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# Aron chpt 9 ed t test independent samples

## by karenjprice on Feb 22, 2011

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## Aron chpt 9 ed t test independent samplesPresentation Transcript

•
• Hypothesis-testing procedure used for studies with two sets of scores
• Each set of scores is from an entirely different group of people and the population variance is not known.
• e.g., a study that compares a treatment group to a control group
• When you have one score for each person with two different groups of people, you can compare the mean of one group to the mean of the other group.
• The t test for independent means focuses on the difference between the means of the two groups.
• The null hypothesis is that Population M 1 = Population M 2
• If the null hypothesis is true, the two population means from which the samples are drawn are the same.
• The population variances are estimated from the sample scores.
• The variance of the distribution of differences between means is based on estimated population variances.
• The goal of a t test for independent means is to decide whether the difference between means of your two actual samples is a more extreme difference than the cutoff difference on this distribution of differences between means.
• With a t test for independent means, two populations are considered.
• An experimental group is taken from one of these populations and a control group is taken from the other population.
• If the null hypothesis is true:
• The populations have equal means.
• The distribution of differences between means has a mean of 0.
• In a t test for independent means, you calculate two estimates of the population variance.
• Each estimate is weighted by a proportion consisting of its sample’s degrees of freedom divided by the total degrees of freedom for both samples.
• The estimates are weighted to account for differences in sample size.
• The weighted estimates are averaged.
• This is known as the pooled estimate of the population variance.
• S 2 Pooled = df 1 (S 2 1 ) + df 2 (S 2 2 )
• df Total df Total
• df Total = df 1 + df 2
• The pooled estimate of the population variance is the best estimate for both populations.
• Even though the two populations have the same variance, if the samples are not the same size, the distributions of means taken from them do not have the same variance.
• This is because the variance of a distribution of means is the population variance divided by the sample size.
• S 2 M 1 = S 2 Pooled / N 1
• S 2 M 2 = S 2 Pooled / N 2
• The Variance of the distribution of differences between means (S 2 Difference ) is the variance of Population 1’s distribution of means plus the variance of Population 2’s distribution of means.
• S 2 Difference = S 2 M 1 + S 2 M 2
• The standard deviation of the distribution of difference between means (S Difference ) is the square root of the variance.
• S Difference = √S 2 Diifference
• Figure the estimated population variances based on each sample.
• S 2 = [∑(X – M) 2 ] / (N – 1)
• Figure the pooled estimate of the population variance.
• S 2 Pooled = df 1 (S 2 1 ) + df 2 (S 2 2 )
• df Total df Total
• df 1 = N 1 – 1 and df 2 = N 2 – 1; df Total = df 1 + df 2
• Figure the variance of each distribution of means.
• S 2 M 1 = S 2 Pooled / N 1
• S 2 M 2 = S 2 Pooled / N 2
• Figure the variance of the distribution of differences between means.
• S 2 Difference = S 2 M 1 + S 2 M 2
• Figure the standard deviation of the distribution of differences between means.
• S Difference =√ S 2 Difference
• Since the distribution of differences between means is based on estimated population variances:
• The distribution of differences between means is a t distribution.
• The variance of this distribution is figured based on population variance estimates from two samples.
• The degrees of freedom of this t distribution are the sum of the degrees of freedom of the two samples.
• df Total = df 1 + df 2
• Figure the difference between your two samples’ means.
• Figure out where this difference is on the distribution of differences between means.
• t = M 1 – M 2 / S Difference
• The comparison distribution is a distribution of differences between means.
• The degrees of freedom for finding the cutoff on the t table is based on two samples.
• Your samples’ score on the comparison distribution is based on the difference between your two means.
• Example from text page 301
• You have a sample of 10 students who were recruited to take part in the study. The researcher wants to see if the group who received the experimental procedure would perform any differently than those who did not receive any experimental procedure.
• 5 students were randomly assigned to the experimental group and 5 were randomly assigned to the control group.
• All of the students rated their overall level of emotional adjustment on a scale from 0 (very poor) to 10 (very positive).
• Population 1: students in the experimental group
• Population 2: students not in the experimental group (control group)
• Research hypothesis: Population 1 students would rate their adjustment differently from Population 2 students (two-tailed tests).
• Null hypothesis: Population 1 students would rate their adjustment the same as Population 2 students.
• The comparison distribution is a distributions of differences between means.
• Its mean = 0.
• Figure the estimate population variances based on each sample.
• S 2 1 = 1.5 and S 2 2 = 2.5
• Figure the pooled estimate of the population variance.
• S 2 Pooled = 2.0
• Figure the variance of each distribution of means.
• S 2 Pooled / N = S 2 M
• S 2 M1 = .40
• S 2 M2 = .40
• Figure the variance of the distribution of differences between means.
• Adding up the variances of the two distributions of means would come out to S 2 Difference = .80
• Figure the standard deviation of the distribution of difference between means.
• S Difference = √S 2 Difference = √.80 = .89
• The shape of the comparison distribution will be a t distribution with a total of 8 degrees of freedom.
• Determine the cutoff sample score on the comparison distribution at which the null hypothesis should be rejected.
• You will use a two-tailed test.
• If you also chose a significance level of .05, the cutoff scores from the t table would be 2.306 and -2.306.
• Determine your sample’s score on the comparison distribution .
• t = (M 1 – M 2 ) / S Difference
• (7 – 4) / .89
• 3 / .89 = 3.37
• Compare your samples’ score on the comparison distribution to the cutoff t score.
• Your samples’ score is 3.37, which is larger than the cutoff score of 2.306.
• You can reject the null hypothesis.
• t Test for a Single Sample
• Population Variance is not known.
• Population mean is known.
• There is 1 score for each participant.
• The comparison distribution is a t distribution.
• df = N – 1
• Formula t = (M – Population M) / Population S M
• t Test for Dependent Means
• Population variance is not known.
• Population mean is not known.
• There are 2 scores for each participant.
• The comparison distribution is a t distribution.
• t test is carried out on a difference score.
• df = N – 1
• Formula t = (M – Population M) / Population S M
• t Test for Independent Means
• Population variance is not known.
• Population mean is not known.
• There is 1 score for each participant.
• The comparison distribution is a t distribution.
• df total = df1 + df2 (df1 = N1 – 1; df2 = N 2 – 1)
• Formula t = (M 1 – M 2 ) / S Difference