Chapter 9. Hypothesis testing. 9.1 Introduction

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1 Chapter 9 Hypothesis testing 9.1 Introduction Confidence intervals are one of the two most common types of statistical inference. Use them when our goal is to estimate a population parameter. The second common type of inference has a different goal: to assess the evidence provided by the data in favor of some claim about the population. Example 9.1 The rapid increases in college tuition over the past few years have been a great concern for college students and their parents. According to Digest of Education Statistics 1997, the average annual cost of in-state tuition and fees for 4-year public colleges in the United States was $3,321 i997. A recent sample of 40 four-year public colleges yielded a mean in-state tuition of $3393. Can we conclude that the mean in-state tuition for 4-year public colleges increases this year? Outline of a test: I. Set hypotheses Null hypothesis H 0 : µ = 3321 Alternative hypothesis H 1 : µ > 3321 A null hypothesis H 0 is a claim (or statement) about a population parameter that is being tested. H 0 can be interpreted as there is no difference An alternative hypothesis H a is a statement that we intend to prove true A test of significance is intended to assess the evidence provided by data against H 0 in favor of an H 1 9-1

2 Both hypotheses are in terms of parameters, never about statistics. Example (a): A person has been indicted for committing a crime and is being tried in a court. The jury will make one of two possible decisions: Null hypothesis: H 0 : The defendant is not guilty Alternative hypothesis: H 1 : The defendant is guilty Example (b): State H 0 and H 1 in each case. Your car averages 32 miles per gallon on the highway. You now switches to a new motor oil that is advertised as increasing gas mileage. After driving 3000 highway miles with new oil, you want to determine if your gas mileage actually has increased A man claims to have extrasensory perception. You don t believe it. The English mathematician John Kerrich tossed a coi0,000 times and got 5067 heads. You suspect that the coin is not balanced. Suppose that H 0 : θ = θ 0. If H 1 : θ > θ 0 or H 1 : θ < θ 0, then it is referred to as a one-tailed test (one-sided test); If H 1 : θ θ 0, then it is referred to as a two-tailed test (two-sided test). II. Choose a test statistic III. Assess if the observed value of the test statistic is surprising when H 0 is true. How unlikely the observed value of the test statistic would be if H 0 were really true? Assume that σ = 150. Under H 0, x N(3321, ) Hence, x 3321 P ( x 3393) = P ( 150/ / 40 ) P (Z 3.04) =.0012 Or we partition the possible values of the test statistic into two subsets: an acceptance region for H 0, and a rejection region for H 0. The rejection region is also called critical region. 9-2

3 The decision procedure could lead to either of two wrong conclusions: Definition 9.1 Type I error: Rejection of the null hypothesis when it is true is called a type I error. Type II error: Acceptance of the null hypothesis when it is false is called a type II error. Level of significance: The probability of committing a type I error is called the level of significance of the test and is denoted by α. Power: The probability of rejecting H 0 given that a specific alternative is true. Let β = P (Type II error). Then Power = 1 β. Most common significance level: 0.05, 0.1, 0.01 Example 9.2 A certain type of cold vaccine is known to be only 25% effective after a period of 2 years. In order to determine if a new and somewhat more expensive vaccine is superior in providing protection against the same virus for a longer period of time, 20 people are chosen at random and inoculated. If 9 or more of those receiving the new vaccine surpass the 2 year period without contracting the virus, the new vaccine will be considered superior to the one presently in use. H 0 : p = 1/4, H 1 : p > 1/4 What is the probability of committing a type I error? What is the probability of committing a type II error when p = 1/2 α = P (type I error) = P (X 9when p = 1/4) = = β = P (type II error) = P (X < 9when p = 1/2) =

4 9.2 Test concerning mean when σ is known Suppose that H 0 : µ = µ 0, H 1 : µ > µ 0. z test statistic: z = x µ 0 σ/ n Critical region: Reject H 0 if z z α Four Steps: 1. State H 0 and H a and specify α 2. Using the sampling distribution of an appropriate test statistic, determine a critical region of size α. 3. Determine the value of the test statistic from the sample data 4. Check whether the value of the test statistic falls into the critical region and accordingly, reject H 0, or accept H 0. Example 9.3 Refer to Example??. Test whether the mean in-state tuition for 4-year public colleges increases this year at the 0.05 level of significance? σ = 150. z = / 40 = 3.04 Reject H 0 at the 0.05 level Definition 9.2 Corresponding to an observed value of a test statistic, the P -value is the lowest level of significance at which the null hypothesis could have been rejected. Decision rule: Reject H 0 if P-value α. The data are statistically significant at level α. Don t reject H 0 if P-value > α. The data are not statistically significant at level α. Or, we don t have sufficient evidence to reject H

