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Chapter10

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Chapter 10

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Chapter10

  1. 1. 1 Chapter 10 Hypothesis Tests Using a Single Sample
  2. 2. 2 BASICS In statistics, a hypothesis is a statement about a population characteristic.
  3. 3. 3 FORMAL STRUCTURE Hypothesis Tests are based on an reductio ad absurdum form of argument. Specifically, we make an assumption and then attempt to show that assumption leads to an absurdity or contradiction, hence the assumption is wrong.
  4. 4. 4 FORMAL STRUCTURE The null hypothesis, denoted H0 is a statement or claim about a population characteristic that is initially assumed to be true. The null hypothesis is so named because it is the “starting point” for the investigation. The phrase “there is no difference” is often used in its interpretation.
  5. 5. 5 FORMAL STRUCTURE The alternate hypothesis, denoted by Ha is the competing claim. The alternate hypothesis is a statement about the same population characteristic that is used in the null hypothesis. Generally, the alternate hypothesis is a statement that specifies that the population has a value different, in some way, from the value given in the null hypothesis.
  6. 6. 6 FORMAL STRUCTURE Rejection of the null hypothesis will imply the acceptance of this alternative hypothesis. Assume H0 is true and attempt to show this leads to an absurdity, hence H0 is false and Ha is true.
  7. 7. 7 FORMAL STRUCTURE Typically one assumes the null hypothesis to be true and then one of the following conclusions are drawn. 1. Reject H0 Equivalent to saying that Ha is correct or true 2. Fail to reject H0 Equivalent to saying that we have failed to show a statistically significant deviation from the claim of the null hypothesis This is not the same as saying that the null hypothesis is true.
  8. 8. 8 AN ANALOGY The Statistical Hypothesis Testing process can be compared very closely with a judicial trial. 1.Assume a defendant is innocent (H0) 2.Present evidence to show guilt 3.Try to prove guilt beyond a reasonable doubt (Ha)
  9. 9. 9 AN ANALOGY Two Hypotheses are then created. H0: Innocent Ha: Not Innocent (Guilt)
  10. 10. 10 Examples of Hypotheses You would like to determine if the diameters of the ball bearings you produce have a mean of 6.5 cm. H0: µ = 6.5 Ha: µ ≠ 6.5 (Two-sided alternative)
  11. 11. 11 The students entering into the math program used to have a mean SAT quantitative score of 525. Are the current students poorer (as measured by the SAT quantitative score)? H0: µ = 525 (Really: µ ≥ 525) Ha: µ < 525 (One-sided alternative) Examples of Hypotheses
  12. 12. 12 Do the “16 ounce” cans of peaches canned and sold by DelMonte meet the claim on the label (on the average)? H0: µ = 16 (Really: µ ≥16) Ha: µ < 16 Examples of Hypotheses Notice, the real concern would be selling the consumer less than 16 ounces of peaches.
  13. 13. 13 Is the proportion of defective parts produced by a manufacturing process more than 5%? H0: π = 0.05 (Really, π ≤ 0.05) Ha: π > 0.05 Examples of Hypotheses
  14. 14. 14 Do two brands of light bulb have the same mean lifetime? H0: µBrand A = µBrand B Ha: µBrand A ≠ µBrand B Examples of Hypotheses
  15. 15. 15 Do parts produced by two different milling machines have the same variability in diameters? or equivalently 0 1 2 a 1 2 H : H : σ = σ σ ≠ σ 2 2 0 1 2 2 2 a 1 2 H : H : σ = σ σ ≠ σ Examples of Hypotheses
  16. 16. 16 Comments on Hypothesis Form The null hypothesis must contain the equal sign. This is absolutely necessary because the test requires the null hypothesis to be assumed to be true and the value attached to the equal sign is then the value assumed to be true and used in subsequent calculations. The alternate hypothesis should be what you are really attempting to show to be true. This is not always possible.
  17. 17. 17 Hypothesis Form The form of the null hypothesis is H0: population characteristic = hypothesized value where the hypothesized value is a specific number determined by the problem context. The alternative (or alternate) hypothesis will have one of the following three forms: Ha: population characteristic > hypothesized value Ha: population characteristic < hypothesized value Ha: population characteristic ≠ hypothesized value
  18. 18. 18 Caution When you set up a hypothesis test, the result is either 1. Strong support for the alternate hypothesis (if the null hypothesis is rejected) 2. There is not sufficient evidence to refute the claim of the null hypothesis (you are stuck with it, because there is a lack of strong evidence against the null hypothesis.
