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Chapter 7 Hypothesis Testing Lesson 7-1 Introduction Introduction Use a random sample to learn something about a larger population. Two ways to learn about a population Confidence Intervals ( Chapter 6) Hypothesis Testing (Chapter 7) Confidence Intervals and Hypothesis Testing Allow us to use sample data to estimate a population value, like the true mean or the true proportion. Example: What is the true average amount students spend weekly on gas? Allows us to use sample data to test a claim about a population, such as testing whether a population proportion or population mean equals some number. Example: Is the true average amount that students spend weekly on gas $20. Lesson 7-2 Basic of Hypothesis Testing Illustrating Hypothesis Testing Population of Light Bulbs Sample of 49 Light Bulbs The manufacture claims that the average lifetime of a bulb is 500 hours? A consumer advocate group would like to know if the mean lifetime of a bulb is less than 500 hours if the population average was 500 hours. Illustrate Hypothesis Testing 42 X 491 518 512 506 494 μ = 500 X 476 488 482 42 s 6 n 49 Illustrating Hypothesis Testing The sample mean of 476 hours is “too far” from the claimed mean (500 hours) that we have reason to believe that the manufacture’s claim is probably not true. Consumer advocate group would conclude that the population mean is likely some value less than 500 hours. The sample mean of 491 hours is “close enough” to the claimed mean (500 hours) that we have reason to believe that the manufacture’s claim is probably true. Consumer advocate group would conclude that the population mean is likely 500 hours. Illustrating Hypothesis Testing This raises an interesting question. How far from the presumed mean of 500 hours can our sample man be before we reject the idea that the bulb lasts 500 hours? To answer this question requires hypothesis testing. Definitions In statistics, a hypothesis is a claim or statement about a characteristic of a population. A hypothesis test (or test of significance) is standard procedure for testing a claim about a characteristic of a population. The claims that we test regard the population mean, population proportion and population standard deviations. Rare Event Rule for Inferential Statistics If, under a given assumption, the probability of a particular observed event is exceptionally small, we conclude that the assumption if probably not correct. Example – Page 385, #2 What do you conclude? Use the rare event rule (Under a given assumption, the probability of a particular observed event is exceptionally small, we conclude that the assumption is probability not correct. Claim: A gender selection method is effective in helping couples have baby girls and among 50 babies, 49 are girls. Gender selection appears to be effective. There is sufficient evidence to support the claim. Under the assumption that the claim is not true, having 49 girls among 50 babies would be a rare event (unusual). Two Types of Hypotheses Null Hypothesis Ho Includes the assumed value of the population parameter Is the statement to be tested. It is assume true until evidence we have indicated otherwise Alternative Hypothesis H1 or Ha The statement that the parameter has a value that somehow differs from the null hypothesis. Is the claim to be tested. Criminal Trial Analogy First, State the 2 hypothesis Then, collect evidence such finger prints, blood spots, hair samples, etc. H0: Defendant is not guilty H1: Defendant is guilty. In statistics, the data is the evidence Then, make initial assumption. Defendant is innocent until proven guilty. In statistics, we always assume the null hypothesis is true. Criminal Trial Analogy Then, make a decision based on the available evidence. If there is sufficient evidence “beyond a reasonable doubt” Behave as if defendant is guilty Reject the null hypothesis If there is not enough evidence Behave as if defendant is not guilty Do not reject