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STATISTICAL INFERENCE PART VII HYPOTHESIS TESTING – APPLICATIONS – ONE POPULATION TESTS 1 HYPOTHESIS TEST FOR POPULATION MEAN, • KNOWN AND X~N(, 2) OR LARGE SAMPLE CASE: Two-sided Test H0: = 0 HA: 0 Test Statistic z Rejecting Area x 0 / n /2 -z/2 • Reject H0 if z < z/2 or z > z/2. /2 1- z/2 Reject H0 Reject H0 Do not reject H0 2 HYPOTHESIS TEST FOR POPULATION MEAN, One-sided Tests 1. H0: = 0 HA: > 0 Test Statistic z x 0 1- / n • Reject H0 if z > z. 2. H0: = 0 x 0 z HA: < 0 / n • Reject H0 if z < z. Rejecting Area z Do not reject H0 Reject H0 1- - z Reject H0 Do not reject H0 3 POWER OF THE TEST AND P-VALUE • 1- = Power of the test = P(Reject H0|H0 is not true) • p-value = Probability of obtaining a test statistics at least as extreme as the one that you observed by chance 4 P-VALUE • http://labstats.net/articles/pvalue.html • Ex: toss a fair coin 20 times and count the number of heads, and then repeat this 10000 times (Or simulate data) • Ho: Coin is fair (i.e. p=0.5) 5 P-VALUE 10 heads out of 20 tosses was the most frequent outcome 6 P-VALUE • Suppose we toss the coin and observe that it lands heads on 16 out of 20 tosses. We expect 10 heads but observe 16, what do we make of this value? Is it very large, or unusual? • P-value=probability of getting the observed results (16 heads) or more extreme results (17, 18, 19, or 20 heads) • 61 out of 10000 trials ended up with 16 or more heads p = 0.0061 • If the unknown coin were fair, then one would expect to obtain 16 or more heads only 0.61% of the time 7 CALCULATION OF P-VALUE x 0 • Determine the value of the test statistics, z 0 / n • For One-Tailed Test: p-value p-value= P(z > z0) if HA: >0 p-value= P(z < z0) if HA: <0 • For Two-Tailed Test p=p-value = 2.P(z>z0) p=p-value = 2.P(z<-z0) z0 p-value z0 p/2 p/2 -z0 z0 8 DECISION RULE BY USING P-VALUES • REJECT H0 IF p-value < p-value • DO NOT REJECT H0 IF p-value 9 Example • Do the contents of bottles of catsup have a net weight below an advertised threshold of 16 ounces? • To test this 25 bottles of catsup were selected. They gave a net sample mean weight of X 15.9 . It is known that the standard deviation is .4 . We want to test this at significance levels 1% and 5%. 10 CALCULATIONS The z-score is: 15.9 16 Z 1.25 .4 25 The p-value is the probability of getting a score worse than this (relative to the alternative hypothesis) i.e., P(Z 1.25) .1056 Compare the p-value to the significance level. Since it is bigger than both 1% and 5%, we do not reject the null hypothesis. 11 P-value for this one-tailed Test • The p-value for this test is 0.1056 0.1056 0.10 0.05 -1.25 reject H0 at • Thus, do not 1% and 5% significance level. We do not have enough evidence to say that the contents of bottles of catsup have a net weight of less than 16 ounces. 12 Test of Hypothesis for the Population Mean ( unknown) • For samples of size n drawn from a Normal Population, the test statistic: x- t s/ n has a Student t-distribution with n1 degrees of freedom 13 EXAMPLE • 5 measurements of the tar content of a certain kind of cigarette yielded 14.5, 14.2, 14.4, 14.3 and 14.6 mg per cigarette. Show the difference between the mean of this sample x 14.4 and the average tar content claimed by the manufacturer, =14.0, is significant at =0.05. 5 ( xi x ) s 2 i 1 n 1 s 0.158 2 ( 14.5 14.4 )2 ... ( 14.6 14.4 )2 0.025 5 1 14 SOLUTION • H0: = 14.0 HA: 14.0 x 0 14.4 14.0 t 5.66 s / n 0.158 / 5 t / 2 ,n1 t0.025 ,4 2.766 Decision Rule: Reject Ho if t<-t/2 or t> t/2. 15 CONCLUSION • Reject H0 at = 0.05. Difference is significant. 0.025 0.95 -2.766 Reject H0 0.025 2.766 5.66 Reject H0 16 P-value of This Test • p-value = 2.P(t > 5.66) = 2(0.0024)=0.0048 Since p-value = 0.0048 < = 0.05, reject H0. Minitab Output T-Test of the Mean Test of mu = 14.0000 vs mu not = 14.0000 Variable C1 N Mean StDev SE Mean 5 14.4000 0.1581 0.0707 T P-Value 5.66 0.0048 17 CONCLUSION USING THE CONFIDENCE INTERVALS MINITAB OUTPUT: Confidence Intervals Variable C1 N Mean StDev SE Mean 95.0 % C.I. 5 14.4000 0.1581 0.0707 ( 14.2036, 14.5964) • Since 14 is not in the interval, reject H0. 18 EXAMPLE Problem: At a certain production facility that assembles computer keyboards, the assembly time is known (from experience) to follow a normal distribution with mean () of 130 seconds and standard deviation () of 15 seconds. The production supervisor suspects that the average time to assemble the keyboards does not quite follow the specified value. To examine this problem, he measures the times for 100 assemblies and found that the sample mean assembly time ( x) is 126.8 seconds. Can the supervisor conclude at the 5% level of significance that the mean assembly time of 130 seconds is incorrect? 