Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Psychometrics wikipedia , lookup
Bootstrapping (statistics) wikipedia , lookup
Foundations of statistics wikipedia , lookup
Taylor's law wikipedia , lookup
Omnibus test wikipedia , lookup
Analysis of variance wikipedia , lookup
Misuse of statistics wikipedia , lookup
Chapter 12 Inferences on Categorical Data Ch 12.1 Goodness-of-Fit Test Objective A : Goodness-of-Fit Test Example 1: The probabilities of getting an A, B, C or D for a science class at a particular university are shown below. Determine the expected counts for each outcome if the sample size is 700. ππππππππππ‘π¦: ππ Expected counts: πΈπ A 0.15 105 B 0.30 210 C 0.35 245 D 0.20 140 Ei ο½ οi ο½ npi E1 ο½ 700 * 0.15 ο½ 105 E 2 ο½ 700 * 0.30 ο½ 210 E3 ο½ 700 * 0.35 ο½ 245 E 4 ο½ 700 * 0.20 ο½ 140 Note: ππ = 0.15 + 0.30 + 0.35 + 0.20 = 1 Check: 105 + 210 + 245 + 140 = 700 Example 2: For example you toss a special rigged coin (binomial since there only two possible outcomes: heads or tails) four times and the number of heads are counted. There are five possible outcomes: 0H, 1H, 2H, 3H, or 4H. The seller of this special coin claims a 0.3 probability of landing on heads each flip. The buyer (who is a statistician) is not sure if he believes the seller. Therefore he conducts his own experiment. How was E1 , E2 ,...... being calculated? P(0 Heads)= P(T) and P(T) and P(T) and P(T) = (0.7)(0.7)(0.7)(0.7) = 0.2401 n = 240.1 + 411.6 + 264 + 75.6 + 8.1 = 1000 (This process was repeated 1000 times.) πΈ1 = π1 π = 0.2401 β 1000 = 240.1(240 times of the 1000 times resulted in zero heads) We can get the rest from Statcrunch: Stat --> Calculator --> Binomial --> Input the following --> Compute (a) Determine the ο£ 2 test statistic. ο£2 ο½ ο₯ ο¨Oi ο Ei ο©2 Ei (260 ο 240.1) 2 (400 ο 411.6) 2 (280 ο 264.6) 2 (50 ο 75.6) 2 (10 ο 8.1) 2 ο« ο« ο« ο« 240.1 411.6 264.6 75.6 8.1 = 11.987 = (b) Determine the degrees of freedom. DF ο½ k ο 1 ο½ 5 ο 1 ο½ 4 (k = 5 since there are 5 possible outcomes) (c) Use StatCrunch to determine the P ο value. Stat --> Calculator --> Chi-Square --> Standard --> The hypothesis tests of this section (categorical data) are always right-tailed; Input the following. P ο value = 0.0174 which is unusual and 0.0174 < 0.05 so reject the null hypothesis Ho. (d) Conclusion There is sufficient evidence to support the claim that the random variable X is not binomial with n = 4, p = 0.3. Therefore, the statistician should not buy the coin. Example 3: Use StatCrunch to perform the hypothesis testing of Example 2 at the ο‘ ο½ 0.01 level of significance. (a) Setup Ho: The random variable X is binomial with n = 4, p = 0.3. H1: The random variable X is not binomial with n=4, p=0.3. (b) P ο value Input Observed Counts in column 1 and Expected Counts in column 2 --> Stat --> Goodness-offit --> Chi-Square test --> Select Var1 for Observed and Var2 for Expected--> Compute P ο value = 0.0174 (same as before) Which is low but the significance level is now 0.01 and 0.0174 is not less than 0.01. Therefore, not unusual at a 0.01 significance level. Can not reject the null. (c) Conclusion There is not sufficient to support the claim that the random variable X is not a binomial with n = 4, p = 0.3. The statistician might want to buy the coin. The significance level makes a difference!!! Example 4: Total = 53+66+38+96+88+59 = 400 a. Set up Ho: p(brown) = 0.12, p(yellow) = 0.15, p(red) = 0.12, p(blue) = 0.23, p(orange) = 0.23, p(green) = 0.15 H1: At least one of the proportions is not equal to or different from the reported claim from the manufacturer. Expected values by hand (what we would expect if what the company claims is true): E(brown) = 0.12*400 = 48, E(yellow) = 0.15*400 = 60, E(red) = 0.12*400 = 48, E(blue) = 0.23*400 = 92, E(orange) = 0.23*400 = 92, E(green) = 0.15*400 = 60. X Observed Expected Brown 53 48 Yellow 66 60 Red 38 48 Blue 96 92 Orange 88 92 Green 59 60 b. p-value (from Statcrunch) Input Observed Counts in column 1 and Expected Counts in column 2--> Stat --> Goodness-of-fit --> Chi-Square test --> Select Var1 for Observed and select Var2 for Expected --> Compute P-value (0.613) which is not unusual and 0.613 is not less than 0.05. Cannot reject the null hypothesis. c. Conclusion: There is not sufficient evidence to support the claim that what the manufacture claims is not true. Thus, the company reports correct proportions. Ch 12.2 Tests for Independence Objective A :Tests for Independence Example 1: Men Women Total (a) Compute the expected values of each cell under the assumption of independence. Pro life Pro choice Total 