the null hypothesis Important Points Neither decision entails proving the null hypothesis or the alternative hypothesis. We merely state there is enough evidence to behave one way or the other. This is also always true in statistics! No matter what decision we make, there is always a chance we made an error. Right-Tailed Test H 0 ____ H1 ____ Critical Region Area of α Critical Value Left-Tailed Test H 0 ____ H1 ____ Two-Tailed Test H 0 ____ H1 ____ Example – Page 386, #6 Examine the given statements, than express the null hypothesis and alternative hypothesis in symbolic form. Be sure to use the correct symbol (μ, σ, p) for the indicated parameter. The mean IQ of statistic students is at least 110. H 0 : 110 H1 : 110 Example – Page 386, #12 Examine the given statements, than express the null hypothesis and alternative hypothesis in symbolic form. Be sure to use the correct symbol (μ, σ, p) for the indicated parameter. Salaries among women business analysts have a standard deviation greater than $3000. H 0 : 3000 H1 : 3000 Test Statistic The test statistic is a value computed from the sample data, and it is used in making the decision about the rejection of the null hypothesis. Proportion z p̂ p pq n Standard Deviation 2 n 1 s 2 2 Mean z t x x s n Example – Page 387, #24 Find the test statistics. The claim is that the proportion of drivers stopped by police in a year is different from the 10.3% reported by the Department of Justice. Sample statistics include n = 800 randomly selected drivers with 12% of them stopped in the past year. p 0.103 n 800 pˆ 0.12 p̂ p z pq n 0.12 0.103 0.103 0.897 800 1.5819 1.58 Definitions The critical region or rejection region is the set of all values of the test statistics that cause us to reject the null hypothesis. The significance level () is the probability that the test statistic will fall in the critical region when the null hypothesis is actually true. If the test statistic falls in the critical region, we reject the null hypothesis, so is the probability of making a mistake of rejecting the null hypothesis when it is true. Definitions A critical value is any value that separates the critical region (where reject the null hypotheses) from the values of the test statistics that do not lead to rejection of the null hypothesis. Example – Page 386, #14 Find the critical values. Two-tailed test; = 0.01 .01 A 0.005 2 0.005 invnorm(1 .005,0,1) 2.575 invnorm(0.005,0,1) 2.575 -2.575 0.005 0 2.575 Example – Page 386, #16 Find the critical values. Left-tailed test; = 0.05 A 0.05 invnorm(0.05,0,1) 1.645 0.05 -1.645 0 P-Values or Probability Values The p-value is the probability of getting a value of the test statistic that is at least as extreme as the one representing the sample data, assuming that the null hypothesis is true. The p-value represents how likely we would be to observe such an extreme sample if the null hypothesis were true. If the p-value is “small” (Typically less than 0.05), we reject the null hypothesis. Example – Page 387, #26 Find the p-value. The test statistic in a left-tailed test is z = -1.72 P ( z 1.72) normalcdf ( E 99, 1.72,0,1) 0.0427 p value 0.0427 0.0427 -1.72 0 Example – Page 387, #30 Find the p-value. With H1 : p = 0.30, test statistic is z = 2.44. P ( z 2.44) normalcdf ( E 99, 2.44,0,1) 0.00734 0.00734 P ( z 2.44) normalcdf (2.44, E 99,0,1) 0.00734 p value 2(0.00734) 0.01469 -2.44 0.00734 0 2.44 Conclusions in Hypothesis Testing We always test the null hypothesis Reject the HO Fail to reject the HO Traditional Method Reject Ho if the test statistics falls within the critical region. Fail to reject Ho if the test statistics does not fall within the critical region. Conclusions in Hypothesis Testing P-value Method Reject the HO if P-value α (where α is the significance level, such as 0.05). Fail to reject the HO if P-value > α Confidence Intervals Because the confidence interval estimate of a population parameter contains the likely values of that parameter, reject a claim