Chapter 7 Inferential Statistics: Estimation and Hypothesis Androgen (ng/ml) at Adult Time of Injection Androgen (ng/ml) 30 Male 4.79 Minutes After Injection 18.53 7.39 13.74 7.91 7.30 0.52 10 4.85 11.78 -2.45 11 11.10 3.90 -0.68 12 3.74 26.00 -0.16 13 94.03 67.48 68.03 14 94.03 17.04 26.55 15 41.70 24.66 1 Null and alternative hypotheses: H: Mp = 0 versus Hi: HD * 0. 2 Significance level: a = 0.05. 3 Test statistic: d = 9.848 and So = 18.474; d- do _ 9.848 - 0 t = sd / Vn 18.474 / 15 = 2.06. 4 Decision rule: Critical region: [t) > to.025.14 = 2. 145, i.e. reject Ho if t 2.145. 5 Decision: Since t = 2.06 2.06) E (0.05, 0.1). As a result, there is some evidence that there is a difference in the mean circulation levels of androgens. 7.4.10 Testing a Hypothesis on Proportions At th Consider the problem of testing the hypothesis that the proportion of secti successes in a binomial experiment equals some specified value. shou Fo Assumption: If the unknown proportion is not expected to be too close to ap 0 or I and n is large, a large sample approximation due to the CLT is given an by : a Hop = Po versus H: p > Po p Zo Z or |z| > za/2.running out 13 One of the most important features of laser printers is the number of pages that can be printed per minute. The average number of pages printed per minute of 25 randomly selected Brand A laser printers was 20 with a standard deviation of 1.5 pages. A random sample of 25 Brand B laser printers produced an average of 12 pages per minute with a standard deviation of 1.25. Assume that the number of pages printed per minute is approximately normally distributed and the population variances of the two brands are the same. Is there strong evidence that Brand A can print more pages per minute than Brand B at 0.05 level of significance? 14 An experiment was conducted to test whether the use of a fuel additive in petrol can increase the traveling distance of cars. Two random samples of 10 cars and 8 cars were selected for the experiment, where all cars were filled with one liter of petrol. For the first sample, the petrol was added with the fuel additive.and Probability TESTS OF HYPOTHESIS 7.4 Hypothesis is a proposition, statement, or assumption based on some of this previous observations about the value of a population parameter for dents testing purposes. And so, hypothesis testing is a process based on some ble to: sample data and probability theory to conclude whether the hypothesis is reasonable and should not be rejected or otherwise. To test the validity of the proposition or assumption, we must select a sample from the esis population, calculate the sample statistics; and based on certain decision tive rules, accept or reject the hypothesis. esis ance 7.4.1 Five-step Procedure for Testing a and Hypothesis There are five systematic steps to facilitate a hypothesis testing. However, esis it is in the last step that we need to decide whether to reject or not the hypothesis. In contrast to a mathematical statement that must be proven, he s o a statistical test of hypothesis does not provide evidence that something a is true. Instead, it is merely an inferential statement based on reliable Type samples and assumptions. In contrast to a mathematical statement that has been proven to be always true, a statistical test of hypothesis provides evidence using the information contained in the sample data supporting the proposition about the population but there is still a small chance that the proposition is actually false. That is, in hypothesis testing, we cannot be 100% sure that the proposition is true. Step 1: The Null Hypothesis and the Alternate Hypothesis lesis The first step is to state the hypothesis to be tested-the null hypothesis. Ho. The capital letter H stands for hypothesis and the subscript 'zero implies 'no difference' or 'no change'. The null hypothesis is a statement fail about the value of a population parameter. For example, 'The mean weight of U.P. students is not different from 70 kg'. The null hypothesis Ho would then be written as Ho: H = 70. Basically, we either reject or 'fail to reject' the null hypothesis. A statement is not rejected if our sample data fail to provide convincing evidence that it is false. It is important to note that accepting the null hypothesis does not prove that He is true. Instead, to prove without any doubt that the null hypothesis is true, the population parameter must be known. However, since it is usually impossible to know the population parameter, it should be noted that we often begin the null hypothesis by between ...'