When evaluated against a \(t\)-distribution with 198 degrees of freedom, we get a \(p\)-value of 0.204. If we use unequal variances, the results are treated the same, but a slightly different formula is used to calculate the values. The 95% Confidence Interval of the Difference is, or with unequal variances.Error Difference is 2.23042 (2.21101 unequal variances) Mean Difference gives the difference between the two sample means, which is -2.84. (2-tailed) gives the \(p\)-value which is 0.204 (0.200 unequal variances) \(df\) gives the degrees of freedom which is 198 (194.268 unequal variances).The \(t\) column gives the \(t\)-statistic which is -1.275 (-1.286 unequal variances).When this is the case, there is a version of the \(t\)-test that adjusts for the unequal variance, which is also displayed. The null hypothesis is that the assumption is met, so a significant result ( \(p<0.05\)) means that the assumption is violated. The Levene’s test for equality of variances allows us to check the assumption. Note that an assumption for the independent samples \(t\)-test is that the two groups have equal variance, but that assumption is often violated. The second table is the result of the \(t\)-test. For females, we had a sample size of \(n = 104\) with a mean IQ of 106.4, a standard deviation of 17.31, and a standard error mean of 1.7. For males, we had a sample of \(n = 95\) with a mean IQ of 103.56, a standard deviation of 13.88 and a standard error mean of 1.42. The first is summary statistics for each sample.
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