CHAPTER 7. Hypothesis Testing

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1 CHAPTER 7 Hypothesis Testing A hypothesis is a statement about one or more populations, and usually deal with population parameters, such as means or standard deviations. A research hypothesis is a conjecture that motivates research. A statistical hypothesis is stated in such a way that it can be evaluated by appropriate statistical techniques. A 6-Step Process for Testing a Hypothesis (1) Understand the data. We will use the SAT Verbal scores. () Make clear your assumptions, including those about normality, equality of variances, independence of samples, etc.. In this case, we are assuming that SAT Verbal scores are normally distributed with a standard deviation of (3) State the hypotheses. Suppose that we believe that the population mean µ of the verbal scores is 600. The null hypothesis H 0 is the hypothesis to be tested. It is the hypothesis of no change, and we assess evidence against this hypothesis in an attempt to discredit it. It is often the hypothesis for which erroneous rejection has the more serious consequences. In a research situation, it may be the complement of the conclusion the researcher is try to make. The null hypothesis is rejected if the data is not compatible with its acceptance. The null hypothesis is not rejected if the data do not provide su cient evidence to cause rejection. 8

2 Our null hypothesis is 7. HYPOTHESIS TESTING 83 H 0 : µ = 600. The alternative hypothesis H A is what we accept as true if H 0 is rejected, since it is the complement of H 0. Our alternative hypothesis is H A : µ 6= 600. Note. Below is a listing of various forms of the null and alternative hypotheses: Two-sided H 0 : µ = 100 H A : µ 6= 100 One-sided H 0 : µ apple 100 H A : µ > 100 H 0 : µ 100 H A : µ < 100 Alternative one-sided H 0 : µ = 100 H A : µ > 100 H 0 : µ = 100 H A : µ < 100 Without knowing which is true, we use sample evidence to accept or reject H 0. accept H 0 reject H 0 H 0 true correct decision Type I error ( ) H 0 false Type II error ( ) correct decision = P (Type I error) = P (reject H 0 H 0 true) = P (Type II error) = P (accept H 0 H 0 false) Typically, a Type I error is the most serious, and so = significance level.

3 84 7. HYPOTHESIS TESTING (4) Get the test statistic where relevant statistic hypothesized parameter test statistic = standard error of the relevant statistic Since we are taking a random sample of size 1 from a normal distribution with a known standard deviation, we have z = x / p 1 = x To select our simple random test sample, we again generate 1 di erent random integers between 1 and 300, inclusively. We get the numbers, 165, 57, 94, 84, 83, 37, 16, 17, 4, 17, 93. These numbers correspond to verbal scores of 540, 640, 400, 450, 60, 540, 450, 640, 530, 660, 600, 540. The mean of this sample is x = with standard deviation s = Thus z = =.131 (5) Make a statistical decision. The typical a priori levels of significance are =.05 and =.01. These are the most commonly used significance levels in published research. Since our alternative hypothesis is two-sided because it contains 6=, these significance levels are divided equally on both sides of the mean. The diagram on the next page shows the case for =.05, with probability =.05 =.05 in each tail. We show the acceptance and rejection regions below using both the x scores and the standardized z scores. The =.05 acceptance region for the x scores is just the 95% confidence interval about the hypothesized mean µ = 600. Press APPS>1:FlashApps>Stats/List Editor>F7:Ints>1:ZInterval and choose Stats as Data Input Method. Then press ENTER.Fill in the table with our information and press ENTER.We get the interval (554.6, 645.4) as shown in the following diagram.

4 7. HYPOTHESIS TESTING 85 Since our test statistics of x = and z =.131 are in the rejection region, we reject the null hypothesis of µ = 600. A test statistic that falls in the rejection region is called significant. The next case we show is for =.01, with probability =.01 =.005 in each tail. In this case the test statistics of x = and z =.131 are in the acceptance region, so there is insu cient evidence to reject the null hypothesis. In general, if H 0 is rejected, we conclude H A is true, and if H 0 is not rejected, we conclude H 0 may be true. (6) Find the p-value, the probability that the test statistic would take a value as extreme or more extreme than that observed if if H 0 were true. The smaller