19 • We want to prove that the time required to do the assembly is different from what experience dictates: H A : 130 X 126.8 • The sample mean is • The standard deviation is 15 • The standardized test statistic value is: 126.8 130 Z 2.13 15 100 20 Two-Tail Hypothesis: H0: 130 Type I Error Probability HA: 130 -z z=test statistic values Reject H0 (z<-z) 1- 0 Do not Reject H0 (-zz<z) z Reject H0 (z>z) 21 Test Statistic: z= X- 126.8 - 130 = = -2.13 15 n 100 Rejection Region .90 .9 0 -z0 1 0 0 z0 1 Z 22 CONCLUSION • Since –2.13<-1.96, it falls in the rejection region. • Hence, we reject the null hypothesis that the time required to do the assembly is 130 seconds. The evidence suggests that the task now takes either more or less than 130 seconds. 23 DECISION RULE • Reject Ho if z < -1.96 or z > 1.96. In terms of X , reject H0 if 15 X 130 1.96 = 127.6 06 100 15 or X 130 1.96 =132.94 100 24 P-VALUE • In our example, the p-value is p value 2.P(Z 2.13) 2(0.0166) 0.0332 So, since 0.0332 < 0.05, we reject the null. 25 Calculating the Probability of Type II Error Ho: = 130 HA: 130 • Suppose we would like to compute the probability of not rejecting H0 given that the null hypothesis is false (for instance =135 instead of 130), i.e. =P(not rejecting Ho|Ho is false). Assuming =135 this statement becomes: P(127.06 x 132.94| 135) 127.06-135 132.94-135 P( Z ) 15/ 100 15/ 100 P(-5.29 Z -1.37) .0853 26 TESTING HYPOTHESIS ABOUT POPULATION PROPORTION, p • ASSUMPTIONS: 1. The experiment is binomial. 2. The sample size is large enough. x: The number of success The sample proportion is x pq p̂ ~ N(p, ) n n approximately for large n (np 5 and nq 5 ). 27 HYPOTHESIS TEST FOR p Two-sided Test H0: p = p0 HA: p p0 Test Statistic Rejecting Area p̂ p z /2 pq / n 1- - z/2 /2 z/2 Reject H0 Do not reject H0 Reject H0 • Reject Ho if z < -z/2 or z > z/2. 28 HYPOTHESIS TEST FOR p One-sided Tests 1. H0: p= p0 HA: p > p0 Test Statistic p̂ p z pq / n Rejecting Area 1- z • Reject Ho if z > z. p̂ p 2. H0: p = p0 z pq / n HA: p < p0 Do not reject H0 Reject H0 1- -z • Reject Ho if z < - z. Reject H0 Do not reject H0 29 EXAMPLE • Mom’s Home Cokin’ claims that 70% of the customers are able to dine for less than $5. Mom wishes to test this claim at the 92% level of confidence. A random sample of 110 patrons revealed that 66 paid less than $5 for lunch. Ho: p = 0.70 HA: p 0.70 30 ANSWER • x = 66, n = 110 and p = 0.70 x 66 p̂ 0.6 n 110 • = 0.08, z/2 = z0.04 = 1.75 • Test Statistic: 0.6 0.7 z 2.289 (0.7)(0.3) /110 31 CONCLUSION • DECISION RULE: Reject H0 if z < -1.75 or z > 1.75. • CONCLUSION: Reject H0 at = 0.08. Mom’s claim is not true. /2 /2 -2.289 -1.75 1.75 32 P-VALUE • p-value = 2. P(z < -2.289) =2(0.011) = 0.022 The smallest value of to reject H0 is 0.022. Since p-value = 0.022 < = 0.08, reject H0. 0.011 -2.289 33 CONFIDENCE INTERVAL APPROACH • Find the 92% CI for p. p̂q̂ (0.6)(0.4) p̂ z / 2 0.6 1.75 n 110 92% CI for p: 0.52 p 0.68 • Since p 0.7 is not in the above interval, reject H0. Mom has overestimated the percentage of customers that pay less than 5$ for a meal. What happens with wider confidence intervals? Exercise: Calculate the 95% and 99% CIs for p. 34 SAMPLING DISTRIBUTION OF s2 • The statistic (n 1)s 2 is chi-squared distributed with n-1 d.f. when the population random variable is normally distributed with variance 2. 2 2 35 CHI-SQUARE DISTRIBUTION f(2) A A 2 0 2 1-A 2 A 36 Inference about the Population Variance (2) • Test statistic (n 1)s 2 2 2 which is chi-squared distributed with n - 1 degrees of freedom 2 LCL = (n - 1) s 2/2 UCL = (n - 1) s2 21 - /2 Confidence interval estimator: 37 Testing the Population Variance (2) EXAMPLE • Proctor and Gamble told its customers that the variance in the weights of its bottles of Pepto-Bismol is less than 1.2 ounces squared. As a marketing representative for P&G, you select 25 bottles and find a variance of 1.7. At the 10% level of significance, is P&G maintaining its pledge of product consistency? H0: 2 = 1.2 HA: 2 < 1.2 38 ANSWER 2 • n=25, s2=1.7, =0.10, 0.90,24 15.659 • Test Statistics: (n 1)s (24)1.7 34 2 1.2 2 2 2 2 • Decision Rule: Reject H0 if ,n 1 15.6587 • Conclusion: Because 2=34 > 15.6587, do not reject H0. • We don’t have enough evidence that suggests the variability in product weights less than 1.2 ounces squared. 39 EXAMPLE • A random sample of 22 observations from a normal population possessed a variance equal to 37.3. Find 90% CI for 2. 90% CI for 2: 2 (n 1)s 2 (n 1)s 2 2 2 0.05,21 0.95,21 (21)37.3 (21)37.3 2 32.6705 11.5913 23.9757 2 67.5765 40