  19. 19. 19 Error Null Hypothesis Decision True False Accept H0 Reject H0 No Error No Error Type I Error Type II Error α β
  20. 20. 20 Error Analogy Consider a medical test where the hypotheses are equivalent to H0: the patient has a specific disease Ha: the patient doesn’t have the disease Then, Type I error is equivalent to a false negative (i.e., Saying the patient does not have the disease when in fact, he does.) Type II error is equivalent to a false positive (i.e., Saying the patient has the disease when, in fact, he does not.)
  21. 21. 21 More on Error The probability of a type I error is denoted by α and is called the level of significance of the test. Thus, a test with α = 0.01 is said to have a level of significance of 0.01 or to be a level 0.01 test. The probability of a type II error is denoted by β.
  22. 22. 22 Relationships Between α and β Generally, with everything else held constant, decreasing one type of error causes the other to increase. The only way to decrease both types of error simultaneously is to increase the sample size. No matter what decision is reached, there is always the risk of one of these errors.
  23. 23. 23 Comment of Process Look at the consequences of type I and type II errors and then identify the largest α that is tolerable for the problem. Employ a test procedure that uses this maximum acceptable value of α (rather than anything smaller) as the level of significance (because using a smaller α increases β).
  24. 24. 24 Test Statistic A test statistic is the function of sample data on which a conclusion to reject or fail to reject H0 is based.
  25. 25. 25 P-value The P-value (also called the observed significance level) is a measure of inconsistency between the hypothesized value for a population characteristic and the observed sample. The P-value is the probability, assuming that H0 is true, of obtaining a test statistic value at least as inconsistent with H0 as what actually resulted.
  26. 26. 26 Decision Criteria A decision as to whether H0 should be rejected results from comparing the P-value to the chosen α: H0 should be rejected if P-value ≤ α. H0 should not be rejected if P-value > α.
  27. 27. 27 Large Sample Hypothesis Test for a Single Proportion In terms of a standard normal random variable z, the approximate P-value for this test depends on the alternate hypothesis and is given for each of the possible alternate hypotheses on the next 3 slides. To test the hypothesis H0: π = hypothesized proportion, compute the z statistic p hypothesized value z hypothesized value(1-hypothesized value) n − =
  28. 28. 28 Hypothesis Test Large Sample Test of Population Proportion p hypothesized value P-value P z hypothesized value(1-hypothesized value) n    −  = <      
  29. 29. 29 p hypothesized value P-value P z hypothesized value(1-hypothesized value) n    −  = >       Hypothesis Test Large Sample Test of Population Proportion
  30. 30. 30 p hypothesized value P-value 2P z hypothesized value(1-hypothesized value) n    −  = >       Hypothesis Test Large Sample Test of Population Proportion
  31. 31. 31 An insurance company states that the proportion of its claims that are settled within 30 days is 0.9. A consumer group thinks that the company drags its feet and takes longer to settle claims. To check these hypotheses, a simple random sample of 200 of the company’s claims was obtained and it was found that 160 of the claims were settled within 30 days. Hypothesis Test Example Large-Sample Test for a Population Proportion
  32. 32. 32 P-value P(z 4.71) 0= < − ≈ 0.8 0.9 0.8 0.9 z 4.71 (0.9)(1 0.9) 0.9(0.1) 200 200 − − = = = − − Hypothesis Test Example 2 Single Proportion continued π = proportion of the company’s claims that are settled within 30 days H0: π = 0.9 HA: π < 0.9 160 p 0.8 200 = =The sample proportion is
  33. 33. 33 Hypothesis Test Example 2 Single Proportion continued The probability of getting a result as strongly or more strongly in favor of the consumer group's claim (the alternate hypothesis Ha) if the company’s claim (H0) was true is essentially 0. Clearly, this gives strong evidence in support of the alternate hypothesis (against the null hypothesis).
  34. 34. 34 Hypothesis Test Example 2 Single Proportion continued We would say that we have strong support for the claim that the proportion of the insurance company’s claims that are settled within 30 days is less than 0.9. Some people would state that we have shown that the true proportion of the insurance company’s claims that are settled within 30 days is statistically significantly less than 0.9.
