t d d sD n See how we got the values below for dbar and the standard deviation

T d d sd n see how we got the values below for dbar

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t = d d 0 sD /√ n See how we got the values below for dbar and the standard deviation of d. = = -2.1 2.34/√ 7 −1.86−0 We compare this with t at alpha = .1, 6 = 1.943 on the table.
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c. Assuming the completion time difference is normally distributed, calculate the value of the test statistic. Since it’s negative, we will compare with -1.943 and see that it is in the rejection region. d. Is the manager’s assertion supported by the data? The manager’s assertion is supported by the data. Section 10.3 47. A recent study claims that girls and boys do not do equally well on math tests taken from 2 nd to 11 th grades. Suppose 344 of 430 girls and 369 of 450 boys score at proficient or advanced levels. a. Construct a 95% confidence level for the difference between the population proportions for scoring at proficient or advanced levels. +/- p 1 p 2 α/2 z * n 1 p (1− p ) 1 1 n 2 p (1− p ) 2 2 p 1 = 430 344 = 8 . 82 p 2 = 450 369 = . (.8-.82) +/- 1.96 430 .8(1−.8) 450 .82(1−.82) -.02 +/- 1.96*.000349 -.02 to -.019 b. Develop the appropriate null and alternate hypotheses. H o : p 1 -p 2 = 0 H a : p 1 -p 2 not equal to 0 c. At the 5% significance level, what is the conclusion? 0 is not within the confidence level. We reject the null hypothesis. There is enough evidence to show that girls and boys may not do equally well on math tests.
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