that the population parameter has value that is not included in the confidence interval. Example – Page 387, #34 State the conclusion. Original claim: The proportion of college graduates who smoke is less than 0.27. Initial conclusions: Reject the null hypothesis H o : p 0.27 H a : p 0.27 Reject H0 There is sufficient evidence to support the claim that the proportion of college graduates who smoke is less than 0.27. Example – Page 387, #36 State the conclusion. Original claim: The proportion of M&Ms that are blue is equal to 0.10. Initial conclusions: Reject the null hypothesis. H o : p 0.10 H a : p 0.10 Reject H0 There is sufficient evidence to reject the claim that the M&M’s are blue is equal to 0.10. Type I and Type II Errors A Type I error is the mistake of rejecting the null hypothesis when it is true. The symbol α (alpha) is used to represent the probability of making a type I error. A Type II error is the mistake of failing to reject the null hypothesis when it is false. The symbol β (beta) is used to represent the probability of a type II error Errors in Criminal Trials Truth Jury Decision Not Guilty (Ho) Guilty Type I Error Guilty (Ha) Putting an innocent person in jail. Reject Ho when Ho is true Not Guilty Type II Error Letting a guilty person go free Fail to reject Ho when Ha is true Controlling Type I and Type II Errors For any fixed , an increase in the sample size n will cause a decrease in β. For any fixed sample size n, a decrease in will cause an increase in β. For any fixed sample size n, an increase in will cause a decrease in β. To decrease both and β, increase the sample size. Example – Page 387, #38 Identify Type I and Type II errors. The proportion of college graduates who smoke is less than 0.27. H o : p 0.27 H1 : p 0.27 Type I Error: Supports the alternative hypothesis when the null hypothesis is true. Type II Error: Supports the null hypothesis when the alternative hypothesis is true. Lesson 7-3 Testing a Claim about a Proportion Assumptions The sample observations are simple random sample. Population size > 10n The conditions np ≥ 5 and nq ≥ 5 are both satisfied. Binomial distribution of sample proportions can be approximated by a normal distribution with np npq Test Statistic p̂ p z pq n n = sample size of number of trials p = population proportion (used in the null hypothesis) q 1 p x pˆ n sample proportion Example – Page 395, #2 In a Gallup survey, 1087 randomly selected adults were asked “Do you have occasion to use alcoholic beverages such as liquor, wine, or beer or are you a total abstainer?” Sixty-two percent of the subjects said that they used alcoholic beverages. Consider a hypothesis test that uses a 0.05 significance level to test the claim that the majority (more than 50%) of adults used alcoholic beverages. n 1087 pˆ 0.62 p 0.50 0.05 H o : p 0.50 H1 : p 0.50 Example – Page 395, #2 n 1087 A). What is the test statistic? p̂ p z pq n 0.62 0.50 0.50 0.50 1087 7.91 pˆ 0.62 p 0.50 0.05 H o : p 0.50 H1 : p 0.50 Example – Page 395, #2 n 1087 B). What is the critical value? pˆ 0.62 p 0.50 0.05 0.05 1.645 invnorm(1 .05,0,1) 1.645 7.91 H o : p 0.50 H1 : p 0.50 z pˆ 7.91 Example – Page 395, #2 C). What is the P-value? n 1087 P ( z 7.91) pˆ 0.62 p 0.50 normalcdf (7.91, E 99,0,1) 1.29 E 15 0.0001 0.05 H o : p 0.50 H1 : p 0.50 z pˆ 7.91 Example – Page 395, #2 D). What is the conclusion? H o : p 0.50 H1 : p 0.50 0.05 z pˆ 7.91 P value 0.0001 1.645 7.91 If P-value ≤ , reject Ho There is sufficient evidence to reject Ho since (p-value = 0.0001 α = 0.05) and conclude that the majority of adult use alcoholic beverages. Example – Page 395, #2 E). Based on the preceding results, can we conclude that 62% is significantly greater than 50% for all such hypothesis tests? No, 62% might not be significantly greater than 50% for some samples, such as one with n = 50 Example – Page 395, #2 Stats/Tests/5:1-PropZTest Example – Page 396, #4 In a recent year, of the 109,857 arrests for Federal offenses, 29.1% were for drug