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Significance TestingStatistical Hypothesis TestingSteps of Hypothesis Tests(cont.)1 & 2-tailed hypothesesSteps 2&3 : Select & Calculate the Appropriate Test StatisticExample (cont.)Step 4 (cont.): DecisionInterpretation2-tailed test interpretationDecision ErrorsErrors (cont.)Lab 17Significance TestingWed, March 24thStatistical Hypothesis TestingProcedure that allows us to make decisions about pop parameters based on sample stats–Example – mean aver salary of all workers = $28,985 (pop mean = y) –Sample of 100 African Amer workers, mean salary = $24,100 (samp mean = ybar)–Is that a significant difference from the population? Or not enough of a difference to be meaningful?Steps of Hypothesis Tests1) State the Research & Null Hypotheses:–Research Hyp (H1): state what is expected & express in terms of pop parameter–Ex) y does not = $28,985 (Af Am salary doesn’t = pop salary; the groups differ)–Null Hyp (Ho): usually states there is no difference/effect (opposite of H1).–Ex) y = $28,985 (Af Am salary = pop salary)(cont.)–Note: Research Hyp (H1) also known as alternative hypothesis (Ha)Null hypothesis is tested; we hope to reject it & find support for H11 & 2-tailed hypotheses(Step 1 cont.) – possible to specify 1 or 2-tailed research hyp (H1)–1-tailed is a directional hyp (expect  > some value or < some value)–2-tailed is nondirectional (specify  is not equal to some value)–Use 1-tailed when you can rely on theory to know what to expect; 2-tailed if no prior expectation–Here, we could expect H1: y < $28,985 (1-tailed test that specifies a lower salary)Steps 2&3 : Select & Calculate the Appropriate Test StatisticHere, a 1-sample z test to compare a sample mean to a known pop mean–Z = (Ybar – y) / ybar–Where ybar is std error and =  y / sqrt N and ybar is sample mean; y is pop mean–Salary example: y = $28,985, y = $23,335ybar = $24,100 and N=100, so…-ybar =23,335 / sqrt(100) = 2,333.5Example (cont.)–Z = 24,100 – 28,985 / 2,333.5 = -2.09Step 4 – use unit normal table to make a probability decision–Look up z score of –2.09 (or 2.09) in column C (proportion beyond z), find .0183.–This is the prob of getting a sample result this extreme ($24,100) if the null hypothesis is true; called p valueStep 4 (cont.): DecisionWe define in advance what is sufficiently improbably to reject the null hypothesis–Find a cutoff point, called  (alpha) below which p must fall to reject null–Usually  = .05, .01, or .001–Reject null when p <= –Here, if choose  = .05, we reject null (p = .0183< .05  reject null)Interpretation–P = .0183 means there is only a 1.83% chance of finding a sample of 100 Afr Amer workers w/mean salary = 24,100 if there is really no difference from overall aver salary. (very unlikely)–Note: if p <  and we reject the null, we can say our findings are ‘statistically significant’; the groups differ significantly2-tailed test interpretationOur example used a 1-tailed test, if we’d made a 2-tailed H1 (y does not = $28,985), we need to adjust the p value–Look up z=-2.09 and find p=.0183, but need to multiple p x 2 if 2-tailed (.0183)x 2 = .0366–P is still < , so still reject nullDecision ErrorsPossible to make 2 types of errors when deciding to reject/fail to reject Ho:Ho is true in realityHo is false in realityYou Reject HoType 1 error ()CorrectYou Don’t Reject HoCorrect Type II errorErrors (cont.)Type 1 error = probability of incorrectly rejecting a true HoType 2 error = probability of failing to reject a false HoSo when  = .05, we have a 5% chance of incorrectly rejecting Ho (Type 1)–Can be more conservative and use  = .01 (1% chance of Type 1), but then increases Type 2 chances…Lab 17Skip


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ISU PSY 138 - Significance Testing

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