# Tuesday, October 22 Interval estimation. Independent samples t-test for the difference between two means. Matched samples t-test.

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Tuesday, October 22 Interval estimation. Independent samples t-test for the difference between two means. Matched samples t-test

Tuesday, October 23 Interval estimation. Independent samples t-test for the difference between two means. Matched samples t-test

Interval Estimation (a.k.a. confidence interval) Is there a range of possible values for  that you can specify, onto which you can attach a statistical probability?

Interval Estimation (a.k.a. confidence interval) Is there a range of possible values for  that you can specify, onto which you can attach a statistical probability?

Confidence Interval X - ts X    X + ts X _ _ Where t = critical value of t for df = N - 1, two-tailed X = observed value of the sample _

Tuesday, October 23 Interval estimation. Independent samples t-test for the difference between two means. Matched samples t-test

Tuesday, October 22 Interval estimation. Independent samples t-test for the difference between two means. Matched samples t-test

H 0 :  1 -  2 = 0 H 1 :  1 -  2  0

X boys =53.75 _ X girls =51.16 _ How do we know if the difference between these means, of 53.75 - 51.16 = 2.59, is reliably different from zero?

X boys =53.75 _ X girls =51.16 _ 95CI: 52.07   boys  55.43 95CI: 49.64   girls  52.68 We could find confidence intervals around each mean...

H 0 :  1 -  2 = 0 H 1 :  1 -  2  0 But we can directly test this hypothesis...

H 0 :  1 -  2 = 0 H 1 :  1 -  2  0 To test this hypothesis, you need to know … …the sampling distribution of the difference between means.  X 1 -X 2 --

H 0 :  1 -  2 = 0 H 1 :  1 -  2  0 To test this hypothesis, you need to know … …the sampling distribution of the difference between means.  X 1 -X 2 -- …which can be used as the error term in the test statistic.

 X 1 -X 2 =    2 X 1 +  2 X 2 The sampling distribution of the difference between means. This reflects the fact that two independent variances contribute to the variance in the difference between the means. -- --

 X 1 -X 2 =    2 X 1 +  2 X 2 The sampling distribution of the difference between means. This reflects the fact that two independent variances contribute to the variance in the difference between the means. -- -- Your intuition should tell you that the variance in the differences between two means is larger than the variance in either of the means separately.

The sampling distribution of the difference between means, at n = , would be: z = (X 1 - X 2 )  X 1 -X 2 -- --

The sampling distribution of the difference between means. Since we don’t know , we must estimate it with the sample statistic s.  X 1 -X 2 =   2 1   2 2 n 1 n 2 + --

The sampling distribution of the difference between means. Rather than using s 2 1 to estimate  2 1 and s 2 2 to estimate  2 2, we pool the two sample estimates to create a more stable estimate of  2 1 and  2 2 by assuming that the variances in the two samples are equal, that is,  2 1 =  2 2.  X 1 -X 2 =   2 1   2 2 n 1 n 2 + --

s X1-X2 = s p 2 s p 2 N 1 N 2 +

s X1-X2 = s p 2 s p 2 N 1 N 2 +

s X1-X2 = s p 2 s p 2 N 1 N 2 + s p 2 = SS w SS 1 + SS 2 N-2 =

Because we are making estimates that vary by degrees of freedom, we use the t-distribution to test the hypothesis. t = (X 1 - X 2 ) - (  1 -  2 )  s X 1 -X 2 …at (n 1 - 1) + (n 2 - 1) degrees of freedom (or N-2)

Assumptions X 1 and X 2 are normally distributed. Homogeneity of variance. Samples are randomly drawn from their respective populations. Samples are independent.

Get district data.

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