5 The P-value is more informative than a statement of significance, because we can then assess significance at any level we choose. Alternative H 1 Reject H 0 at the α level P-value H 1 : µ > µ 0 z z α P (Z z) H 1 : µ < µ 0 z z α P (Z z) H 1 : µ µ 0 z z α/2 2P (Z z ) Example 9.4 A car manufacturer advertises that its new compact models get 47 miles per gallon (mpg) on high way. A random sample of size 64 gives x = 46.5 mpg. Does this indicate that the average mpg of the compact models is overrated at 0.05 level of significance? at 0.01 level of significance? (assume σ = 2) z = / = 2 P value = Example 9.5 A computer has a random digit generator designed to produce random digits that are uniformly distributed on the digits from 0 to 9. If this is true, the digits generated come from the population with µ = 4.5, and σ = A command to generate 10,000 random numbers gives outcomes with mean We want to test if the mean of random digits generated by the computer is equal to 4.5 at the 5% level. z = ( ) 10000/ = Confidence intervals and two-sided tests: A level α two-sided significance test rejects H 0 : level 1 α confidence interval for µ. µ = µ 0 exactly when µ 0 falls outside a Example 9.6 Refer to Example??. Find a 95% confidence interval for µ. Does 4.5 fall outside the confidence interval? 9-5

6 4.443 ± 1.96( )/100 = ± = (4.3867, ) 9.3 Test concerning mean when σ is unknown Null hypothesis: H 0 : µ = µ 0 t statistic: t = x µ 0 s/ n Alternative H 1 Reject H 0 at the α level P-value H 1 : µ > µ 0 t t α,n 1 P (T n 1 t) H 1 : µ < µ 0 t t α,n 1 P (T n 1 t) H 1 : µ µ 0 t t α/2,n 1 2P (T n 1 t ) Example 9.7 In the American Heart Association journal Hypertension, researchers report that individuals who practice Transcendent Meditation (TM) lower their blood pressure significantly. If a random sample of 225 male TM practitioners meditate for 8.5 hours per week with According to a dietary study, a high sodium intake may be related to ulcers, stomach cancer, and migraine headaches. The human requirement for salt is only 220 milligrams per day, which is surpassed in most single servings of ready-to-eat cereals. If a random sample of 20 similar servings of certain cereal has a mean sodium content of 224 milligrams and a standard deviation of 24.5 milligrams, does this suggest at the 0,05 level of significance that the average sodium content for a single serving of such cereal is greater tha20 milligrams? Example 9.8 It is claimed that a new diet will reduce a person s weight by 4.5 kilograms on the average in a period of 4 weeks. The weights of 9 women who followed this diet were recorded before and after a 4-week period: 9-6

7 Weight before Weight after Difference (a) Test the manufacturer s claim. (b) Find a 90% C.I. for the mean weight loss in a 4-week period. x = 3.278, s = % C.I. is ± 1.86(2.475/3) = ± = (1.743, 4.813) t = 1.488, P value = Hypothesis tests for p Null hypothesis: H 0 : p = p 0 z-test statistic: z = ˆp p 0 p 0 (1 p 0 ) n Alternative Reject H 0 at α level if P-value H 1 : p > p 0 z z α P (Z z) H 1 : p < p 0 z z α P (Z z) H 1 : p p 0 z z α/2 2P (Z z ) Example 9.9 The English mathematician John Kerrich tossed a coi0,000 times and obtained 5067 heads. 9-7

8 (a) Is this significant evidence at the 5% level that the coin is not balanced? (b) Find a 95% confidence interval for the probability that Kerrich s coin comes up heads. (a) H 0 : p = 1/2 H 1 : p 1/2 ˆp =.5067, n = 10, 000 z = ( )/(.5.5/10000) = 1.34 P value = =.1802 (b).5067 ± 1.96(.5067(1.5067)/10000).5 =.5067 ± =.5067 ±.0098 = (.4969,.5165) Example 9.10 A method currently used by doctors to screen women for possible breast cancer fails to detect cancer i5% of the women who actually have the disease. A new method has been developed that researchers hope will be able to detect cancer more accurately. A random sample of 100 women known to have breast cancer were screened using the new method. Of these, the method failed to detect cancer in ten. Do the data provide sufficient evidence to indicate that the new screening method is better than the one currently in use? Let p be the failing rate of the new method. H 0 : p =.15, H 1 : p <.15 n = 100, ˆp = z = = 1.4 P-value= (.85)/