offenses (based on the data from the U.S. Department of Justice). Use a 0.01 significance level to test the claim that the drug offense is equal 30%. How can the result be explained, given that 29.1% appears to be so close to 30%. H o : p 0.30 n 109,857 H1 : p 0.30 p 0.30 x 109857 0.291 31969 0.01 Example – Page 396, #4 n 109,857 p 0.30 x 109857 0.291 31969 0.01 There is sufficient evidence to reject Ho since p-value = 0.0001 ≤ = 0.01 and conclude that arrests for drugs offenses is less than 30%. The sample size was so large that we were able to distinguish between 29.1% and 30%. Example – Page 397, #10 In 1990, 5.8% of job applicants who were test for drugs failed the test. At the 0.01 significance level, test the claim that the failure rate is not lower if a simple random sample of 1520 current job applicants results in 58 failures (based on the data from the American Management Association). Does the result suggest that fewer job applicants now use drugs? n 1520 H o : p 0.058 p 0.058 H1 : p 0.058 x 58 0.01 Example – Page 397, #10 n 1520 p 0.058 x 58 H o : p 0.058 H1 : p 0.058 0.01 There is sufficient evidence to reject Ho since p-value = 0.0004 ≤ = .01 and conclude that fewer job applicants now use drugs. Example – Page 397, #16 A simple random sample of households with TV sets in use show that 1024 of them were tuned to 60 Minutes while 3836 were tuned to some other show. Use a 0.025 significance level to test the claim of a CBS executive that “60 Minutes gets more than a 20 share,” which means that more than 20% of the sets in use are turned to 60 Minutes. If you are a commercial advertiser and you are trying to negotiate lower costs, what would you argue? Step 1 – Identify the population of interest and the parameter you want to draw conclusions about. p = proportion of households that were tuned to 60 Minutes. n 4860 H o : p 0.20 x 1024 H1 : p 0.20 p 0.20 0.025 Example – Page 397, #16 Step 2 – Choose the appropriate inference procedure. Verify conditions for using the selected procedure. n 4860 Use a one proportion z-test Question stated SRS Population size > 10(4860) = 48600 np = (4860)(0.20) = 972 ≥ 5 and nq = (4860)(.80) = 3888 ≥ 5 x 1024 p 0.20 0.025 H o : p 0.20 H1 : p 0.20 Example – Page 397, #16 Step 3 – Carry out the inference procedure. n 4860 x 1024 p 0.20 0.025 H o : p 0.20 H1 : p 0.20 Example – Page 397, #16 p̂ p 0.2107 0.20 z 0.20(.80) pq 4860 n = 1.865 p-value = 0.0311 n 4860 H o : p 0.20 x 1024 H1 : p 0.20 p 0.20 0.025 0.025 1.96 Example – Page 397, #16 Step 4 – Interpret your results in the context of the problem. There is sufficient evidence to fail to reject Ho since P-value = 0.0311 > = 0.025 and are unable to conclude that more than 20% of households are watching 60 Minutes. Even though p̂ 0.211 happens to be greater than 20%, that is not enough evidence to 97.5% (1 – 0.025 = 0.975) certain that the true population proportion is greater than 20%. Lesson 7-4 Testing a Claim About a Mean: σ Known Assumptions A simple random sample is obtained. The population standard deviation σ is known The population from which the sample is drawn is normally distribution or n > 30. Mean X Standard Deviation X n Test Statistic z x X n Example – Page 404, #2 Determine whether the given conditions justify using the methods of this section when testing a claim about the population mean μ. The sample size n = 7, σ is not know, and the original population is normally distributed. No Example – Page 404, #4 Determine whether the given conditions justify using the methods of this section when testing a claim about the population mean μ. The sample size n = 47, σ = 12.6 and the original population is not normally distributed. yes Example – Page 404, #6 Find the test statistic, P-value, critical values and state the final conclusion. Claim: The mean body temperature of healthy adults is less than 98.6° F. Sample Data: n 106 H o : μ 98.60 x 98.20 σ 0.62 H1 : μ 98.60 α 0.01 Example – Page 404, #6 z x X n H o : μ 98.60 n 106 H1 : μ 98.60 