Download Statistical Analysis: Mean, Root MSE, R-Square, and Hypothesis Testing and more Exams Data Analysis & Statistical Methods in PDF only on Docsity! Summary of Fit Mean of Response 36.4545 Root MSE 6.7441 R-Square 0.8900 Adj R-Sq 0.8845 Analysis of Variance Source Model Error C Total DF 1 20 21 Sum of Squares 7361.7866 909.6679 8271.4545 Mean Square 7361.7866 45.48340 F Stat 161.8566 Pr > F <.0001 STAT 515 – Fall 2002 - Final Exam Solutions (modified slightly for Spring 2003 practice) Part I: 1) The probability of observing a test statistic at least as extreme as the one observed if the null hypothesis is true. 2) The automaker would likely prefer that the median fuel economy for the entire line be reported. Because the distribution is very skewed to the left, if we were given the mean and standard we would be able to say that at least 75% of the mileages should be within two standard deviations of the mean. 3) A biased coin (probability of a head on one flip = 0.4) is flipped 9 times. What are the mean and standard deviation of the number of heads that will be observed? mean=np=3.6 standard deviation=square root of(np(1-p))= 1.469694 4) P(20<X<25)=P((20-20)/4 < (X-mean)/sd < (25-20)/4)= P(0<Z<1.25)=0.3944 Part II: 1A) H0: p=0.4 HA: p<0.4 where p=proportion of consumers deterred by web-site complexity B) p-value=0.5-0.4911=0.0089. We reject the null hypothesis because the p-value is less than α. The percentage who are deterred is less than 40. C) D) The sample needs to be random and 5ˆ ≥pn and .5)ˆ1( ≥− pn 2) A) B) If x (# observed geese) = 100 then the estimated y (actual # geese)is 2.9101+1.1714(100)= 120.0501 C) Because it is extrapolating (we don’t have any observations that big). D) On average, how far from the estimated regression line do you expect the observed values to be? Square root of MSE is 6.7441. E) Residual vs. Predicted Plot is for checking: Errors have mean 0 at each x and errors have equal variance at each x Q-Q plot of the Residuals is for checking: Errors are normally distributed 3A) Test of independence because we know the total number of heart patients in advance, but not the row and column totals. B) Write out the tables of expected values for conducting this test. Abstain 7 or fewer 7 or more Con. 146 281*896/1913=131.6 106 281*696/1913=102.2 29 281*321/1913=47.2 281 Not 750 1632*896/1913=764.4 590 1632*696/1913=593.8 292 1632*321/1913=273.8 1632 896 696 321 1913 C) Give the formula for X2 for this problem (plugging the values in, but not needing to simplify). 8.273 2 )8.273292( 8.593 2 )8.593590( 4.764 2 )4.764750( 2.47 2 )2.4729( 2.102 2 )2.102106( 6.131 2 )6.131146( − + − + − + − + − + − D) What is the rejection region for conducting this test at α=0.01? Checking df=(3-1)(2-1)=2 we get 9.21034 or greater. E) Why is, or why isn't, the sample size of this experiment large enough for performing this hypothesis test? It is large enough because all of the expected values are greater than 5. 37.2 0632.0 15.0 60 )4.01(4.0 4.025.0 )1( ˆ = − = − − = − − = n pp pp z )36.0,14.0(11.025.0 60 )25.01(25.0 96.125.0 )ˆ1(ˆ 2 ˆ ⇒±⇒ − ±⇒ − ± n pp zp α