9 In many problems in applied research, we are interested in hypotheses concerning differences between the means of two populations. 9.5 Two sample z-test with known variances Consider two independent random samples of size and from two normal populations having the means µ 1 and µ 2 and the known variances σ 2 1 and σ 2 2. Recall that X 1 X 2 (µ 1 µ 2 ) N(0, 1) σ1 2 + σ2 2 Assume that the null hypothesis is H 0 : µ 1 µ 2 = d 0 and let be the z-statistic. Then z = x 1 x 2 d 0 σ σ2 2 H 1 P value Reject H 0 at the α-level if µ 1 µ 2 d 0 2P (Z > z ) z > z α/2 µ 1 µ 2 > d 0 P (Z > z) z > z α µ 1 µ 2 < d 0 P (Z < z) z < z α 9.6 Two-Sample t Test for Independent Samples with Equal Variances Consider two independent random samples of size and from two normal populations having the means µ 1 and µ 2 and the unknown common variance σ 2. Recall that where X 1 X 2 (µ 1 µ 2 ) t( + 2) 1 S p + 1 Sp 2 = ( 1)S1 2 + ( 1)S Assume that the null hypothesis is H 0 : µ 1 µ 2 = d 0 and let t = x 1 x 2 d 0 1 s p

10 be the t-statistic. Then H 1 P value Reject H 0 at the α-level if µ 1 µ 2 d 0 2P (T n1 + 2 > t ) t > t n1 + 2,α/2 µ 1 µ 2 > d 0 P (T n1 + 2 > t) t > t n1 + 2,α µ 1 µ 2 < d 0 P (T n1 + 2 < t) t < t n1 + 2,α Example 9.11 An experiment is performed to determine whether the average nicotine content of one kind of cigarette exceeds that of another kind by.2 milligram. If = 50 cigarettes of the first kind had an average nicotine content of x 1 = 2.61 milligrams with s 1 =.12 milligram, whereas = 40 cigarettes of the other kind had an average nicotine content of x 2 = 2.38 milligrams with s 2 =.14 milligram. Test H 0 : µ 1 µ 2 =.2 against H 1 : µ 1 µ 2 >.2 at the.05 level of significance. Don t reject H 0. s 2 p = 49(.12)2 + 39(.14) 2 =.0167 s p = t =.129 1/50 + 1/40 = 1.09 Example 9.12 In the comparison of two kinds of paint, a consumer testing service finds that four 1-gallon cans of one brand cover on the average 546 square feet with a standard deviation of 31 square feet, whereas four 1-gallon cans of another brand cover on the average 492 square feet with a standard deviation of 26 square feet. Assuming that the two populations sampled are normal and have equal variances, test H 0 : µ 1 µ 2 = 0 against H 1 : µ 1 µ 2 > 0 at the 0.05 level of significance. s p = , t = 2.67 t 6,.05 = Reject H Testing for Equality of Two Variances It is noted that we assume σ 1 = σ 2 in the previous two sample t test, but how can we know σ 1 = σ

11 The F distribution: Let U and V be independent random variables, U χ 2 d 1, V χ 2 d 2. variable F = U/d 1 V/d 2 Then the random is said to have an F distribution with d 1, d 2 degrees of freedom, denoted by F F d1,d 2. The density function of an F random variable with d 1 and d 2 degrees of freedom is given by for x > 0 x d 1/2 1 f(x) = c (1 + x(d 1 /d 2 )) (d 1+d 2 )/2 If S 2 1 and S 2 2 are the variances of independent random samples of size and from normal populations with the variances σ 2 1 and σ 2 2, then In particular, if σ 1 = σ 2, then F = S2 1/σ 2 1 S 2 2/σ 2 2 F = S2 1 S 2 2 F n1 1, 1 F n1 1, 1 Hypothesis testing for equality of variances: H 0 : σ 1 = σ 2 Test statistic: f = s 2 1/s 2 2 Alternative Reject H 0 at α level if p-value H 1 : σ 1 > σ 2 f F n1 1, 1,1 α P (F n1 1, 1 f) H 1 : σ 1 < σ 2 f F n1 1, 1,α P (F n1 1, 1 f) H 1 : σ 1 σ 2 f F n1 1, 1,1 α/2 or f F n1 1, 1,α/2 min{2p (F n1 1, 1 f), 2P (F n1 1, 1 f)} 9-11