  35. 35. 35 A county judge has agreed that he will give up his county judgeship and run for a state judgeship unless there is evidence at the 0.10 level that more then 25% of his party is in opposition. A SRS of 800 party members included 217 who opposed him. Please advise this judge. Hypothesis Test Example Single Proportion
  36. 36. 36 Hypothesis Test Example Single Proportion continued π = proportion of his party that is in opposition H0: π = 0.25 HA: π > 0.25 α = 0.10 Note: hypothesized value = 0.25 217 n 800, p 0.27125 800 = = = 0.27125 0.25 z 1.39 0.25(0.75) 800 − = =
  37. 37. 37 Hypothesis Test Example Single Proportion continued At a level of significance of 0.10, there is sufficient evidence to support the claim that the true percentage of the party members that oppose him is more than 25%. Under these circumstances, I would advise him not to run. P-value=P(z 1.39) 1 0.9177 0.0823> = − =
  38. 38. 38 1. Describe (determine) the population characteristic about which hypotheses are to be tested. 2. State the null hypothesis H0. 3. State the alternate hypothesis Ha. 4. Select the significance level α for the test. 5. Display the test statistic to be used, with substitution of the hypothesized value identified in step 2 but without any computation at this point. Steps in a Hypothesis-Testing Analysis
  39. 39. 39 Steps in a Hypothesis-Testing Analysis 6. Check to make sure that any assumptions required for the test are reasonable. 7. Compute all quantities appearing in the test statistic and then the value of the test statistic itself. 8. Determine the P-value associated with the observed value of the test statistic 9. State the conclusion in the context of the problem, including the level of significance.
  40. 40. 40 x hypothesized mean z n − = σ Hypothesis Test (Large samples) Single Sample Test of Population Mean In terms of a standard normal random variable z, the approximate P-value for this test depends on the alternate hypothesis and is given for each of the possible alternate hypotheses on the next 3 slides. To test the hypothesis H0: µ = hypothesized mean, compute the z statistic
  41. 41. 41 x hypothesized mean P-value P Z n   − = <  σ    Hypothesis Test Single Sample Test of Population Mean H0: µ = hypothesized mean HA: µ < hypothesized mean
  42. 42. 42 x hypothesized mean P-value P Z n   − = >  σ    Hypothesis Test Single Sample Test of Population Mean H0: µ = hypothesized mean HA: µ > hypothesized mean
  43. 43. 43 x hypothesized mean P-value 2P Z n   −  = > σ     Hypothesis Test Single Sample Test of Population Mean H0: µ = hypothesized mean HA: µ ≠ hypothesized mean
  44. 44. 44 It is not likely that one would know σ but not know µ, so calculating a z value using the formula would not be very realistic. x hypothesized mean z n − = σ Reality Check For large values of n (>30) it is generally acceptable to use s to estimate σ, however, it is much more common to apply the t-distribution.
  45. 45. 45 x hypothesized mean t s n − = Hypothesis Test (σ unknown) Single Sample Test of Population Mean The approximate P-value for this test is found using a t random variable with degrees of freedom df = n-1. The procedure is described in the next group of slides. To test the null hypothesis µ = hypothesized mean, when we may assume that the underlying distribution is normal or approximately normal, compute the t statistic