x 98.20 σ 0.62 98.20 98.60 6.64 0.62 106 α 0.01 p value normalcdf ( E 99, 6.64,98.20,0.62) 1.577 E 11 0.0001 Example – Page 404, #6 Critical Value n 106 x 98.20 σ 0.62 α 0.01 0.01 H o : μ 98.60 2.33 invnorm(0.01) 2.33 H1 : μ 98.60 Example – Page 404, #6 Final Conclusion Reject Ho since the P-value ≤ , there is sufficient evidence to support the claim that the mean is less than 98.6°F. Example – Page 404, #6 n 106 Stats/Tests/1:Z-Test X 98.20 σ 0.62 α 0.01 H o : μ 98.60 H1 : μ 98.60 Example – Page 405, #10 The health of the bear population in Yellowstone National Park is monitored by periodic measurements take from anesthetized bears. A sample of 54 bears has a mean weight of 182.9 lb. Assuming that σ is known to be 121.8 lb, use a 0.10 significance level to test the claim that the population mean of all such bear weights is less than 200 lb. Step 1 – Identify the population of interest and the parameter you want to draw conclusions about. µ = mean weight of bears in Yellowstone National Park Ho: μ = 200 H1: < 200 Example – Page 405, #10 Step 2 – Choose the appropriate inference procedure. Verify the conditions for using the selected procedure. Use a one sample z-test Not sure if it’s a SRS – should be representative the population. Standard deviation of the population is known Approximately normal distributed since n > 30 Example – Page 405, #10 Step 3 – Carry out the inference procedure. z x X n 182.9 200 1.032 121.8 54 p value 0.1511 0.10 1.282 Example – Page 405, #10 Step 4 – Interpret your results in the context of the problem There is sufficient evidence to fail to reject Ho since (p-value = 0.151 > α = 0.10) and are unable to conclude that weight of bears in Yellowstone National Park is less than 200 lbs. Underlying Rational of Hypothesis Testing If the sample results can easily occur when assumption (null hypothesis) is true, we attribute the relatively small discrepancy between the assumption and the sample results to chance. If the sample results cannot easily occur when that assumption (null hypothesis) is true, we explain the relatively large discrepancy between the assumption and the sample by concluding that the assumption is not true. Example – Page 406, #14 Analysis of the last digits of sample data values sometimes reveals whether the data have been accurately measured and reported. When single digits 0 through 9 are randomly selected with replacement, the mean should be 4.50 and the standard deviation should be 2.87. Reported data (such as weights or heights) are often rounded so that the last digits include disproportionately more 0s and 5s. The last digits in the reported lengths (in feet) of the 73 home runs hit by Barry Bonds in 2001 are used to test the claim that they come from a population with mean 4.50 (based on data from USA Today). When Minitab is used to test the claim, the display is as shown here. Using a 0.05 significance level, interpret the Minitab results. Does it appear that the distances were accurately measured? Example – Page 406, #14 Test of mu = 4.5 vs mu not = 4.5 The assume sigma = 2.87 Variable BONDS Variable BONDS N 73 Mean 1.753 95.00% CI (1.095, 2.412) StDev 2.650 Z – 8.18 Test statistics = – 8.18, P-value = 0.000. Reject Ho SE Mean 0.336 P 0.000 Lesson 7-5 Testing a Claim about a Mean: σ Not Known Assumptions for Testing Claims About a Population Mean: Not Known Simple random sample Population standard deviation σ is not known Population is normally distributed or n > 30 Test Statistic: σ Not Known. x x t s n Choosing the Appropriate Distribution σ is known, use a z-distribution σ is unknown, use a t-distribution Example – Page 415, Problem #2 Determine whether the hypothesis test involves a normal distribution, t-distribution or neither. Claim: 75 . Sample Data: n 25, x 102, s 15.3 The sample data appear to come from a population with a distribution that is very far from normal, and σ is unknown. neither Distribution is not normal and n 30 Example –Page 415, Problem #4 Determine