. stating 'The mean ... is not different from ...', or 'There is no difference The alternative hypothesis, on the other hand, describes what you will conclude if you reject the null hypothesis. It is a statement that is acceptedif the sample data provide us with statistically significant evidence that the null hypothesis is false; written as H, or H.. It is also referred to as the research hypothesis. The null hypothesis represents the current or reported condition. The alternative hypothesis is that the statement is not true. For example, if Hot U = 70, then the alternative hypothesis would be written as H: u # 70 is 'The mean weight is different from 70 kg; or H: u > 70 is 'The mean weight is greater than 70 kg; or H: u 70. 2 Significance level: a = 0.05. 3 Test statistic: z= z= X - Mo_71.8-70 = 2.02. own 8.9/ 100 4 Decision rule: Critical region: z> 1.645, i.e. reject Ho if 2 > 1.645 . 5 Decision: Since z = 2.02 > 1.645, reject Ho and conclude that the mean score is more than 70 . The p-value corresponding to z = 2.02 is given by the area of the shaded region. .9 for Example 7. P Z Ass 0 2.02 un Using Table A4, we have p = P(Z > 2.02) = 0.0217. As a result, the evidence in favor of H, is even stronger than that suggested by a 0.05 level of significance. Example 22 A mill produces flour in small bags before distributing them to wholesalers. The weight of each bag is 8 kg with a standard deviation of 0.5 kg. A random sample of 50 bags was taken and found that the average weight is 7.8 kg. Using a significance level of 0.01, test the hypothesis that / = 8 kg against the alternative where u # 8 kg. 1 Null and alternative hypotheses: Ho: / = 8 kg versus H: ! = 8 kg. 2 Significance level: a = 0.01. of cla 3 Test statistic: z = X -Mo _ 7.8-8 an o/Vn 0.5/ 50 =-2.83. Sic 4 Decision rule: Critical region: (z| > Zo:005 = 2.576, i.e. reject Ho if 2 2.576. 5 Decision: Since z= -2.83 Zo.os = 1.645, i.e. reject H, if 4 2 > 1.645. 5 Decision: Since z= 2.49 > 1.645, reject H, and agree with the claim that a car is driven on average more than 20,000 kilometers per year. Using Table A4, the p-value corresponding to z= 2.49 is p = P(Z > 2.49) = 0.0064. As a result, the evidence in favor of His even stronger than that suggested by a 0.05 level of significance. 7.4.5 Testing for Population Mean: Small Sample, Population Variance Unknown Assumption: A sample of size n (small) from a normal population with of this unknown mean and unknown variance. dents able to: Ho: H= H versus H: u> Ho- H to t to/2; Degrees of freedom = n -1. ation ce is ed to Example 24 own MERALCO has listed a variety of home appliances each with annual tion total electricity consumption in kilowatt-hours. For a medium-sized ce is refrigerator, the annual electricity consumption is said to be an average ed to of 46 kilowatt-hours. A random sample of 12 homes included in the nown; study showed that medium-sized refrigerators consume electricity at ntral an average of 42 kilowatt-hours per year with a standard deviation eorem of 11.9 kilowatt-hours. At a significance level of 0.05 and assuming used that the population is normal, does this sample show that medium- per year? sized refrigerators consume, on average, less than 46 kilowatt-hours 1 Null and alternative hypotheses: Ho: H = 46 kwh versus Hi: H -1.796, do not reject Ho and conclude that the average kilowatt-hours consumed annually by medium-sized refrigerators is not significantly less than 46.The first sample yielded an average traveling distance of 18 km with a standard deviation of 3.5 km. while the average traveling distance from the second sample was 15 km with a standard deviation of 4.0 km. Is there evidence to believe that the fuel additive causes an increase in the traveling distance of the cars by more than 2 km? Assume normal populations and use the 0.02 level of significance. 15 A new treatment was proposed to fight breast cancer. Six randomly selected breast cancer patients were treated with the new treatment. For comparison, five patients treated with the old treatment were also selected at random. The survival periods, in years, from the time the treatments started are recorded as follows: New Treatment N J UI Old Treatment A 2 NJ Assuming normal populations with equal variances, can the new treatment be said to be more effective at a 0.05 level of significance? 16 Five runners were randomly selected to determine if there is a difference in the number of calories burned per minute between running outdoor and running indoor. Each runner was asked to do the outdoor and indoor running of the same duration but on two different days. The results (number of calories burned per minute) were recorded as follows:9 A random sample of 49 bags of potato chips weighed. on average, 5.23 ounces with a standard deviation of 0.24 ounce. Test the hypothesis that # = 5.5 ounces against the alternative hypothesis 8. Ho: p= 1/4 and H,: p > V/4. As mentioned earlier, the probability of committing a 'Type | error', called the level of significance, is a= P(Type | error) = P(X > 8 when p = 1/4) = 1 - _b x; 20,- = 1-0.9591 = 0.0409. x= 0 However, the probability