  46. 46. 46 x hypothesized mean P-value P t s n   − = <       Hypothesis Test Single Sample Test of Population Mean H0: µ = hypothesized mean HA: µ < hypothesized mean
  47. 47. 47 x hypothesized mean P-value P t s n   − = >       Hypothesis Test Single Sample Test of Population Mean H0: µ = hypothesized mean HA: µ > hypothesized mean
  48. 48. 48 x hypothesized mean P-value 2P t s n   −  = >      Hypothesis Test Single Sample Test of Population Mean H0: µ = hypothesized mean HA: µ ≠ hypothesized mean
  49. 49. 49 The t statistic can be used for all sample sizes, however, the smaller the sample, the more important the assumption that the underlying distribution is normal. Typically, when n >15 the underlying distribution need only be centrally weighted and may be somewhat skewed. Hypothesis Test (σ unknown) Single Sample Test of Population Mean
  50. 50. 50 Tail areas for t curves t df 1 2 3 4 5 6 7 8 9 10 11 12 0.0 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.1 0.468 0.465 0.463 0.463 0.462 0.462 0.462 0.461 0.461 0.461 0.461 0.461 0.2 0.437 0.430 0.427 0.426 0.425 0.424 0.424 0.423 0.423 0.423 0.423 0.422 0.3 0.407 0.396 0.392 0.390 0.388 0.387 0.386 0.386 0.385 0.385 0.385 0.385 0.4 0.379 0.364 0.358 0.355 0.353 0.352 0.351 0.350 0.349 0.349 0.348 0.348 0.5 0.352 0.333 0.326 0.322 0.319 0.317 0.316 0.315 0.315 0.314 0.313 0.313 0.6 0.328 0.305 0.295 0.290 0.287 0.285 0.284 0.283 0.282 0.281 0.280 0.280 0.7 0.306 0.278 0.267 0.261 0.258 0.255 0.253 0.252 0.251 0.250 0.249 0.249 0.8 0.285 0.254 0.241 0.234 0.230 0.227 0.225 0.223 0.222 0.221 0.220 0.220 0.9 0.267 0.232 0.217 0.210 0.205 0.201 0.199 0.197 0.196 0.195 0.194 0.193 1.0 0.250 0.211 0.196 0.187 0.182 0.178 0.175 0.173 0.172 0.170 0.169 0.169 1.1 0.235 0.193 0.176 0.167 0.161 0.157 0.154 0.152 0.150 0.149 0.147 0.146 1.2 0.221 0.177 0.158 0.148 0.142 0.138 0.135 0.132 0.130 0.129 0.128 0.127 1.3 0.209 0.162 0.142 0.132 0.125 0.121 0.117 0.115 0.113 0.111 0.110 0.109 1.4 0.197 0.148 0.128 0.117 0.110 0.106 0.102 0.100 0.098 0.096 0.095 0.093 1.5 0.187 0.136 0.115 0.104 0.097 0.092 0.089 0.086 0.084 0.082 0.081 0.080 1.6 0.178 0.125 0.104 0.092 0.085 0.080 0.077 0.074 0.072 0.070 0.069 0.068 1.7 0.169 0.116 0.094 0.082 0.075 0.070 0.066 0.064 0.062 0.060 0.059 0.057 1.8 0.161 0.107 0.085 0.073 0.066 0.061 0.057 0.055 0.053 0.051 0.050 0.049 1.9 0.154 0.099 0.077 0.065 0.058 0.053 0.050 0.047 0.045 0.043 0.042 0.041 2.0 0.148 0.092 0.070 0.058 0.051 0.046 0.043 0.040 0.038 0.037 0.035 0.034 2.1 0.141 0.085 0.063 0.052 0.045 0.040 0.037 0.034 0.033 0.031 0.030 0.029 2.2 0.136 0.079 0.058 0.046 0.040 0.035 0.032 0.029 0.028 0.026 0.025 0.024 2.3 0.131 0.074 0.052 0.041 0.035 0.031 0.027 0.025 0.023 0.022 0.021 0.020 2.4 0.126 0.069 0.048 0.037 0.031 0.027 0.024 0.022 0.020 0.019 0.018 0.017 2.5 0.121 0.065 0.044 0.033 0.027 0.023 0.020 0.018 0.017 0.016 0.015 0.014 2.6 0.117 0.061 0.040 0.030 0.024 0.020 0.018 0.016 0.014 0.013 0.012 0.012 2.7 0.113 0.057 0.037 0.027 0.021 0.018 0.015 0.014 0.012 0.011 0.010 0.010 