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Hypothesis Testing Introduction Always about a population parameter Attempt to prove (or disprove) some assumption Setup: alternate hypothesis: What you wish to prove Example: Person is guilty of crime null hypothesis: Assume the opposite of what is to be proven. The null is always stated as an equality. Example: Person is innocent The test 1. Take a sample, compute statistic of interest. The evidence gathered against defendent 2. How likely is it that if the null were true, you would get such a statistic? (the p-value) How likely is it that an innocent person would be found at the scene of crime, with gun in hand, etc. 3. If very unlikely, then null must be false, hence alternate is proven beyond reasonable doubt. 4. If quite likely, then null may be true, so not enough evidence to discard it in favor of the alternate. Types of Errors Null is really Null is really True False reject null, Type I Error Good Decision assume alternate is (convict the proven innocent) do not reject null, Good Decision Type II Error evidence for alternate (let guilty go free) not strong enough Hypothesis Testing Roadmap Hypothesis Testing Continuous Attribute Normal, Non-Normal, c2 Contingency Interval Scaled Ordinal Scaled Tables Means Variance Medians Variance Correlation Z-tests c2 Correlation Levene’s Same tests as Non-Normal t-tests F-test Sign Test Medians ANOVA Bartlett’s Wilcoxon Kruskal- Correlation Wallis Regression Mood’s Friedman’s Parametric Tests Use parametric tests when: 1. The data are normally distributed 2. The variances of populations (if more than one is sampled from) are equal 3. The data are at least interval scaled One sample z - test Used when testing to see if sample comes from a known population. A sample of 25 measurements shows a mean of 17. Test whether this is significantly different from a the hypothesized mean of 15, assuming the population standard deviation is known to be 4. One-Sample Z Test of mu = 15 vs not = 15 The assumed standard deviation = 4 N Mean SE Mean 95% CI Z P 25 17.0000 0.8000 (15.4320, 18.5680) 2.50 0.012 Z-test for proportions 70% of 200 customers surveyed say they prefer the taste of Brand X over competitors. Test the hypothesis that more than 66% of people in the population prefer Brand X. Test and CI for One Proportion Test of p = 0.66 vs p > 0.66 95% Lower Sample X N Sample p Bound Z-Value P-Value 1 140 200 0.700000 0.646701 1.19 0.116 One sample t-test BP Probability Plot of BP Reduction Normal - 95% CI Reduction 99 % Mean StDev 13.82 3.925 10 95 N AD 17 0.204 90 P-Value 0.850 12 80 9 70 Percent 60 50 8 40 30 7 20 12 10 5 14 13 1 0 5 10 15 20 25 30 BP Reduction 15 16 18 12 The data show reductions in Blood Pressure in a 18 sample of 17 people after a certain treatment. We 19 wish to test whether the average reduction in BP 20 was at least 13%, a benchmark set by some other 17 15 treatment that we wish to match or better. One Sample t-test – Minitab results One-Sample T: BP Reduction Test of mu = 13 vs > 13 95% Lower Variable N Mean StDev SE Mean Bound T P BP Reduction 17 13.8235 3.9248 0.9519 12.1616 0.87 0.200 The p-value of 0.20 indicates that the reduction in BP could not be proven to be greater than 13%. There is a 0.20 probability that it is not greater than 13%. Two Sample t-test You realize that though the overall reduction is not proven to be more than 13%, there seems to be a difference between how men and women react to the treatment. You separate the 17 observations by gender, and wish to test whether there is in fact a significant difference between genders. M F Test for Equal Variances for BP Reduction F-Test 10 15 F Test Statistic P-Value 0.96 0.941 12 16 Lev ene's Test Gender Test Statistic 0.14 P-Value 0.716 9 18 M 8 12 1 2 3 4 5 6 95% Bonferroni Confidence Intervals for StDevs 7 18 12 19 F 14 20 Gender 13 17 M 15 6 8 10 12 14 16 18 20 BP Reduction Two Sample t-test The test for equal variances shows that they are not different for the 2 samples. Thus a 2-sample t test may be conducted. The results are shown below. The p-value indicates there is a significant difference between the genders in their reaction to the treatment. Two-sample T for BP Reduction M vs BP Reduction F N Mean StDev SE Mean BP Red M 8 10.63 2.50 0.89 BP Red F 9 16.67 2.45 0.82 Difference = mu (BP Red M) - mu (BP Red F) Estimate for difference: -6.04167 95% CI for difference: (-8.60489, -3.47844) T-Test of difference = 0 (vs not =): T-Value = -5.02 P-Value = 0.000 DF = 15 Both use Pooled StDev = 2.4749 Basics of ANOVA Analysis of Variance, or ANOVA is a technique used to test the hypothesis that there is a difference between Obs. Type A Type B the means of two or more populations. It is used in Regression, as well as to analyze a factorial experiment design, and in Gauge R&R studies. 1 2 6 The basic premise of ANOVA is that differences in the 2 3 7 means of 2 or more groups can be seen by partitioning the Sum of Squares. Sum of Squares 3 4 8 (SS) is simply the sum of the squared deviations of the observations from their means. Consider the following Mean 3 7 example with two groups. The measurements show the thumb lengths in centimeters of two types of SS 2 2 primates. Overall Total variation (SS) is 28, of which only 4 (2+2) is within the two groups. Thus 24 of the 28 is due to the Mean = 5 differences between the groups. This partitioning of SS into ‘between’ and ‘within’ is used to test the SS = 28 hypothesis that the groups are in fact different from each other. See www.statsoft.com for more details. Results of ANOVA The results of One-way ANOVA: Type A, Type B running an ANOVA on the sample data from the previous slide are shown Source DF SS MS F P here. The hypothesis test Factor 1 24.00 24.00 24.00 0.008 computes the F-value as the Error 4 4.00 1.00 ratio of MS ‘Between’ to Total 5 28.00 MS ‘Within’. The greater the ___________________________________ value of F, the greater the likelihood that there is in fact S = 1 R-Sq = 85.71% R-Sq(adj) = 82.14% a difference between the groups. looking it up in an F-distribution table shows a p-value of 0.008, indicating a 99.2% confidence that the difference is real (exists in the Population, not just in the sample). Minitab: Stat/ANOVA/One-Way (unstacked) Two-Way ANOVA Strength Temp Speed Is the strength of steel produced different 20.0 Low Slow for different temperatures to which it is 22.0 Low Slow heated and the speed with which it is 21.5 Low Slow cooled? Here 2 factors (speed and temp) 23.0 Low Fast 24.0 Low Fast are varied at 2 levels each, and strengths 22.0 Low Fast of 3 parts produced at each combination 25.0 High Slow are measured as the response variable. 24.0 High Slow 24.5 High Slow 17.0 High Fast Two-way ANOVA: Strength versus Temp, Speed 18.0 High Fast 17.5 High Fast Source DF SS MS F P Temp 1 3.5208 3.5208 5.45 0.048 Speed 1 20.0208 20.0208 31.00 0.001 The results show Interaction 1 58.5208 58.5208 90.61 0.000 significant main effects Error 8 5.1667 0.6458 as well as an Total 11 87.2292 interaction effect. S = 0.8036 R-Sq = 94.08% R-Sq(adj) = 91.86% Two-Way ANOVA The box plots give an indication of the interaction effect. The effect of speed on the response is different for different levels of temperature. Thus, there is an interaction effect between temperature and speed. Boxplot of Strength by Temp, Speed 25 24 23 22 Strength 21 20 19 18 17 16 Speed Fast Slow Fast Slow Temp High Low

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posted: | 3/21/2013 |

language: | English |

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