently run at feed rate 0.30, 40 tests are independently run at feed rate 0.35, the respective empirical means are x = 41.5 and y = 40.2 minutes, and the respective empirical standard deviations are s X = 3.92 and s Y = 4.51, can it be concluded at the α = 0.01 level that the two feed rates result in different tool lives? 17. A large supermarket chain is interested in determining whether a difference exists between the mean shelf life in days) of brand S bread and brand H bread. Random samples of 50 freshly baked loaves of each brand were tested, with the results shown in the table Brand S Brand H x 1 = 4.1 x 2 = 5.2 s 1 = 1.2 s 2 = 1.4 a) Is there sufficient evidence to conclude that a difference does exist between the mean shelf lives of brand S and brand H bread? Test at the α = 0.05 level. b) Find the observed significance level for the test and interpret its value. c) Let µ 1 and µ 2 represent the mean shelf lives for brands S and H, respectively. Construct a 90% confidence interval for µ 1 µ 2 ). Give an interpretation of your confidence interval. 39

40 18. High nitrate intake in food consumption has been shown to have a number of deleterious effects, including lower thyroxin production, increased incidence of cyanosis in newborns, and lower milk production in dairy cows. The following data is the result of an experiment to measure the percentage of weight gain for young laboratory mice given a standard diet and mice given ppm nitrate in their drinking water. Nitrate Control Use the test procedure with statistic T to decide whether the data indicates at level 0.01 that a heavy dose of nitrate retards true average percentage weight gain in mice. 19. It is desired to compare the rigor of two quality inspectors. A and B, each making determinations regarding the acceptability of final bicycle assemblies. Independent random samples of 100 are taken from the bicycles inspected by the respective inspectors. Of the 100 inspected by A, 4 are declared unsatisfactory, while of the 100 inspected by B, 8 are declared unsatisfactory. Can it be concluded at the α = 0.10 level that the inspectors differ in their manner of inspection? 20. In a studty on the fertility of married women conducted by Martin O Connell and Carolyn C. Rogers for the Census Bureau in 1979, two groups childless wives aged 25 to 29 were selected at random and each wife was asked if she eventually planned to have a child. One group was selected from among those wives married less than two years and the other from among those wives married five years. Suppose that 240 of 300 wives married less than two years planned to have children some day compared to 288 of the 400 wives married five years. Can we conclude that the proportion of wives married less than two years who planned to have children is significantly higher than the proportion of wives married five years? Make use of a P -value. 21. Two types of instruments for measuring the amount of sulfur monoxide in the atmosphere are being compared in an air-pollution experiment. It is desired to determine whether the two types of instruments yield measurements having the same variability. The following reading were recorded for the two instruments: Sulfur Monoxide Instrument A Instrument B Assuming the populations of measurements to be approximately normally distributed, test the hypothsis that σ 2 A = σ 2 B against the alternative that σ 2 A σ 2 B. Use α =

41 22. The paper Evaluating Variability in Filling Operations described two different filling operations used in a ground-beef packing plant. Both filling operations were set to fill packages with 1400 g of ground beef. In a random sample of size 30 taken from each filling operation the resulting means and standard deviations were g and g for operation 1 and g and 9.96 g for operation 2. a) Using a 0.05 significance level, is there sufficinet evidence to indicate that the true mean weight of the packages differs for the two operations? b) Does the data from operation 1 suggest that the true mean weight of packages produced by operation 1 higher than 1400 g? Use a 0.05 significance level. 23. To compare two methods of teaching reading, randomly selected groups of elementary scholl children were assigned to each of the two teaching methods for a 6-month period. The criterion for measuring achievement was a reading comprehension test. The results are shown in the accompanying table. Assume that scores are normally distributed. Method 1 Method 2 n 1 = 11 n 2 = 14 x 1 = 64 x 2 = 69 s 2 1 = 52 s 2 = 71 a) Does the two methods produce the same variability in score? Test using α = b) Do the data provide sufficient evidence to indicate a difference in mean score on the comprehension test for the two teaching methods? Test using α =