2.8 0.109 0.054 0.034 0.024 0.019 0.016 0.013 0.012 0.010 0.009 0.009 0.008 2.9 0.106 0.051 0.031 0.022 0.017 0.014 0.011 0.010 0.009 0.008 0.007 0.007 3.0 0.102 0.048 0.029 0.020 0.015 0.012 0.010 0.009 0.007 0.007 0.006 0.006 3.1 0.099 0.045 0.027 0.018 0.013 0.011 0.009 0.007 0.006 0.006 0.005 0.005 3.2 0.096 0.043 0.025 0.016 0.012 0.009 0.008 0.006 0.005 0.005 0.004 0.004 3.3 0.094 0.040 0.023 0.015 0.011 0.008 0.007 0.005 0.005 0.004 0.004 0.003 3.4 0.091 0.038 0.021 0.014 0.010 0.007 0.006 0.005 0.004 0.003 0.003 0.003 3.5 0.089 0.036 0.020 0.012 0.009 0.006 0.005 0.004 0.003 0.003 0.002 0.002 3.6 0.086 0.035 0.018 0.011 0.008 0.006 0.004 0.003 0.003 0.002 0.002 0.002 3.7 0.084 0.033 0.017 0.010 0.007 0.005 0.004 0.003 0.002 0.002 0.002 0.002 3.8 0.082 0.031 0.016 0.010 0.006 0.004 0.003 0.003 0.002 0.002 0.001 0.001 3.9 0.080 0.030 0.015 0.009 0.006 0.004 0.003 0.002 0.002 0.001 0.001 0.001 4.0 0.078 0.029 0.014 0.008 0.005 0.004 0.003 0.002 0.002 0.001 0.001 0.001
  51. 51. 51 t df 25 26 27 28 29 30 35 40 60 120 ∞(=z) 0.0 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.1 0.461 0.461 0.461 0.461 0.461 0.461 0.460 0.460 0.460 0.460 0.460 0.2 0.422 0.422 0.421 0.421 0.421 0.421 0.421 0.421 0.421 0.421 0.421 0.3 0.383 0.383 0.383 0.383 0.383 0.383 0.383 0.383 0.383 0.382 0.382 0.4 0.346 0.346 0.346 0.346 0.346 0.346 0.346 0.346 0.345 0.345 0.345 0.5 0.311 0.311 0.311 0.310 0.310 0.310 0.310 0.310 0.309 0.309 0.309 0.6 0.277 0.277 0.277 0.277 0.277 0.277 0.276 0.276 0.275 0.275 0.274 0.7 0.245 0.245 0.245 0.245 0.245 0.245 0.244 0.244 0.243 0.243 0.242 0.8 0.216 0.215 0.215 0.215 0.215 0.215 0.215 0.214 0.213 0.213 0.212 0.9 0.188 0.188 0.188 0.188 0.188 0.188 0.187 0.187 0.186 0.185 0.184 1.0 0.163 0.163 0.163 0.163 0.163 0.163 0.162 0.162 0.161 0.160 0.159 1.1 0.141 0.141 0.141 0.140 0.140 0.140 0.139 0.139 0.138 0.137 0.136 1.2 0.121 0.120 0.120 0.120 0.120 0.120 0.119 0.119 0.117 0.116 0.115 1.3 0.103 0.103 0.102 0.102 0.102 0.102 0.101 0.101 0.099 0.098 0.097 1.4 0.087 0.087 0.086 0.086 0.086 0.086 0.085 0.085 0.083 0.082 0.081 1.5 0.073 0.073 0.073 0.072 0.072 0.072 0.071 0.071 0.069 0.068 0.067 1.6 0.061 0.061 0.061 0.060 0.060 0.060 0.059 0.059 0.057 0.056 0.055 1.7 0.051 0.051 0.050 0.050 0.050 0.050 0.049 0.048 0.047 0.046 0.045 1.8 0.042 0.042 0.042 0.041 0.041 0.041 0.040 0.040 0.038 0.037 0.036 1.9 0.035 0.034 0.034 0.034 0.034 0.034 0.033 0.032 0.031 0.030 0.029 2.0 0.028 0.028 0.028 0.028 0.027 0.027 0.027 0.026 0.025 0.024 0.023 2.1 0.023 0.023 0.023 0.022 0.022 0.022 0.022 0.021 0.020 0.019 0.018 2.2 0.019 0.018 0.018 0.018 0.018 0.018 0.017 0.017 0.016 0.015 0.014 2.3 0.015 0.015 0.015 0.015 0.014 0.014 0.014 0.013 0.012 0.012 0.011 2.4 0.012 0.012 0.012 0.012 0.012 0.011 0.011 0.011 0.010 0.009 0.008 2.5 0.010 0.010 0.009 0.009 0.009 0.009 0.009 0.008 0.008 0.007 0.006 2.6 0.008 0.008 0.007 0.007 0.007 0.007 0.007 0.006 0.006 0.005 0.005 2.7 0.006 0.006 0.006 0.006 0.006 0.006 0.005 0.005 0.004 0.004 0.003 2.8 0.005 0.005 0.005 0.005 0.004 0.004 0.004 0.004 0.003 0.003 0.003 2.9 0.004 0.004 0.004 0.004 0.004 0.003 0.003 0.003 0.003 0.002 0.002 3.0 0.003 0.003 0.003 0.003 0.003 0.003 0.002 0.002 0.002 0.002 0.001 3.1 0.002 0.002 