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Chapter 8 The t Test for Independent Means Part 1: March 6, 2008.

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Presentation on theme: "Chapter 8 The t Test for Independent Means Part 1: March 6, 2008."— Presentation transcript:

1 Chapter 8 The t Test for Independent Means Part 1: March 6, 2008

2 t Test for Independent Means Comparing two samples –e.g., experimental and control group –Scores are independent of each other Focus on differences betw 2 samples, so comparison distribution is: –Distribution of differences between means

3 The Distribution of Differences Between Means If null hypothesis is true, the 2 populations (where we get sample means) have equal means If null hypothesis true, the mean of the distribution of differences equals 0

4 Pooled Variance Start by estimating the population variance –Assume the 2 populations have the same variance, but sample variance will differ… – so pool the sample variances to estimate pop variance = Pooled estimate of pop variance Sample 1 variance Sample 2 variance df total = total N-2 df2 = Group2 N 2 -1

5 Variance (cont.) Note – check to make sure S 2 pooled is between the 2 estimates of S 2 We’ll also need to figure S 2 M for each of the 2 groups:

6 The Distribution of Differences Between Means Use these to figure variance of the distribution of differences between means (S 2 difference ) Then take sqrt for standard deviation of the distribution of differences between means (S difference )

7 T formula and df –t distribution/table – need to know df, alpha –Where df 1 = N 1 -1 and df 2 = N 2 -1 t observed for the difference between the two actual means = Compare T observed to T critical. If T obs is in critical/rejection region  Reject Null

8 Example Group 1 – watch TV news; Group 2 – radio news. –Is there a significant difference in knowledge based on news source? –Research Hyp? –Null Hyp?

9 Example (cont.) –M1 = 24, S 2 = 4N1 = 61 –M2 = 26, S 2 = 6N2 = 21 –Alpha =.01, 2-tailed test, df tot = N-2 = 80 –S 2 pooled = –S 2 M1 = –S 2 M2 = –S 2 difference = –S difference =

10 (cont.) t criticals, alpha =.01, df=80, 2 tailed –2.639 and –2.639 t observed = Reject or fail to reject null? –Conclusion? –APA-style sentence:

11 Assumptions 1) Each of the population distributions (from which we get the 2 sample means) follows a normal curve 2) The two populations have the same variance –This becomes important when interpreting Ind Samples t using SPSS –SPSS provides 2 sets of results for ind samples t-test: 1 st assumes equal variances in 2 groups 2 nd assumes unequal variances –You have to check output to see which of these is true –SPSS provides “Levine’s test” to indicate whether the 2 groups have equal variance or not. –Then, use the results for either equal or unequal variances (depending on results of Levine’s test…)


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