5 86 7. HYPOTHESIS TESTING p is, the more evidence there is against H 0. In the past p-values were di cult to compute. That is why researchers used set significance values like =.05 and =.01 which did not require them to find p-values. But p-values are now readily available to us by calculator and computer. For our problem, take the TI and enter APPS>1:FlashApps>Stats/List Editor>F6:Tests>1:ZTest and choose Stats as Data Input Method. Then press ENTER.Fill in the table with our information and press ENTER. Put in 600 for µ0, for, for x, and 1 for n. Choose µ 6= µ0 for Alternate Hyp, and Draw for Results. We see that p = Note that this is smaller than.05 and greater than.01, causing us to reject the null hypothesis at the =.05 significance level but not at the =.01 significance level. A more modern approach to hypothesis testing is, instead of using preset significance levels, to find the p-value and then make decisions based on it. (3 0 ) Suppose we believe that there is no way the population mean could be above 600 for the SAT Verbal scores. Then we would use a one-sided test. H 0 : µ 600 H 1 : µ < 600 (4 0 ) To find the boundaries of the one-sided rejection region for an =.05 significance level, we note that we want the boundary to be the point where the area

6 7. HYPOTHESIS TESTING 87 (probability) to the left of it is.05. On the TI for the x-score, assuming we are in the Data/List Editor, we can find this by F5:Distr>:Inverse>1:Inverse Normal, and in the window that opens put in.05 for Area, 600 for µ, and 3.16 for. Pressing Enter, we get x = If we repeat the process and enter 0 for µ and 1 for, we get z = (5 0 ) Since our test statistics of x = and z = region, we reject the null hypothesis of µ = are in the rejection For =.01, we change Area to.01 for each of the two previous cases to get x = and z =.363. Since our test statistics of x = and z =.131 are in the acceptance region, we accept the null hypothesis of µ = 600.

7 88 7. HYPOTHESIS TESTING (6 0 ) Again, for the p-value, take the TI and enter APPS>1:FlashApps>Stats/List Editor>F6:Tests>1:ZTest and choose Stats as Data Input Method. Then press ENTER.Fill in the table with our information and press ENTER. Put in 600 for µ0, for, for x, and 1 for n. Choose µ < µ0 for Alternate Hyp, and Draw for Results. We see that the p-value p = is one-half what is was for the two-sided test. Testing a Single Population Mean Case 1: Normally distributed population, known. For H 0 : µ = µ 0, the test statistic is z = x µ 0, p n which has the standard normal distribution when H 0 is true. Problem (Page 87 #30). (1) n = 5, x = 73 () Assume a normal population with = 16. (3) H 0 : µ apple 70, H A : µ > 70 or H 0 : µ = 70, H A : µ > 70 (4) z = x µ p n = p 5 =.9375

8 (5) Assuming H 0 : µ = 70, and since for µ is ( 1, ). 7. HYPOTHESIS TESTING p 5 = , a 95% CI Since x = 73 is in the acceptance range, H 0 is not rejected and we continue to use µ = 70 as the population mean. (6) Since p =.1741 >.05, there is insu cient evidence to reject H 0. Case : Normally distributed population with for H 0 : µ = µ 0, is t = x µ 0 s p n unknown. The test statistic, with n 1 degrees of freedom.

9 90 7. HYPOTHESIS TESTING Problem (7..1). (1) n = 15. From TI, x = and s = () We assume a normal population. (3) H 0 : µ = 1 H A : µ 6= 1 (4) t = p 15 = The 95% CI for µ, given µ = 1, is 1 ± p 15 = (11.9, 1.71). t x rejection acceptance rejection (5) Since t =.1448 < 4.311, we reject H 0. Also, notice that x = is in the rejection region of the CI. (6) We find p = Prob(x = µ = 1) =.0007 as strong evidence of rejecting H 0 : µ = 1. Note. (1) In the above problem, we have given 3 di erent ways of rejecting H 0. One is su cient. The same can be said for not rejecting H 0. () The slight di erences in computation are due to di ering round-o schemes.

10 7. HYPOTHESIS TESTING 91 Case 3: Sampling from a nonnormal population. For large n, we use the Central Limit Theorem. For H 0 : µ = µ 0, the test statistic is One may use s to estimate. Problem (Page 94 #47). z = x µ 0 p n. pre x = , s = , n = 66 post x = , s = , n = 66 We shall work only with the post numbers at this time. We can assume the sampling distribution is not approximately normally distributed, and we estimate with s. We test H 0 : µ = 39 H A : µ < 39 with =.01. The 99% CI for µ, given µ = 39 and p = , is (361.8, 1). x z rejection acceptance Clearly, x = is in the rejection region. Also, the test statistic is z = = p 66 is less than.363 and We clearly reject H 0. p = Prob(x = µ = 39) =