196 (actual) 199 395 239 249 488 435 448 883 By assuming an individual opinion and gender are independent, (row total)(column total) (395)(435) πΈ11 = 883 β 194.6 (expected) Eο½ ( grand total) E11 ο½ (395)( 435) (395)( 448) ο½ 194.592 E12 ο½ ο½ 200.408 883 883 E 21 ο½ (488)( 435) (488)( 448) ο½ 240.408 E 22 ο½ ο½ 247.592 883 883 Summarize the observed counts and expected counts in a table where the expected counts are expressed in a parenthesis. Gender Men Women Column Total Pro Life 196 (194.592) 239 (240.408) 435 Pro Choice 199 (200.408) 249 (247.592) 448 Row Total 395 488 883 (Grand Total) (b) Verify that the requirements for performing a chi-square test of independence are satisfied. (1) Are all expected frequencies are greater than or equal to 1? Yes. (2) No more than 20% of the expected frequencies are less than 5: true. (c) Determine the ο£ 2 test statistic. ο£ ο½ο₯ 2 ο¨Oi ο Ei ο©2 Ei (196 ο 194.592) 2 (199 ο 200.408) 2 (239 ο 240.408) 2 (249 ο 247.592) 2 ο« ο« ο« 194.592 200.408 240.408 247.592 = 0.0374 = (d) Determine the degrees of freedom. DF = (r - 1)(c - 1) = (2 - 1)(2 - 1) = 1 (e) Test whether an individual's opinion regarding abortion is independent of gender at the ο‘ ο½ 0.10 level of significance. Set up: Ho: An individual opinion regarding abortion is independent of gender. H1: An individual opinion regarding abortion is dependent of gender. Or Ho: There is no difference between gender and opinion on abortion. H1: There is a difference between gender and opinion on abortion. P value: from statcrunch: Stat --> Calculator --> Chi-Square --> Standard --> The hypothesis tests of this section are always right-tailed; Input the following. P-value = 0.847 which is not unusual and 0.847 is not less the significance level of 0.10. Cannot reject the null hypothesis. Conclusion: There is not enough evidence to support the claim that abortion opinion and gender are dependent. Or There is not enough evidence to show that there is a difference between gender and position on abortion. Example 2: Use StatCrunch to redo example 1 of testing whether an individual's opinion regarding abortion is independent of gender at the ο‘ ο½ 0.10 level of significance. (a) Setup Same as before (b) P ο value from Statcrunch Input the data as shown. Stat --> Tables --> Contingency --> With Summary --> Under Select Column(s), click Pro Life and Ctrl click Pro choice --> Under Row Labels, select Gender --> Under Display, select Expected count --> Compute StatCrunch Results: (c) Conclusion P-value = 0.8489 which is the same as before. Example 3: Test whether prenatal care and the wantedness of pregnancy are independent at the ο‘ ο½ 0.05 level of significance. Note: df = (r-1)(c-1) = (3-1)(3-1) = 4 (a) Setup Ho: Prenatal care and βwantednessβ of pregnancy are independent. H1: Prenatal care and βwantednessβ of pregnancy are dependent. Or Ho: Prenatal care and βwantednessβ of pregnancy are not associated. H1: Prenatal care and βwantednessβ of pregnancy are associated. (b) P ο value Input the data into Statcrunch (see below) Stat --> Tables --> Contingency --> With Summary --> Under Select Column(s), click Less than 3 months, 3 to 5 Months, and More Than 5 Months (use the Ctrl) --> Under Row Labels, select Wantedness of Pregnancy --> Under Display, select Expected count --> Compute StatCrunch Results: P-value (0.0003) which is unusual and 0.0003 is less than 0.05. Reject the null hypothesis. (c) Conclusion There is enough evidence to show that prenatal care and βwantednessβ of pregnancy are dependent (associated). Ch 12.3 Comparing Three or More Means - One-Way Analysis of Variance, ANOVA (Supplemental Materials) Objective A :One-Way ANOVA Test Condition: π ππππππ π‘ < 2π π ππππππ π‘ (the largest standard deviation has to be less than twice the smallest) π₯Μ 1(πππ’π) = 5 π₯Μ 2(πππ) = 8 π₯Μ 3(πππππ) = 11 π¦Μ 1(πππ’π) = 5 π¦Μ 2(πππ) = 8 π¦Μ 3(πππππ) = 11 There is small variability βwithinβ each sample compared to the variability between the means (notice there is no overlap of the intervals). Here there is large variability βwithinβ each sample compared to the variability between the means. In this case it is harder to determine if there is a significant difference between the means. Notice the ratio uses the βbetweenβ variability to βwithinβ variability. Example 1: (a) Setup: Ho: The mean reaction times are the same for all three groups. H1: At least one of the groupβs mean reaction time is different. or Ho: π1 = π2 = π2 H1: π1 β π2 or π2 β π3 or π1 β π3 (b) 1. 2. 3. 