0.002 0.002 0.002 0.002 0.002 0.002 0.001 0.001 0.001 3.2 0.002 0.002 0.002 0.002 0.002 0.002 0.001 0.001 0.001 0.001 0.001 3.3 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.000 3.4 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.000 0.000 3.5 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.000 0.000 0.000 3.6 0.001 0.001 0.001 0.001 0.001 0.001 0.000 0.000 0.000 0.000 0.000 3.7 0.001 0.001 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 3.8 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 3.9 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 4.0 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 Tail areas for t curves
  52. 52. 52 An manufacturer of a special bolt requires that this type of bolt have a mean shearing strength in excess of 110 lb. To determine if the manufacturer’s bolts meet the required standards a sample of 25 bolts was obtained and tested. The sample mean was 112.7 lb and the sample standard deviation was 9.62 lb. Use this information to perform an appropriate hypothesis test with a significance level of 0.05. Example of Hypothesis Test Single Sample Test of Population Mean - continued
  53. 53. 53 µ = the mean shearing strength of this specific type of bolt Τhe hypotheses to be tested are H0: µ = 110 lb Ha: µ > 110 lb The significance level to be used for the test is α = 0.05. Example of Hypothesis Test Single Sample Test of Population Mean - continued x 110 t s n − =The test statistic is
  54. 54. 54 Example of Hypothesis Test Single Sample Test of Population Mean - continued x 112.7, s 9.62, n 25, df 24= = = = 112.7 110 P-value P t 9.62 25 P(t 1.4) 0.087   − = >       = > =
  55. 55. 55 Because P-value = 0.087 > 0.05 = α, we fail to reject H0. At a level of significance of 0.05, there is insufficient evidence to conclude that the mean shearing strength of this brand of bolt exceeds 110 lbs. Example of Hypothesis Test Single Sample Test of Population Mean - conclusion
  56. 56. 56 Using the t table t df 13 14 15 16 17 18 19 20 21 22 23 24 0.0 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.1 0.461 0.461 0.461 0.461 0.461 0.461 0.461 0.461 0.461 0.461 0.461 0.461 0.2 0.422 0.422 0.422 0.422 0.422 0.422 0.422 0.422 0.422 0.422 0.422 0.422 0.3 0.384 0.384 0.384 0.384 0.384 0.384 0.384 0.384 0.384 0.383 0.383 0.383 0.4 0.348 0.348 0.347 0.347 0.347 0.347 0.347 0.347 0.347 0.347 0.346 0.346 0.5 0.313 0.312 0.312 0.312 0.312 0.312 0.311 0.311 0.311 0.311 0.311 0.311 0.6 0.279 0.279 0.279 0.278 0.278 0.278 0.278 0.278 0.277 0.277 0.277 0.277 0.7 0.248 0.248 0.247 0.247 0.247 0.246 0.246 0.246 0.246 0.246 0.245 0.245 0.8 0.219 0.219 0.218 0.218 0.217 0.217 0.217 0.217 0.216 0.216 0.216 0.216 0.9 0.192 0.192 0.191 0.191 0.190 0.190 0.190 0.189 0.189 0.189 0.189 0.189 1.0 0.168 0.167 0.167 0.166 0.166 0.165 0.165 0.165 0.164 0.164 0.164 0.164 1.1 0.146 0.145 0.144 0.144 0.143 0.143 0.143 0.142 0.142 0.142 0.141 0.141 1.2 0.126 0.125 0.124 0.124 0.123 0.123 0.122 0.122 0.122 0.121 0.121 0.121 1.3 0.108 0.107 0.107 0.106 0.105 0.105 0.105 0.104 0.104 0.104 0.103 0.103 1.4 0.092 0.092 0.091 0.090 0.090 0.089 0.089 0.088 0.088 0.088 0.087 0.087 1.5 0.079 0.078 0.077 0.077 0.076 0.075 0.075 0.075 0.074 