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Solutions to homework 6 for the statistical methods in nursing course offered in spring, 2008. The solutions include calculations for chi-square tests, regression analysis, and anova. Students can use this document to check their understanding of these statistical methods.

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Download Statistical Analysis in Nursing: Homework Solutions for Spring, 2008 and more Assignments Statistics in PDF only on Docsity! 960-682– Statistical Methods in Nursing– Spring, 2008 Homework 6 Solutions, 4 May 2009 1. Question 12.04 , page 522. (10 pts) Expected values are 132, 132, 132, 132. The chi-square statistic is 10.9697 with 3 degrees of freedom. Hence the p-value is .01189. Reject the null hypothesis. Births are not evenly distributed over quarters. Total for this question: 10. 2. Question 12.24 , page 538. (15 pts) The expected values are 139.073 135.927 42.986 42.014 38.941 38.059 The χ2 statistic is 10.0905. Looking 10.09 up in the χ2 table for 2 degrees of freedom, we find that the p value is between .01 and .005. Hence reject the null hypothesis that age of children doesn’t influence effectiveness of talking to children about drinking. Total for this question: 15. 3. Question 13.22 , page 571. (12 pts) Recall that the estimated regression coefficient is b = ( ∑ j xjyj − nX̄Ȳ )/( ∑ j x2j − nx̄2) = (1156.8 − 15 × (609/15) × (33.1/15))/(28037 − 15 ∗ (609/15)2) = −0.0565, the total slum of squares is ∑ j y 2 j − nȲ 2 = 84.45 − 15 × (33.1/15)2 = 11.40933, R2 = ( ∑ j xjyj − nX̄Ȳ )2/(( ∑ j x 2 j − nx̄2) × ( ∑ j y 2 j − nȲ 2)) = (1156.8 − 15 × (609/15) × (33.1/15))2/((28037 − 15 × (609/15)2) × (84.45 − 15 × (33.1/15)2)) = 0.926, and se = √ 11.40933 × (1 − 0.926)/13 = 0.255. Standard error of b is 0.255/ √ 3311.6 = 0.00443. Hence a 95% confidence interval for β, the mean change associated with an increase in distance of 1 cm, is −0.0565 ± 2.16 × 0.00443 = (−0.0661,−0.0469). Any properly constructed confidence interval of a different confidence level is also acceptable. Total for this question: 12. 4. Question 13.36 , page 589. a (3 pts) sa+2b = √ 1/20 + (2 − 2.5)2/2516.486 = 4.04. b (3 pts) Since sa+bx depends on x only through (x− 2.5)2, and since (3− 2.5)2 = (2− 2.5)2, sa+3b = sa+2b = 4.04. c (3 pts) sa+2.8b = √ 1/20 + (2.8 − 2.5)2/2516.486 = 3.82. d (3 pts) sa+bx is minimized where x = 2.5, since this causes (x − 2.5)2/25 = 0. Total for this question: 12. 1 960-682– Statistical Methods in Nursing– Spring, 2008 5. Question 14.34 , page 633. The data for this question are on the CD that came with your text book, and on line. See http://stat.rutgers.edu/~kolassa/960-532/dpdatasets.html for more informa- tion. a. The table of fitted coefficients is Estimate Std. Error t value Pr(> |t|) (Intercept) 76.4371 9.0818 8.4165 0 x1 -7.3452 10.7995 -0.6801 0.5061 x2 9.6131 10.7995 0.8901 0.3866 x3 -0.9149 1.0676 -0.8569 0.4041 x4 0.0963 0.0983 0.9795 0.3419 x1x1 -13.4524 6.5994 -2.0384 0.0584 x1x2 3.75 8.8225 0.425 0.6765 x1x3 -0.75 0.8823 -0.8501 0.4078 x1x4 0.1417 0.0588 2.4086 0.0284 x2x2 2.7976 6.5994 0.4239 0.6773 x2x3 2 0.8823 2.2669 0.0376 x2x4 -0.125 0.0588 -2.1252 0.0495 x3x3 0.028 0.066 0.4239 0.6773 x3x4 0.0033 0.0059 0.5667 0.5788 x4x4 -0.0003 0.0003 -1.0914 0.2913 and so the fitted regression equation is y = 76.4371− 7.3452× x1 + 9.6131× x2− 0.9149× x3 + 0.0963×x4− 13.4524× x1x1 + 3.75×x1x2− 0.75× x1x3 + 0.1417×x1x4 + 2.7976× x2x2 + 2 × x2x3 − 0.125 × x2x4 + 0.028 × x3x3 + 0.0033 × x3x4 − 0.0003 × x4x4. b. The output from the above regression also includes the information that F -statistic is 8.755 on 14 and 16 degrees of freedom, and the p-value is 4.968×10−5. Reject the null hypothesis that all independent variables and their interactions have zero coefficient in favor of the alternative hypothesis that at least one of these coefficients is non-zero. c. The output from the above regression also notes that se = 0.3529, indicating that the standard deviation of the random component added to the regression equation to obtain the actual dependent variable values have standard deviation 0.3529, and that R2 = 0.8845, indicating that the regression equation explains 88% of the variation seen in the dependent variable. Furthermore, the analysis of variance table shows that the sum of squared residuals is 1.9926; this represents the sum of the squared residuals. Total for this question: 0. 6. Question 15.18 , page 679. Here is the ANOVA output: Df Sum Sq Mean Sq F value Pr(> F ) Group 2 1.348 0.674 3.5089 0.0517 Residuals 18 3.4574 0.1921 The p value is .052, greater than the cutoff we’ve used for any of the tests. Do not reject the null hypothesis of no difference between groups. 2