of committing a 'Type II error' is impossible to compute unless we have a specific alternative hypothesis. If we test the null hypothesis that p= 1/4 against the alternative hypothesis that p= 1/2, then we are able to compute B. We simply find the probability of obtaining 8 or fewer in the group that have recovered when p= 1/2. In this case, B = P(Type ll error) = P(X b x;20, =0.2517. x=0 Even if we want to develop a test that will minimize both probabilities, this is not possible because they are inversely related. So what we do is to set the value of P(Type | error) at a then look for a test that will give the smallest P(Type II error). The tests presented in this book are such that this probability can be reduced by increasing the sample size. Step 3: The Test Statistic tic is mined There are many test statistics; z, t, F and c'. The test statistic is a value, determined from sample information, used to determine whether or not sed to to reject the null hypothesis. ether t the Suppose we have a random sample from a normal distribution whose S. mean u is unknown but its variance o is known. In testing Ho: M = to the test statistic is Z=_. When the null hypothesis is true, that is u, the test statistic Z follows a standard normal distribution based on the sampling distribution of X. The distribution of the test statistic when the null hypothesis is true is called the null distribution.s and Probability of significance, selects a sample, and makes a decision whether or not to reject the null hypothesis. The statistician makes a recommendation based on this sample evidence. We say that the data provide us with sufficient evidence supporting the alternative hypothesis when the decision is to reject Ho. When we decide to reject Hy the only possible error we can commit is a Type I error and we have set P(Type I error) = a to be small. We are more prudent in our interpretation when we decide not to reject H. We simply say that the data do not provide us with sufficient evidence supporting //, instead of saying that the data provide us with sufficient evidence supporting H. When we decide not to reject Ho, the only possible error we can commit is a Type II error. Unfortunately, we cannot compute P(Type II error) unless the alternative hypothesis states a specific value for the parameter. The only thing that we know is that the test that we are using will give the smallest P(Type II error) for a fixed a, but this can still be large particularly if the true value of the parameter is very close to the hypothesized value in Ho or when the sample size is not large enough. 7.4.2 One-tailed and Two-tailed Tests of this nts Significance to: There are two different types of tests of significance-a one-tailed test looks for an increase or decrease in the parameter whereas a two-tailed test looks for any change in the parameter (increase or decrease). We fe can perform the test at any level (usually 1%, 5% or 10%). For example, performing the test at a 5% level means that there is a 5% chance of wrongly rejecting the null hypothesis. Refer to Figure 7.6, it indicates that a one-tailed test is being applied. The region of rejection is only in the right (upper) tail of the curve. Figure 7.7 portrays a situation where the rejection region is in the left (lower) tail of the normal curve. le- Region of ce rejection Do not reject Ho -1.645 0 Critical value scale of z One way to determine the location of the rejection region is to look at the direction in which the inequality sign in the alternative hypothesisExercise 7.3 I Suppose a researcher wishes to test the hypothesis that at least 30% of the public is allergic to some cheese products. Explain how the researcher could commit a Type I error and a Type II error. 2 After going through a long trial period, a judge must decide whether the person charged is guilty or not. (a) State the hypothesis being tested if the judge commits a Type I error by deciding the person charged as guilty. (b) State the hypothesis being tested if the judge commits a Type II error by deciding the person charged as guilty. 3 The proportion of male students in a public university is known to be p= 0.3. To test this hypothesis, a sample of 15 students is chosen at random. If the number of males is between 6 and 12, the null hypothesis will be accepted, otherwise p # 0.3. (a) Evaluate a assuming that p=0.3. Use the binomial distribution. (b) Evaluate B for the alternative p = 0.2 and p = 0.4.Imation and Hypothe It is believed that the proportion of lecturers with PhD at a public university is p = 0.4. The hypothesis p = 0.4 should be rejected in favor of the alternative p ). If, for example, H,: U 70 (right tail). If no direction is specified under the alternative hypothesis, a two- tailed test is being applied, for example, H,: u # 70. In a two-tailed test, the region of