Key Concept: Hypothesis Testing with Single Sample t Test
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A single sample t test was conducted to determine whether job satisfaction ratings in the HR department differed from job satisfaction ratings in the company as a whole. The results were not significant, t(9) = -2.12, p> .05. Based on these results we fail to reject the null hypothesis that there is no difference in mean ratings, and conclude that HR Employees' job satisfaction ratings are not significantly different, on average, from the job satisfaction ratings of employees in the rest of the company.
Effect Size
Cohen's d: d= (M-μ)/s = d=2-3.1)/1.63 = -0.67 According to Cohen's conventions, this is around a medium effect. Nonsignificant test with a medium effect? It's possible with low power in the study.
Single Sample t Test
Compare sample mean to a known population mean. We don't know the population standard deviation so it has to be estimated. When to use a Single-sample t Test. For example: Do 20 young drivers have a different number of yearly traffic violation when compared to general population of drivers overall? Sample with M= 3.1 <-> Population with μ = 2.5, σ unknown.
Sample t Test Research questions
Do HR dept. job satisfaction ratings differ from ratings of company overall? μ = 3.1 HR Sample: N = 10 Alpha = .05 Estimate population... Step 1: Identify populations, comparison distribution and assumptions Population 1: All employees working in HR Population 2: All employees working in company overall excluding HR COMPARISON DISTRIBUTION: Dist. of Means Does not obviously violate three assumptions: 1. DV is scale 2. Not randomly selected, but test should be robust as long as we acknowledge limitations. 3. Data do not suggest a skewed distribution, so no reason to doubt that ratings are normally distributed. Step 2: State the null and research hypotheses Determine direction of test: test is 2 tailed because it's looking for any difference, not whether the ratings are higher or lower. State the null and research hypothesis: HO: μ 1 = μ 2: there is no difference between the mean job satisfaction ratings of HR employees and the company overall. HO: μ 1 ≠ μ 2: there is some kind of difference between the mean job satisfaction ratings of HR employees and the company overall. Step 3: Determine the characteristics of the comparison distribution μ m = s m μ = μ= 3.1 (Used for t Test first) FIND s: s= √Σ(X-M)^2/(N-1) estimating a population THEN FIND sm: = s/√N to find standard error * See Finding s using sample score below. Step 4: Determine critical values (cut-offs): use the t table Calculate degrees of freedom: N-1 = 10-1 = 9. Determine p level (=alpha): set at 0.05. Determine if test is one or two-tailed: 2 tailed Look in corresponding column and row of t table for critical value =- these are listed as positive values. Remember for two-tailed res, critical t is "+ or -" the t value given in the table: t= +/- 2.26. If sample t statistic is more extreme than either -2.26 or 2.26 it is significant. Step 5: Calculate the test statistic: t t= (M-μm)/s m t= (2-31.1/.52 = -1.1/.52 = -2.12 Final statistic.. Step 6. Make a decision: Compare sample t score to critical t value and make a decision about the hypothesis. Critical t = either -2.26 or 2.26. Sample t: -2.12 (Negative so look in left tail). -2.12 is not more extreme than -.2.26; it lies outside the rejection region: -3 _______________-2.26___-2.12_______-2 <rejection region---Crit t ---Sample t FAIL TO REJECT null hypothesis.
Finding s using Sample Score
to estimate population standar deviation: Scores 5, 4, 0, 2, 0, 1, 2, 1, 3, 2; M=2 Find s: with DENOMINATOR: df = 10-1 = 9 [(5-2)^2 + (5-2)^2 +(0-2)^2 + (2-2)^2 + (0-2)^2 + (1-2)^2 + (2-2)^2 + (1-2)^2 + (3-2)^2 + (2-2)^2 = 24/9 = 2.67 √2.67 = 1.63 so, s= 1.63 NOW: plug this figure into: s m = 2/√N therefore: sm = 1.63/√10 = 1.63/3.16 = 0.52