12 Example 9.13 A hypothesis of ongoing clinical interest is that vitamin C prevents the common cold. A study is organized to test the hypothesis using 20 prisoners as participants. In the study, 10 are randomly allocated to receive vitamin C capsules and 10 to receive placebo capsules. The number of colds over a 12-month period for each participant is given in table below. i Vitamin C x i1 Placebo x i2 d i = x i1 x i We wish to test the hypothesis that vitamin C prevents the common cold. (i) Is a one-sample or two-sample test needed here? (ii) Is a one-sided or two-sided test needed here? (iii) Which of the following test procedures should be used to test this hypothesis? (more than one may be necessary) (a) One-sample t test (b) Paired t test (c) Two-sample t test with equal variances (d) Two-sample t test with unequal variances (e) F test for the equality of two variances (iv) Carry out the test procedure(s) in part (iii), and report the range of p-value. (v) Drive a 95% CI for the mean difference (vitamin C - placebo) in number of colds per year between the two groups. 9-12

13 9.8 Two-Sample t Test for Independent Samples with Unequal Variances Recall that where T = ( X 1 X 2 ) (µ 1 µ 2 ) v = S S2 2 ( s2 1 + s2 2 ) ( s2 1 ) ( s2 2 ) 2 t(v), Assume that the null hypothesis is H 0 : µ 1 µ 2 = d 0 and let t = x 1 x 2 d 0 be the t-statistic. Then approximately we have s s2 2 H 1 P value Reject H 0 at the α-level if µ 1 µ 2 d 0 2P (T v > t ) t > t α/2,v µ 1 µ 2 > d 0 P (T v > t) t > t α,v µ 1 µ 2 < d 0 P (T v < t) t < t α,v Example , p.341 A 1980 study was conducted whose purpose was to compare the indoor air quality in offices where smoking was permitted with that in offices where smoking was not permitted. Measurements were made of carbon monoxide (CO) at 1:20pm in 40 work areas where smoking was permitted and 40 work areas where smoking was not permitted. Where smoking was permitted, the mean CO was 11.6 parts per million (ppm) and the standard deviation CO was 7.3 ppm. Where smoking was not permitted, the mean CO was 6.9 parts per million (ppm) and the standard deviation CO was 2.7 ppm. (i) Test for whether or not the standard deviation of CO is significantly different in the two types of working environments. (ii) Test for whether or not the mean CO is significantly different in the two types of working environments. (iii) Provide a 95% CI for the difference in mean CO between the smoking and nonsmoking working environments. 9-13

14 (i) f = 7.31, p value < 0.05 (ii) t = 3.82, d = 49.5, k = 49, t 49,.975 < t 40,0.975 = (iii) 4.7 ± 2.009(1.232) = 4.7 ± 2.47 = (2.2, 7.2) 9.9 Tests on Two Proportions Consider two population proportions, p 1 and p 2. We d like to know whether p 1 and p 2 are equal. So we get two independent random samples. Population proportion Sample size Sample proportion p 1 ˆp 1 p 2 ˆp 2 The sampling distribution of ˆp 1 ˆp 2 : E(ˆp 1 ˆp 2 ) = p 1 p 2 Var(ˆp 1 ˆp 2 ) = p 1(1 p 1 ) + p 2(1 p 2 ) ˆp 1 ˆp 2 (p 1 p 2 ) N(0, 1) p 1 (1 p 1 ) + p 2(1 p 2 ) Hypothesis testing for p 1 = p 2 Null hypothesis: H 0 : p 1 = p 2 z-statistic: z = ˆp 1 ˆp 2 ( 1 ˆp(1 ˆp) + 1 ), 9-14

15 where ˆp is the pooled sample proportion ˆp = ˆp 1 + ˆp 2 + count of successes in both samples = + Alternative Reject H 0 at α level if p-value H 1 : p 1 > p 2 z z 1 α P (Z z) H 1 : p 1 < p 2 z z 1 α P (Z z) H 1 : p 1 p 2 z z 1 α/2 2P (Z z ) Example 9.15 Does taking aspirin regularly help prevent heart attacks? A double-blind randomized comparative experiment assigned 11,037 male doctors to take aspirin and another 11,034 to take a placebo. After 5 years, 104 of the aspirin group and 189 of the control group had died of heart attacks. (i) Is this difference large enough to convince us that aspirin works? (ii) Find a 99% CI for the proportion difference. = 11037, ˆp 1 = 104/11037 = , = 11034, ˆp 2 = 189/11034 = , ˆp = ( )/( ) = 293/22071 = H 0 : p 1 = p 2, H 1 : p 1 < p 2 z = ( )/ ( )(1/ /11037) = / = P-value= %C.I.is(SE =.00154) ± = ± = ( , ) 9-15