0.074 0.074 0.073 1.6 0.067 0.066 0.065 0.065 0.064 0.064 0.063 0.063 0.062 0.062 0.062 0.061 1.7 0.056 0.056 0.055 0.054 0.054 0.053 0.053 0.052 0.052 0.052 0.051 0.051 1.8 0.048 0.047 0.046 0.045 0.045 0.044 0.044 0.043 0.043 0.043 0.042 0.042 1.9 0.040 0.039 0.038 0.038 0.037 0.037 0.036 0.036 0.036 0.035 0.035 0.035 2.0 0.033 0.033 0.032 0.031 0.031 0.030 0.030 0.030 0.029 0.029 0.029 0.028 2.1 0.028 0.027 0.027 0.026 0.025 0.025 0.025 0.024 0.024 0.024 0.023 0.023 2.2 0.023 0.023 0.022 0.021 0.021 0.021 0.020 0.020 0.020 0.019 0.019 0.019 2.3 0.019 0.019 0.018 0.018 0.017 0.017 0.016 0.016 0.016 0.016 0.015 0.015 2.4 0.016 0.015 0.015 0.014 0.014 0.014 0.013 0.013 0.013 0.013 0.012 0.012 2.5 0.013 0.013 0.012 0.012 0.011 0.011 0.011 0.011 0.010 0.010 0.010 0.010 2.6 0.011 0.010 0.010 0.010 0.009 0.009 0.009 0.009 0.008 0.008 0.008 0.008 2.7 0.009 0.009 0.008 0.008 0.008 0.007 0.007 0.007 0.007 0.007 0.006 0.006 2.8 0.008 0.007 0.007 0.006 0.006 0.006 0.006 0.006 0.005 0.005 0.005 0.005 2.9 0.006 0.006 0.005 0.005 0.005 0.005 0.005 0.004 0.004 0.004 0.004 0.004 3.0 0.005 0.005 0.004 0.004 0.004 0.004 0.004 0.004 0.003 0.003 0.003 0.003 3.1 0.004 0.004 0.004 0.003 0.003 0.003 0.003 0.003 0.003 0.003 0.003 0.002 3.2 0.003 0.003 0.003 0.003 0.003 0.002 0.002 0.002 0.002 0.002 0.002 0.002 3.3 0.003 0.003 0.002 0.002 0.002 0.002 0.002 0.002 0.002 0.002 0.002 0.002 3.4 0.002 0.002 0.002 0.002 0.002 0.002 0.002 0.001 0.001 0.001 0.001 0.001 3.5 0.002 0.002 0.002 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 3.6 0.002 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 3.7 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 3.8 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.000 0.000 0.000 3.9 0.001 0.001 0.001 0.001 0.001 0.001 0.000 0.000 0.000 0.000 0.000 0.000 4.0 0.001 0.001 0.001 0.001 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 t = 1.4 n = 25 df = 24 Tail area = 0.087
  57. 57. 57 Revisit the problem with α=0.10 What would happen if the significance level of the test was 0.10 instead of 0.05? At the 0.10 level of significance there is sufficient evidence to conclude that the mean shearing strength of this brand of bolt exceeds 120 lbs. Now P-value = 0.087 < 0.10 = α, and we reject H0 at the 0.10 level of significance and conclude
  58. 58. 58 Comments continued Many people are bothered by the fact that different choices of α lead to different conclusions. This is nature of a process where you control the probability of being wrong when you select the level of significance. This reflects your willingness to accept a certain level of type I error.
  59. 59. 59 Another Example A jeweler is planning on manufacturing gold charms. His design calls for a particular piece to contain 0.08 ounces of gold. The jeweler would like to know if the pieces that he makes contain (on the average) 0.08 ounces of gold. To test to see if the pieces contain 0.08 ounces of gold, he made a sample of 16 of these particular pieces and obtained the following data. 0.0773 0.0779 0.0756 0.0792 0.0777 0.0713 0.0818 0.0802 0.0802 0.0785 0.0764 0.0806 0.0786 0.0776 0.0793 0.0755 Use a level of significance of 0.01 to perform an appropriate hypothesis test.
