Inferential Stat Week 13

T-test Procedure The t-test for independent means is the appropriate test statistic Small number of cases Testing significance between two independent means

Data The researcher collected the data using a survey. One question on the survey asked the person’s gender Another question asked the person how many lost workdays they experienced due to a MSD over the past year. There were 15 respondents, 10 males and 5 females

Data Table Females Males 2 3 5 6 7 4 Average 1.00 2.70 2 3 5 6 7 4 Average 1.00 2.70 Sample Standard Deviation (s) 1.41 2.71 N 10

T-test Calculation

T-table To determine the cutoff score, there are Na+Nb-2=13 degrees of freedom. At .05 (use .025 column), the critical score is 2.16.

Decision Your calculated t-test score is -1.30. The cutoff score is 2.16. In this example, look at the negative side of the curve, so you use -2.16. Your t-score would fall into the unshaded area of the curve so you do not reject the Null hypothesis and conclude the means are not significantly different from one another.

Z-test Can be used to test hypotheses involving two means similar to the t- test, but the sample size is > 25. At a very large sample size: t-test = z-test value.

Z-test An industrial hygienist collected readings from a CO meter for a process that was running in the plant. The hygienist collected 30 samples over a two-day period. He wanted to determine if the average CO readings for the two-day period were significantly different from the manufacturer's published average for the system. The manufacturer indicates that the process should have an average reading of 35 ppm with a variance of 55 ppm. These values were based on a sample of 40 readings. The data was collected and analyzed as follows:

Z-test 23 21 33 22 26 15 32 35 24 19 44 37 14 34 29 30 25 18

Z-test Hypothesis: The researcher wishes to determine if the obtained average is significantly different than the manufacturer's obtained average. The null and alternative hypotheses are as follows: Null Hypothesis: The sample mean is equal to the population mean of 35 ppm Alternative Hypothesis: The sample mean is not equal to the population mean of 35 ppm

Z-test Population mean = 35 variance = 55 sample size = 40 Sample mean = 27.4 variance = 61.49 sample size = 30 z = -4.1 Make a decision: since -4.1 < -1.96 (z at α/2 = 0.025)  Significant

Chi Square (χ2) Chi square is a non-parametric test of significance appropriate when the data are in the form of frequency counts occurring in two or more mutually exclusive categories. Expected value: the null mean, all groups will have the same mean. I.E. toss a coin. Face = Tail = 0.5 probability. Elections: 3 nominees: 33.33% each. If there is a 90 voters, then each is expected to get 30.

Chi Square (χ2) For example, EA = 1,1 = (39x20)/67 = 11.6 Hand-on   Hand-on Classroom Both Types Total Male A = 15 B = 4 C = 20 nr=1 = A+B+C = 39 Female D = 5 E =12 F = 11 nr=2 = C+D+E = 20 nc=1 = A+D = 20 nc=2 = B+E = 16 nc=3 = C+F = 31 n = 67

Chi Square (χ2) A researcher wants to determine if there is a relationship between gender and the type of training received (Hand-on, classroom, combination). The gender question is male/female (categorical) and the training is categorical. To set up the Chi Square Test: Set up the table with the frequencies of the responses in the format below: For example, those males that said, they received hands-on training totaled 15. Those females that said they received hands-on training totaled to 5, and so on.   Hand-on Classroom Both Types Total Male 15 4 20 39 Female 5 12 11 28 16 31 67

Chi Square (χ2) The next step is to determine the expected number of cases in the cells. The expected cases table would be as follows:   Hand-on Classroom Both Types Total Male 11.6 9.3 18 39 Female 8.4 6.7 13 28 20 16 31 67

Chi Square (χ2) Next, using the Chi Square Test, you will take the (observed minus the expected)2 / expected and sum the results from each of the cells. In this example, there will be 6 parts that will be added to get the Chi Square answer. Now compare to critical value from tables at α = 0.05 and df = (2-1)x(3-1) = 2  X2 = 5.99, NOT SIGNIFICANT.

One-way Analysis of Variance (ANOVA) Can be used to test hypotheses that compare more than two means. T-test is a special case of ANOVA makes comparisons about two different estimates of a population variance to test a hypothesis concerning the population mean. Use the F test. The variances are the within group variance and the between group variance. F ratio = Between the groups/within the groups The null hypothesis (Ho) is: Ave Grp 1 = Ave Grp 2 = Ave Grp 3 etc. The alternative hypothesis (H1) is: Ave Grp 1 ≠ Ave Grp 2 ≠ Ave Grp 3 etc.

ANOVA Group 1 Group 2 Group 3 10 12 18 14 11 19 13 16 22 17 Group Mean = = 11.6 12.0 18.4 Grand Mean = = 14.0 ni = 5 (each group) n = 3 N = 15

ANOVA Calculate the sum of squares (SS)

ANOVA SS Within = SS total - SS Between

ANOVA Calculate the df Calculate the Mean Squares Mean squares between = SS between/df between = 145.6 / 2 = 72.8 Mean squares within = SS within /df within = 38.4 / 12 = 3.2 df between = n -1 = (3 - 1) = 2 df within = nixn – n = (5x3 - 3) = 12 df total = (nixn - 1) = 14

ANOVA Calculate the F Ratio Complete the ANOVA summary table F ratio = MS between/MS Within F ratio = 72.8 / 3.2 = 22.75 Complete the ANOVA summary table

ANOVA Conclusions Compare the calculated F ratio with the critical value from the table. To use the F table you need df between (nominator = df1) and df within (dominator = df2) F2,12 = 3.885 Make a decision. F calculated > F critical , then reject Ho and there is a significant difference between groups.

ANOVA Conduct the post-hoc analysis to decide which one is significantly different than the other. Tukey test Scheffe’s test.