rejection is in both (upper and lower) tails. The area of rejection is divided equally into the two tails of the sampling distribution. Figure 7.8 shows the two areas and the critical values. Figure Regions rejection Rejection Region of Do not Region of Two-Tail rejection reject Ho rejection Level of 0.025 0.025 -1.96 0 1.96 Scale of z Critical value Critical value 7.4.3 Testing for Population Mean: At the Population Variance Known sectio should Assumption: A sample of size n from a normal population with unknown - Forn mean but known variance. app and Ho: U= Ho versus H: u> Ho- hyp a po H Zo Z or |z| > za/2- By the CLT, this test could also be used even if the sample does not come from a normal distribution so long as the sample size is sufficiently large (at least 30). Example 21 From a random sample of 100 students who have passed a statistics course, the average score was 71.8. Assuming that the population standard deviation is 8.9, with a significance level of 0.05, does it seem to signify that the average score is more than 70?Five runners were randomly selected to determine if there is a difference in the number of calories burned per minute between running outdoor and running indoor. Each runner was asked to do the outdoor and indoor running of the same duration but on two different days. The results (number of calories burned per minute) were recorded as follows: Running Runner 1 2 3 4 Outdoor 14 5 15 13 16 14 Indoor 12 11 12 10 11 Assuming normal populations, test at 0.05 level of significance whether the outdoor running and the indoor running give, on average, the same result. 17 A researcher has to decide the type of tires for heavy vehicles that give better fuel economy. Two types of tires were tested, radial tires and belted tires. In this test, twelve trucks equipped with radial tires were driven for a certain distance, and without changing drivers, the same trucks were then equipped with belted tires and driven for the same distance. The fuel consumptions, in kilometers per liter, were recorded as follows:g a Type II error. Tate, determine the 6 A manufacturer has produced a new type of bungee jumping cords which has a mean breaking strength of 500 kg with a standard deviation of 20 kg. A random sample of 60 cords will be used to test the hypothesis that / = 500 against the alternative that u 0.876) = 0 " and P(t > 1.363) - 0.1 then we say p E (0.1, 0.2). 7.4.6 Testing for Two Population Means: Population Variances Known At the all Assumption: Two samples of size n, and n, from two normal populations with unknown means but known variances. section should nown Ho: H - 12 = do versus H: H - M2 > dj. - Comp for the HI - H2 Zo Z Za/2. value (7.4.31 rejectic By the CLT, this test could also be used even if the samples do not come - Solve from a normal distribution so long as the sample sizes are sufficiently involvin large (both at least 30). hypoth the po mean. Example 25 nua A random sample of size n, = 25, taken from a normal population with ized a standard deviation of = 5.2, has a mean x, = 81. A second random sample of size n2 = 36, taken from a different normal population with a standard deviation 62 = 3.4, has a mean x2 = 76. Test the at hypothesis that /1 = /2 against the alternative / # /2. Use a 0.02 level of significance. 1 Null and alternative hypotheses: Ho: My - /2 = 0 versus Hi: My - H2 # 0. 2 Significance level: a = 0.02. 3 Test statistic: 81-76 = 5.92. ( x 1 - X2 ) - do Z = ( 15.2)2/ 25 ) + ( 13.4) ? / 36) ((OR /m ) + (03 / n2 ) 4 Decision rule: Critical region: (z) > Zoo1 =2.326, i.e. reject Ho if z2.326. 5 Decision: Since z = 5.92 > 2.33, reject Ho and conclude that the two population means are not equal. Using Table A4, the p-value corresponding to z - 5.92 is p = P( - 5.92) = P(7 5.92) = 2P(Z >5.92) 2 0 (not equal to O)atistics and Probability 7.4.9 Testing for Two Population Means: Paired Observations There are situations in which the samples are not independent. A random sample of size n (small) is taken from a population and the values are recorded. After a certain period of time, the values of the same sample will again be recorded. Thus, there will be a pair of values for each member of the sample. The set of sample pairs is aptly called a paired sample. The test of hypothesis to be conducted to find out if there is a difference between the two sets of values is called a paired difference test. For the test hypothesis to be conducted there is essentially only one sample, not two. We are testing the hypothesis that the distribution of the differences has a mean of do, designated up Ha: Up > do, Up to t ta/2; Degrees of freedom = n - 1. Example 28 A drug has been claimed to influence the circulation levels of androgens in the blood. A sample of 15 adult males was tested. Each of them had been injected with the drug, and the levels of androgens in the blood at the time of injection and 30 minutes after the injection were recorded in nanograms per milliliter (ng/ml) as in Table 7.2. Assuming that the populations of the levels of androgens at time of injection and 30 minutes later are normally distributed, test at the of androgens. 