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Assignment 2 Answers 1- (a) The null and the alternative hypothesis can be represented as follows; 0 : 0 : 1 0 - = - girls boys girls boys H H μ μ μ μ (b) The difference in height and the standard error of the difference are as follows; inches Y Y SE inches Y Y girls boys girls boys 77 . 0 57 ) 2 . 4 ( 55 ) 9 . 3 ( ) ( 6 . 0 2 2 _ _ _ _ = + = - - = - (c) 95% confidence interval for the difference in height: ). 10 . 2 , 90 . 0 ( 77 . 0 96 . 1 6 . 0 - = × ± - (d) 78 . 0 77 . 0 6 . 0 ) ( ) ( - = - = - - = - - - - - girls boys girls boys Y Y SE Y Y stat t Based on the above, the |-0.78|<2.58. The 2.58 represents the critical value at the 1% level. Hence you cannot reject the null hypothesis. The critical value for the one-sided hypothesis would have been 2.33. Assuming a one-sided hypothesis implies that you have some information about the problem at hand, and, as a result, can be more easily convinced than if you had no prior expectation. 2- (a) The confidence interval for the mean weekly earnings is; ] 69 . 452 , 29 . 416 [ 20 . 18 49 . 434 1744 67 . 294 58 . 2 49 . 434 = ± = × ± Based on the sample at hand, the best guess for the population mean is \$434.49. However, because of random sampling error, this guess is likely to be wrong. Instead, the interval estimate for the average earnings lies between \$416.29 and \$452.69. Committing to such and interval repeatedly implies that the resulting statement is incorrect 1 out of 100 times. For a 90% confidence interval is smaller. A smaller interval implies, given the same average earnings and the standard deviation, that the statement will be false more often. The larger the confidence interval, the more likely it is to contain the population value.

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