whether the hypothesis test involves a normal distribution, t-distribution or neither. Claim: 2.80. Sample Data: n 150, x 2.88, s 0.24 The sample data appear to come from a population with a distribution that is not normal, and σ is unknown. t-distribution σ is unknown and n 30 Example - Page 415, Problem #8 Find the P-value. Two-Tailed test with n 9 and test statistic t 1.577 Use Table A-3 df 8 1.577 Falls between: 1.860 and 1.397 P-value is between: 0.10 and 0.20 Using the T-83 2nd/VARS/5:tcdf tcdf (E 99, 1.577,8) 0.07672 two tailed : (0.07672 2) 0.1535 0.07672 – 1.577 Example – Page 415, Problem #10 Find the test statistic, P-value, critical values and state the final conclusion. Claim: The mean body temperature of healthy adults is less than 98.6F. Sample data: n 35, x 98.20F , s 0.62. The significance level is 0.01. Ho : 98.6 H1 : 98.6 Page 415, Problem #10 Find the test statistic and P-Value: 98.20 98.6 x x 3.812 t 0.62 s 35 n n 35 x 98.20 s 0.62 α 0.01 Ho : 98.6 H1 : 98.6 p value tcdf (E99, 3.812,34) 2.73E 4 0.000273 Page 415, Problem #10 Final Conclusion Reject Ho since P-value ≤ α, there is sufficient evidence to conclude that μ < 98.6 n 35 X 98.20 s 0.62 α 0.01 Ho : 98.6 H1 : 98.6 pv 0.000273 Page 415, Problem #10 n 35 x 98.20 s 0.62 α 0.01 Ho : 98.6 H1 : 98.6 STAT/TESTS/2:T-Test Example – Page 415, Problem #18 Heather Carielli is a former student of the author who earned a master’s degree in statistics at the University of Massachusetts. When she randomly selected 16 new textbooks in the college bookstore, she found that they had prices with a mean of $70.41 and a standard deviation of $19.70. Is there sufficient evidence to warrant rejection of a claim in the college catalog that the mean price of a textbook at this college is less than $75? μ 75 x 70.41 s 19.7 n 16 α 0.05 H o : μ 75 H1 : μ 75 Example – Page 415, Problem #18 75 x 70.41 s 19.7 n 16 0.05 Test Statistic t 0.932 P-value p value 0.1831 Ho : 75 H1 : 75 Example – Page 415, Problem #18 Fail to reject Ho since P-value > α, there is not sufficient evidence to conclude that μ < 75 75 x 70.41 s 19.7 n 16 0.05 Ho : 75 H1 : 75 p value 0.1831 Example – Page 416, Problem #20 In previous tests, baseballs were dropped 24 ft onto a concrete surface, and they bounced an average of 92.84 in. In a test of a sample of 40 new balls, they bounced heights had a mean of 92.67 in. and standard deviation of 1.79 in. Use a 0.05 significance level to determine whether there is sufficient evidence to support the claim that the new balls have bounce heights with a mean different from 92.84 in. Does it appear that the new baseballs are different. 92.84 x 92.67 s 1.79 n 40 0.05 Ho : 92.84 H1 : 92.84 Page 416, Problem #20 92.84 x 92.67 s 1.79 n 40 0.05 Fail to reject Ho since P-value > α, there is not sufficient evidence to conclude that μ ≠ 92.84. No, it does not appear that new baseballs are significantly different in this respect. Ho : 92.84 H1 : 92.84 P value 0.552 Example – Page 418, Problem #26 Patients with chronic fatigue syndrome were tested, than retested after being treated with fludrocortisone. Listed below are the changes in fatigue after the treatment. A standard scale from -7 to 7 was used, with positive values representing improvements. Use a 0.01 significance level to test the claim that the mean change is positive. Does the treatment appear to be effective? 6 5 0 5 6 7 3 3 2 6 5 5 0 6 3 4 3 7 0 4 4 Example – Page 418, Problem #26 Step 1 – Identify the population of interest and the parameter you want to draw conclusions about. µ = improvements of chronic fatigue syndrome after being treated with fludrocortisone Ho: μ = 0 H1: > 0 Example – Page 418, Problem #26 Step 2 – Choose the appropriate inference procedure. Verify conditions for using the selected procedure. Use a one-sample t-test Not sure if its Random Sample (SRS). Assume its representative of the population The population standard deviation σ is not known Sampling distribution is approximately normal since the normal