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Lecture

Comparing two populations.doc

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Department
Administrative Studies
Course
ADMS 2320
Professor
All Professors
Semester
Winter

Description
Case 1: 1 For each variable measured in this study, success would mean a reduction over time. Excel is used to measure the difference scores between the before and after measures, and since it performs this calculation as BEFORE – AFTER, a positive number indicates a reduction. This means that in general, to decide if a variable is reduced significantly, the following hypotheses are tested: H 0 μ D = 0 program makes no diff H 1μ >D0 program is successful Rejection region: Degrees of freedom are ν = n D1 = 33 – 1 = 32. Assume a 5% significance level. This is a one tailed test. t > tα ,ν t0.05 ,321.6939 The test statistic for matched pairs is t = xD−μ D s D nD Excel is used to calculate the results: Weight ~ t-Test: Paired Two Sample for Means Weight 1 Weight 2 Mean 79.8879 78.6212 Variance 255.5367 251.0892 Observations 33 33 Pearson Correlation 0.9876 Hypothesized Mean Difference 0 df 32 t Stat 2.8969 P(T<=t) one-tail 0.0034 t Critical one-tail 1.6939 P(T<=t) two-tail 0.0067 t Critical two-tail 2.0369 t=2.8969 exceeds the critical value and so the null hypothesis is rejected. This means that there is sufficient evidence to infer that the program leads to positive reductions – in other words people lost weight. ((repeat for Chl lvl, fat, …)) all get rejected – significant reductions in each variable.. overall the program is successful. 2. To determine if gender affects reduction in each of the variables measured in the study, the hypotheses to be tested are: H 0 μ F μ M = 0 Gender makes no diff to reduction H 1 μ F μ M ≠ 0 Gender makes a diff to reduction To perform this test properly, it must first be determined whether or not equal variances can be assumed. This is determined by testing (for each variable), the following hypotheses: 2 H : σ 1 =1 Equal variances can be assumed 0 σ 2 2 2 σ 1 H 1 2 ≠1 Equal variances cannot be assumed σ 2 (where population 1 represents females and population 2 males) Rejection Region: Assume a 5% significance level. Degrees of freedom are v1= (n 11) = 16 – 1 = 15 v2= (n 21) = 17 – 1 = 16 This is a two tailed test and so the null hypothesis is rejected if 1 1 F < F = F = 0.3497 or F > F α / 2 ,ν 1,ν0.025 ,15 ,1679 α / 2,ν 2,ν 10.025 ,16 ,15 The test statistic is calculated in Excel using the formula: 2 s1 F = 2 s2 Results: Weight ~ F-Test Two-Sample for Variances Variable 1 Variable 2 Mean 1.31875 1.217647059 Variance 1.542958333 11.16654412 Observations 16 17 df 15 16 F 0.13817689 F=0.1382 falls within the rejection region, and so the null hypothesis is rejected. This means that equal variances cannot be assumed for this variable. Cholesterol Level~ F-Test Two-Sample for Variances Variable 1 Variable 2 Mean 0.481875 0.717058824 Variance 0.48061625 0.551822059 Observations 16 17 df 15 16 F 0.870962373 F=0.8710 does not fall within the rejection region, and so the null hypothesis is not rejected. This means that equal variances can be assumed for this variable. Fat Intake~ F-Test Two-Sample for Variances Variable 1 Variable 2 Mean 14.5 24.87647059 Variance 229.9146667 906.7619118 Observations 16 17 df 15 16 F 0.253555717 F=0.2536 falls within the rejection region, and so the null hypothesis is rejected. This means that equal variances cannot be assumed for this variable. Cholesterol Intake~ F-Test Two-Sample for Variances Variable 1 Variable 2 Mean 38.4563 90.5588 Variance 5683.4920 46983.5176 Observations 16 17 df 15 16 F 0.1210 F=0.1210 falls within the rejection region, and so the null hypothesis is rejected. This means that equal variances cannot be assumed for this variable. Percent Daily Calories from Fat~ F-Test Two-Sample for Variances Variable 1 Variable 2 Mean 6.875 4.623529412 Variance 18.694 34.28691176 Observations 16 17 df 15 16 F 0.545222624 F=0.5452 does not fall within the rejection region, and so the null hypothesis is not rejected. This means that equal variances can be assumed for this variable. Equal variance can be assumed for both Cholesterol Level and Percent Daily Calories from Fat. The two variables will be dealt with first. H 0 μ −Fμ M = 0 Gender makes no diff to reduction H 1 μ −Fμ M ≠ 0 Gender makes a diff to reduction Rejection Region: Assume a 5% significance level. Degrees of freedom are v = n1+ n 2 2 = 16 + 17 – 2 = 31 This is a two tailed test so the rejection region is t < −t α / 2 ,ν−t0.025 ,31-2.040 or t > α / 2,ν t0.025 ,312.040 The test statistic is calculated in Excel as (x1− x 2 − (μ 1 μ ) 2 t = (n −1)s + (n −1)s 2 2  1 1  sp= 1 1 2 2 sp  +  n1+ n 2 2 n1 n 2 Cholesterol Level~ t-Test: Two-Sample Assuming Equal Variances Variable 1 Variable 2 Mean 0.481875 0.717058824 Variance 0.48061625 0.551822059 Observations 16 17 Pooled Variance 0.517367635 Hypothesized Mean Difference 0 df 31 - t Stat 0.938718334 P(T<=t) two-tail 0.355132326 t Critical two-tail 2.039514584 t = 0.9387 does not fall within the rejection region so the null is not rejected. There is not enough evidence to infer that gender affect cholesterol level reduction. Percent Daily Calories from Fat~ t-Test: Two-Sample Assuming Equal Variances Variable 1 Variable 2 Mean 6.875 4.623529412 Variance 18.694 34.28691176 Observations 16 17 Pooled Variance 26.74195446 Hypothesized Mean Difference 0 df 31 t Stat 1.249963086 P(T<=t) two-tail 0.220667338 t Critical two-tail 2.039514584 t = 1.250 does not fall within the rejection region so the null is not rejected. There is not enough evidence to infer that gender affect percent daily calories from fat reduction. With the other variables, unequal variances must be assumed. The degrees of freedom will be different in each case since the formula is dependent on more than just the sample size 2 s2 s2   1 + 2   n1 n 2 v= 2 2 s1  s2   n   n   1  2 + n1−1 n2−1 It is left to Excel to calculate the degrees of freedom for each case and as a result, the critical values also need to be determined by Excel. Weight~ t-Test: Two-
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