11 9 7. HYPOTHESIS TESTING The Di erence between Two Population Means Case 1: Normally distributed populations with population variances known. We are testing H 0 : µ 1 µ = (µ 1 µ ) 0 H A : µ 1 µ 6= (µ 1 µ ) 0, etc., where (µ 1 µ ) 0 is the hypothesized di erence. If (µ 1 µ ) 0 = 0, we could also use H 0 : µ 1 = µ H A : µ 1 6= µ. For H 0 : µ 1 µ = (µ 1 µ ) 0, the test statistic is z = (x 1 x ) (µ 1 µ ) s n 1 Case : Normally distributed populations with population variances unknown, but assumed equal. For H 0 : µ 1 µ = (µ 1 µ ) 0, the test statistic is n with pooled variance t n 1+n = (x 1 x ) (µ 1 µ ) s 0 s p + s p n 1 n s p = (n 1 1)s 1 + (n 1)s. n 1 + n

12 7. HYPOTHESIS TESTING 93 Problem (7.3.6). We have the following data: Sample n x s We assume the samples come from normally distributed populations with equal variances. We test H 0 : µ 1 µ apple 0 H A : µ 1 µ > 0 with =.05. We find t = We also have s p = (14)1.0 + (1)1.5 = Assuming H 0 : µ 1 µ apple 0, and noting r =.7484, a 95% CI for µ 1 µ is ( 1,.7484). x1-x.7484 t acceptance rejection Clearly, x 1 x = 1.75 is in the rejection region. Also, the test statistic (4.75 3) 0 t = r = is greater than and We clearly reject H 0. p = Prob(x 1 x = 1.75 µ 1 µ apple 0) =

13 94 7. HYPOTHESIS TESTING Case 3: Normally distributed population, variances unknown and not known to be equal. For H 0 : µ 1 µ = (µ 1 µ ) 0, the test statistic is t = (x 1 x ) (µ 1 µ ) s 0 s 1 + s n 1 n where = apple s 1 n 1 + s " s 1 n 1 n n s n n 1 #. Note. We again di er from the text here, but align with the TI and SPSS. Problem (7.3.6). We again have the following data: Sample n x s We now assume the samples come from normally distributed populations with unequal variances. We test with =.05. We find H 0 : µ 1 µ apple 0 H A : µ 1 µ > 0 = apple 1 " # = 35

14 7. HYPOTHESIS TESTING 95 and t = Assuming H 0 : µ 1 µ apple 0, and noting r =.6945, a 95% CI for µ 1 µ is ( 1,.6945). x1-x.6945 t acceptance rejection Clearly, x 1 x = 1.75 is in the rejection region. Also, the test statistic (4.75 3) 0 t = r = is greater than and We clearly reject H 0. p = Prob(x 1 x = 1.75 µ 1 µ apple 0) = Case 4: Nonnormal populations. If the sample sizes are large, for the test statistic is H 0 : µ 1 µ = (µ 1 µ ) 0, z = (x 1 x ) (µ 1 µ ) s 0, + n 1 n using s 1 and s as estimates for 1 and, if necessary. 1

15 96 7. HYPOTHESIS TESTING Problem (Page 84 #0). We have the following data: Treatment n x s x (1) Etanercept () Etanercept+methotrexate We assume the samples come from nonnormally distributed populations. Since s x = p s p, s = s x n. Thus s1 =.84 p 40 and s =.57 p 57. For n H 0 : µ 1 µ = 0 with =.05, the test statistic is ( ) 0 z = s (.84 p 40) + (.57p 57) = Assuming H 0 : µ 1 µ = 0 a 95% CI for µ 1 µ is s (.84 p 40) 0 ± (.57p 57) = ( , ) z x rejection acceptance rejection x 1 x = 1.16 is in the acceptance region, as is the test statistic z = Also, p = Prob(x 1 x = 1.16 µ 1 µ = 0) = >.05. We do not reject H 0 and continue to treat the population means as equal.