probability plot show as linear trend with no outliers Example – Page 418, Problem #26 Step 3 – Carry out the inference procedure 40 x x 8.455 t 2.168 s n 21 p value 2.45E 8 0.0001 0.01 2.53 Example – Page 418, Problem #26 Step 4 – Interpret your results in the context of the problem There is sufficient evidence to reject Ho since (p-value = 0.0001 α = 0.01) and conclude that there was improvements for patients with chronic fatigue syndrome that was treated with fludrocortisone. Lesson 7-6 Testing a Claim about a Standard Deviation or Variance Main Idea of Standard Deviation and Variance The main idea is that the standard deviation and the variance are measures of consistency. The less consistent the values of a variable, the higher the standard deviation of the variable Assumptions for Testing Claims about σ and σ² The sample is simple random The population has a normal distribution. Test Statistic for Testing Claim about σ and σ² 2 (n 1)s 2 2 Characteristics of the Chi-Square Distribution It is not symmetric The shape depends upon the degrees of freedom. As the number of degrees of freedom increases, the distribution becomes more symmetric. 2 The values of are greater than or equal to 0. Chi-Square Distribution for 10 and 20 Degrees of Freedom There is a different distribution for each number of degrees of freedom. Example – Page 423, #2 Find the critical values, test statistic, P-value, and state the final conclusion. H1 : 12, 0.01, n 5, s 18 Find the critical value or critical region: Use table A-4 DF = 4 and 0.01 2 4,0.01 13.277 Example – Page 423, #2 H1 : 12, 0.01, n 5, s 18 Find the test statistic: 2 (n 1)s 2 2 (5 1)(18)2 12 2 9.00 Example – Page 423, #2 H1 : 12, 0.01, n 5, s 18 Find the P-value: Use Table A-4 df 4 and 9.00 0.05 p value 0.10 Use TI-83 2cdf (9, E 99,4) 0.0611 Example – Page 423, #2 Critical region = 13.277 Test statistic = 9.00 P-value = 0.0611 13.277 4 9.00 0.01 χ 42 Fail to reject Ho since P-value > α, and conclude that σ = 12. Example – Page 424, #4 Find the critical values, test statistic, P-value, and state the final conclusion. H1 : 4, 0.05, n 81, s 4.7 Find the critical value or critical region: Use table A-4 DF = 80 and = 0.05 χ χ 2 R 2 .025 106.629 χ χ 2 L 2 .975 57.153 Example – Page 424, #4 H1 : 4, 0.05, n 81, s 4.7 Find the test statistic: 2 (n 1)s 2 2 (81 1)(4.7)2 4 2 110.45 Find the P-value: 2cdf (110.45, E 99,80) 0.0137 Example – Page 424, #4 Critical region = 57.153 and 106.629 Test statistic = 110.450 P-value = 0.0137 57.153 80 106.629 110.450 0.025 0.025 χ802 Reject Ho since P-value < , and conclude that σ ≠ 4 (in fact , that σ > 4) . Example – Page 424, #8 Tests in the author’s past statistics classes have scores with a standard deviation equal to 14.1. One of his recent classes has 27 test scores with a standard deviation of 9.3. Use 0.01 significance level to test the claim that this current class has less variation than past classes. Does lower standard deviation suggest that current class is doing better? Identify the population of interest and parameter you want to Draw conclusions about. State the hypothesis in words and symbols Ho : 14.1 H1 : 14.1 σ = standard deviation of class test scores Example – Page 424, #8 Choose the appropriate inference procedure. Verify the Conditions for using the selected procedure. Use a σ test Conditions Simple Random Sample Population is normally distributed Example – Page 424, #8 If the conditions are met, carry out the inference procedure n 27 2 2 ( n 1)s (27 1)(9.3) 2 11.311 14.1 2 2 14.1 s 9.3 p value 2cdf ( E 99,11.31,26) 0.0056 0.01 Example – Page 424, #8 Interpret your results in the context of the problem Reject Ho since P-value ≥ α, there and conclude that the standard deviation of test scores is less than 14.1. No, a lower standard deviation does not suggest that the current class is doing better, but they are more similar to each other and show less variation. Standard deviation speaks only about the spread of the scores not their location.