0.05 level of significance whether the drug has influenced the levels of Androgens Adult of Injection Male Androgen (ng/ml) at Androgen (ng/ml) 30 Minutes Later Time of Injection Minutes After Injection 2.76 N 7.02 5.18 4.26 3.10 2.68 -2.08 4 5.44 3.05 2.76 en 4.10 3.99 0.94 6 7 7.05 5.21 1.11 6.60 10.26 3.21 13.91 7.31atistics and Probability 7.4.7 Testing for Two Population Means: Large Samples, Population Variances Unknown Assumption: Two samples of size n, and n, (large) from two normal populations with unknown means and unknown variances. Hjill, - /2 = do versus H: 1, - H2 > do- HI - H2 Zo Z or z| > Za/2. Example 26 A manufacturer of fiber optic cables who produces two types of cables have been claiming that the average strength of type A cable exceeds the average strength of type B cable by more than 12 kg. To test this claim, 50 cables of each type were tested under the same conditions. Type A has an average strength of 86.7 kg with a standard deviation of 6.28 kg, and type B has an average strength of 77.8 kg with a standard deviation of 5.61 kg. At a significance level of 0.05, test the manufacturer's claim. 1 Null and alternative hypotheses: Ho: /, - /2 = 12 versus HI: HI - H2 > 12. 2 Significance level: a = 0.05. 3 Test statistic: Z = (x1 - x2) - do (86.7-77.8)-12 V(silmy ) + ( 52 / 12 ) = -2.603. (16.28)2 /50) + (15.61)2/50) Z> 1.645. 4 Decision rule: Critical region: z> Zo.05 = 1.645, i.e. reject Ho if 5 Decision: Since z= -2.603 -2.603) = 0.9957. It is not right to say that the non-rejection of Ho implies that the average strength of Type A will never exceed the average strength of Type B by more than 12 kg. We can only say that the data do not provide us with sufficient evidence to conclude that the average than 12 kg . strength of Type A exceeds the average strength of Type B by moreChapter 7 > Inferential Statistics: Estimation and Hypothesis Testing 229 Figure 7.10 p-value for Example 22 p/2 -2.83 0 p/2 Z 2.83 Therefore, using Table A4, we have p = P D= P(1Z > 2.83) = P(Z 2.83) = 2P(Z > 2.83) = 0.0046, which allows us to reject the null hypothesis, u = 8 kg at a level of significance smaller than 0.01. 7.4.4 Testing for Population Mean: Large Sample, Population Variance Unknown Assumption: A sample of size n (large) from a normal population with unknown mean and unknown variance. At the end of this section, students Ho: M= Ho versus H: u> Ho- should be able to: H Zo Z Or |z| > za/2. assumed to be known ( b ) the Example 23 population it is claimed that a car is driven on average more than 20,000 variance is assumed to kilometers per year. To test this claim, a random sample of 30 car be unknown; owners has been selected. The owners were asked to keep records and of the number of kilometers they traveled. Would you agree with this (c) the Central claim if the random sample showed an average of 22,500 kilometers Limit Theorem and a standard deviation of 5,500 kilometers? Use a 0.05 level of is to be used. significance to justify your answer. 1 Null and alternative hypotheses: Ho: H = 20,000 km versus H,: H > 20,000 km. 2 Significance level: a = 0.05. X - Mo 22,500-20,000 = 2.49. 3 Test statistic: Z = 5,500 / V30 s/ VnChapter 7 Inferential Statistics: Estimation and Hypothesis Testin 14.8 Testing for Two Population Means: Small Samples, Population Variances Unknown Assumption: Two independent samples of size n, and n, (small) from a formal population with unknown mean and unknown variance, i.e. 01 = 0. Ho: - M2 = do versus H: M - M2 > do. HI - H2 to t ta/2; Degrees of freedom = n, + n2 - 2. Example 27 An experiment was conducted to compare the strength of two different brands of denim trousers. A random sample of twelve pairs of trousers of brand A and a random sample of ten pairs of brand B were similarly tested. Brand A gave an average strength of 85 kg with a sample standard deviation of 4, while brand B gave an average of 81 kg and a standard deviation of 5. Can we conclude at the 0.05 level of significance that the strength of brand A trousers exceeds that of brand B by more than 2 kg? Assume the populations to be approximately normal with equal variances. 1 Null and alternative hypotheses: Ho: M - H2 = 2 versus Hi: My - H2 > 2. 2 Significance level: a = 0.05. 3 Test statistic: ( n1 - 1) (57 ) + ( n2 - 1)(52 ) (1 1)(16) +(9)(25) =4.478, = 12+10-2 my + 12 - 2 (85 -81)-2 = 1.04. t = ( X1 - X2) - do 4.478 (1/ 12) + (1/10) spy ( ( 1/ m, ) + ( 1/ n2 ) ) 4 Decision rule: Critical region: t> to.05,20 = 1.725, i.e. reject Ho if t > 1.725 . 5 Decision: Since t= 1.04 1.04) E (0.1, 0.2)