  60. 60. 60 Another Example 1. The population characteristic being studied is µ = true mean gold content for this particular type of charm. 2. Null hypothesis: H0:µ = 0.08 oz 3. Alternate hypothesis: Ha:µ ≠ 0.08 oz 4. Significance level: α = 0.01 5. Test statistic: x hypothesized mean x 0.08 t s s n n − − = =
  61. 61. 61 Another Example 6. Minitab was used to create a normal plot along with a graphical display of the descriptive statistics for the sample data. The result of this display is that it is reasonable to assume that the population of gold contents of this type of charm is normally distributed P-Value: 0.396 A-Squared: 0.363 Anderson-Darling Normality Test N: 16 StDev: 0.0025143 Average: 0.0779813 0.0820.0770.072 .999 .99 .95 .80 .50 .20 .05 .01 .001 Probability Gold Normal Probability Plot
  62. 62. 62 Another Example We can see that with the exception of one outlier, the data is reasonably symmetric and mound shaped in shape, indicating that the assumption that the population of amounts of gold for this particular charm can reasonably be expected to be normally distributed. 0.0820.0800.0780.0760.0740.072 95% Confidence Interval for Mu 0.07950.07850.07750.0765 95% Confidence Interval for Median Variable: Gold 7.71E-02 1.86E-03 7.66E-02 Maximum 3rd Quartile Median 1st Quartile Minimum N Kurtosis Skewness Variance StDev Mean P-Value: A-Squared: 7.95E-02 3.89E-03 7.93E-02 8.18E-02 8.00E-02 7.82E-02 7.66E-02 7.13E-02 16 2.23191 -1.10922 6.32E-06 2.51E-03 7.80E-02 0.396 0.363 95% Confidence Interval for Median 95% Confidence Interval for Sigma 95% Confidence Interval for Mu Anderson-Darling Normality Test Descriptive Statistics
  63. 63. 63 Another Example 8. P-value: This is a two tailed test. Looking up in the table of tail areas for t curves, t = 3.2 with df = 15 we see the table entry is 0.003 so P-Value = 2(0.003) = 0.006 n 16, x 0.077981, s 0.0025143 0.077981 0.08 t 3.2 0.0025143 16 = = = − = = − 7. Computations: n 16, x 0.077981, s 0.0025143 0.077981 0.08 t 3.2 0.0025143 16 = = = − = = − 7. Computations:
  64. 64. 64 Another Example 9. Conclusion: Since P-value = 0.006 ≤ 0.01 = α, we reject H0 at the 0.01 level of significance. At the 0.01 level of significance there is convincing evidence that the true mean gold content of this type of charm is not 0.08 ounces. Actually when rejecting a null hypothesis for the ≠ alternative, a one tailed claim is supported. In this case, at the 0.01 level of significance, there is convincing evidence that the true mean gold content of this type of charm is less than 0.08 ounces.
  65. 65. 65 Power and Probability of Type II Error The power of a test is the probability of rejecting the null hypothesis. When H0 is false, the power is the probability that the null hypothesis is rejected. Specifically, power = 1 – β.
  66. 66. 66 Effects of Various Factors on Power 1. The larger the size of the discrepancy between the hypothesized value and the true value of the population characteristic, the higher the power. 2. The larger the significance level, α, the higher the power of the test. 3. The larger the sample size, the higher the power of the test.
  67. 67. 67 Some Comments Calculating β (hence power) depends on knowing the true value of the population characteristic being tested. Since the true value is not known, generally, one calculates β for a number of possible “true” values of the characteristic under study and then sketches a power curve.
  68. 68. 68 Example (based on z-curve) Consider the earlier example where we tested H0: µ = 110 vs. Ha: µ > 110 and furthermore, suppose the true standard deviation of the bolts was actually 10 lbs.
  69. 69. 69 Example (based on z-curve) Power Curves Different α's 0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00 118 119 120 121 122 123 124 125 126 127 128 True Value of µ Power(1-β) α = 0.10 α = 0.05 α = 0.01 H0: µ = 120 Ha: µ > 120 σ = 10
  70. 70. 70 Example (based on z-curve) Power Curves Different n's 0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00 118 119 120 121 122 123 124 125 126 127 128 True Value of µ Power(1-β) n = 45 n = 90 n = 180 n = 360 H0: µ = 120 Ha: µ > 120 σ = 10

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