16 7. HYPOTHESIS TESTING 97 Paired Comparisons the samples are not independent here. The idea is to remove extraneous factors, keeping as many variables as possible the same. Examples are before and after, left and right, identical twins, etc. But: (1) It takes time and money to create matches. () You are left with only n 1 df instead of n. We compute di erences d, so this is really a one-sample test. For the test statistic is H 0 : µ d = µ d0, t n 1 = d µ d 0 s d, where d is the sample mean di erence, µ d0 is the hypothesized mean di erence, and s d = s d p. n Problem (Page 94 #47). We assume a normally distributed population of di erences. We can assume the sampling distribution is approximately normally distributed. We have d = , s d = 9.197, n = 66, s d = We test H 0 : µ d 0 H A : µ d < 0 with =.05. The 95% CI for µ d, given µ d = 0 and (11.349) = , is ( , 1). d t rejection acceptance Clearly, d = is in the rejection region. Also, the test statistic t = =

17 98 7. HYPOTHESIS TESTING is less than and We clearly reject H 0. A Single Population Proportion p = Prob(d = µ d 0) = For H 0 : p = p 0 H A : p 6= p 0, if np 0 5, and n(1 p 0 ) 5 so that the Central Limit Theorem applies, use bp p 0 z = r p0 (1 p 0 ) as the test statistic. Note. We use p 0 rather than bp in the formula since p 0 is hypothesized as the true population proportion. n

18 7. HYPOTHESIS TESTING 99 Problem (7.5.6). We have n = 50, p 0 =.9, bp = =.95. Also, np 0 = 50(.9) = 5, n(1 p 0 ) = 50(.1) = 5, so the Central Theorem applies. We test H 0 : p apple.9 H A : p >.9 r.9(.1) with =.05. Assuming p 0 =.9 and for p is ( 1,.9313). p.9313 z =.9313, a 95% CI acceptance rejection Clearly, bp =.95 is in the rejection region. Also, the test statistic.95.9 z = r = (.1) is greater than and We clearly reject H p = Prob(bp =.95 p apple.9) =

19 HYPOTHESIS TESTING The Di erence between Two Population Proportions For H 0 : p 1 p = (p 1 p ) 0, based on the hypothesis that (p 1 p ) 0 = 0, we pool the samples to compute s p = x 1 + x p(1 p) p(1 p) and b bp1 bp n 1 + n = +. n 1 n Our test statistic is then z = (bp 1 bp ) (p 1 p ) 0 b bp1 bp, which is approximately normally distributed if H 0 is true. Problem (7.6.4). We have the following data n Number Overweight (x) (1) Male () Female Total We test bp 1 = =.14, bp = =.4, p = =.1971 H 0 : p 1 p = 0 H A : p 1 p 6= 0 with =.05. With z.975 = 1.96, and assuming H 0 is true, a 95% CI for p 1 p is r.1971(.809) 0 ± (.809) = ±.0843 = (.0843,.0843) z p1-p rejection acceptance rejection

20 7. HYPOTHESIS TESTING 101 Our test statistic is (.14.4) 0 z = r.1971(.809) (.809) = =.356, which lies in the rejection region as does bp 1 bp =.1. We also have p =.000 <.05 The Type II Error and the Power of a Test accept H 0 reject H 0 H 0 true correct decision Type I error ( ) H 0 false Type II error ( ) correct decision Type II Error failing to reject H 0 when H 0 is false. We set, the probability of a Type I error. Power of a Test = 1 depends on = Prob(rejecting H 0 H 0 is false). (1) The true value of the parameter of interest. () The hypothesized value of the parameter. (3) (4) n and 1 are testing. may be computed for any alternative value of the parameter we Example (7.9.1). A two-sided example. = 3.6, n = 100, =.05 H 0 : µ = 17.5 H A : µ 6= 17.5

21 10 7. HYPOTHESIS TESTING Upper and lower critical x-values: xu = µ0 + z p n xl = µ0 zp n 3.6 = = = = Suppose H0 is false with true mean = µ1 = This gives the normal sampling distribution f (x1). The hypothesized mean gives the sampling distribution f (x0). = the area under the curve of f (x1) that overlaps the nonrejection region of f (x0).

22 7. HYPOTHESIS TESTING 103 = P (16.79 apple x apple 18.1 µ = 16.5, =.36) = Note. nmcdf(16.79, 18.1, 16.5,.36) =.10. Power of the test = 1 = (1) The further µ 1 is from µ 0, the smaller is. () To make smaller, increase n. Example (7.9.). A one-sided example. = 15, n = 0, =.01 H 0 : µ 65 H 1 : µ < x L = p = Suppose H 0 is false with a true mean of µ 1 = 55. = P x > µ = 55, = p 15 = nmcdf( , 1E99, 55, 15 p ) =.56. Power of the test = 1 =.7438 Note. The text s numbers di er from ours due to excessive rounding.