4. There are 3 simple random samples. The 3 samples are independent of each other. Normal probability plots indicate that the sample data come from a normal population. If the largest sample standard deviation is no more than twice the smallest sample standard deviation, we can assume the populations have the same variance. We can use Statcrunch to find the standard deviations for each group. Input the data into the first three columns as shown.: Stat ο Summary Stats ο Columns ο Input the following ο Compute Verify: largest SD < 2 smallest SD 0.1100 < 2(0.0638) 0.1100< 0.1276 true We can assume the populations have the same variance. (C) P value from Statcrunch: Stat ο ANOVA ο One Way ο Select all three columns to compare (Simple, Go/No Go, Choice) ο Compute Note: We could have obtained each sample standard deviation using ANOVA instead of using Summary Stats. P-value = 0.0681 which is not unusual and 0.0681 is not less than 0.05. Do not reject the null hypothesis. Conclusion: There is not sufficient evidence to support the claim that at least one mean reaction time is different from the others. (d) Draw boxplots of the three stimuli. Stat ο Graph ο Boxplots ο Select all three columns ο Check draw boxes horizontally and select the box to show the means ο Compute d) Compute the 95 % CI for each group. T stats β one sample β with data (do this for each group): Simple (0.352 0.486) Go no go (0.354, 0.586) Choice (0.455, 0.647) Diagram of the CIβs: e) Does the output for the 95% CI support the conclusion from the hypothesis test? All intervals overlap. Therefore, there is no difference between the three groups. This is the same conclusion as before. f) Test the hypothesis at a 0.10 significance level. In this case, the p value of 0.681 < 0.10 so it is unusual at this significance level. Therefore, reject the null. Conclusion: There is enough evidence to support the claim that there is a difference in reaction time in at least one group. (However, we donβt necessarily which group was the different one.) g) Find the corresponding 90% CI and show how it supports the conclusion above. T stats β one sample β with data (do this for each group): Simple (0.367 0.472) Go no go (0.380, 0.561 Choice (0.475, 0.626) Diagram of the CIβs: Since group 1 and group 3 do not overlap, then there is a difference among the three groups. This is the same conclusion as above. Furthermore, we know that the two groups that were significantly different were group 1 & 3. Ch 12.4 Two - Way Analysis of Variance (Supplemental Materials) Objective A : Two - Way ANOVA Test An example of interaction effect is sleeping pills and alcohol. They are usually not fatal when taken alone, but can be fatal when combined. Example 1: (a) Verify that the largest sample standard deviation is no more than twice the smallest standard deviation. If this is true, we can assume the populations have the same variance. Input the data in StatCrunch. Stat ο Summary Stats ο Columns ο Input the following, then click Compute The largest SDis 2.6457513 and the smallest SD is 1.5275252. Verify: largest SD < 2 smallest SD 2.64< 2 (1.53) 2.64< 3.06 true We can assume the populations have the same variance. (b) Test whether there is an interaction effect between the drug dosage and age. Inputting the data is a bit tricky for a two-way ANOVA analysis. Use cut and paste or type the data in to produce three columns. ο Response (HDL), Row Factor (Age), Col Factor (Dosage) ο see below. Stat ο ANOVA ο Two Way ο Input the following ο Compute Click on β >β for the next page of the StatCrunch outputs: The above graph is called the interaction plots. We look at the level of parallelism among the lines. Since the lines are roughly parallel, we conclude there is no interaction between age and drug dosages. (c) If there is no interaction between age and drug dosages, determine whether there is sufficent evidence to conclude that the mean increase in HDL cholesterol is different (i) among each drug dosage group, (ii) for each age group. (i) the mean increase in HDL cholesterol among each drug dosage group is different The P-value for dosage is <0.0001 which is unusual and 0.0001 is less than 0.05. Reject Ho. There is sufficient evidence to support the claim that HDL cholesterol level changes with dosage. Hit > for the next page of the StatCrunch outputs: The above graph indicates the mean of HDL increases as the drug dosage increases. (ii) the mean increase in HDL cholesterol among each age group is different P-value for age is 0.0838 which is not unusual and 0.0838 is not less than 0.05. Do not reject Ho. There is not sufficient evidence to support the claim that HDL cholesterol changes with age.