Stats_Class,_Wk_6

La Tonya Brooks Wk 6 Stat Problem Set Chap. 3 Ex. 88 A. Mean: 73.06 Mode: 66.20 Std: 34.23 B. Mean: 28 Mode: 18 Std: 26 C. Mean: 45913 Mode: 44174 Std: 5894 Chap. 5 Ex. 56 A. p^n 0.9^4 = 0.6561 B. (1-p)^n (1-0.9)^4 0.1^4 = 0.0001 C. at least one = 1 – prob(all arrived) = 1 - 0.6561 = 0.3439 Chap. 6 Ex. 64 A. 0.270670566473225   B. = 1 - 4 or less = 1 - 0.947346982656289 = 0.052653017343711 C. 0.135335283236613 Chap. 7 Ex. 50 A. z(65200) = (65200-60000)/2000 = 2.6 prob(z > 2.6) = 0.466% B. z(57060) = (57060-60000)/2000 = -1.47 z(58280) = (58280-60000)/2000 = -0.86 prob(-1.47 < z < -0.86) = 12.41%   C. z(62000) = (62000-60000)/2000 = 1 prob(z < 1) = 84.13% D. z(70000) = (70000-60000)/2000 = 5 prob(z > 5) = about 0% (it's greater than 0, but the z table shows it as 0) It is not reasonable, since the proportion of a population that's more than 5 standard deviations above the mean is basically 0. Chap. 8 Ex. 38 A. z = (25-23.5)/(5/sqrt(50)) z = 2.12132 prob(z > 2.12132) = 0.0169 B. z(22.5) = (22.5-23.5)/(5/sqrt(50)) = -1.4142 prob(-1.4142 < z < 2.12132) = 0.9044 C. z = +/-1.6449 The interval goes from: mean - z*sd/sqrt(N) to mean + z*sd/sqrt(N) 23.5 - 1.6449*5/sqrt(50) to 23.5 + 1.6449*5/sqrt(50) 22.3369 to 24.6631 Chap. 9 Ex. 54 A. The t value for df = n-1 = 19, with 90% confidence is: 1.7291 The interval goes from: mean - t*sd/sqrt(N) to mean + t*sd/sqrt(N) 10979 - 1.7291*1000/sqrt(20) to 10979 - 1.7291*1000/sqrt(20)   B. The z value for 99% confidence is 2.5758 The formula for sample size is: N = (z*sd/E)^2 N = (2.5758*1000/250)^2 N = 106.16 Round up to: N = 107 Chap. 10 Ex. 42 H0: game length is >= 3.5 hours Ha: game length is < 3.5 hours   mean = 2.9553 stdev = 0.5596   Get the t test statistic: t = (x-mu)/(stdev/sqrt(N)) t = (2.9553-3.5)/(0.5596/sqrt(17)) t = -4.0133   Get the critical value for df = N-1 = 16, one tail, alpha is 0.05: -1.7459   Since our test statistic is much lower than the critical value, we reject the null hypothesis. There is enough evidence to conclude that games are shorter than 3.50 hours. Chap. 11 Ex. 58 Since the test-statistic exceeds the critical value, we conclude that the percent is less now than it was 5 years ago. Std=12.4778 Avg=6.3 test-statistic is (6.3-0)/(12.4778/sqrt(20)) = 2.258 The critical value for this test is found from the t-table to be 1.645 Chap. 12 Ex. 42 A. No B. No Chap. 13 Ex. 37 t = [r/sqrt(1 - r^2)] * sqrt(n - 2) = [0.94/sqrt(1 - 0.94^2)] * sqrt(25 - 2) = 13.21   The critical value of t for 23 dof and a = 0.05 is 2.0686   Since 13.21 > 2.0686, we conclude that the correlation between the variables is significant. Ex. 40 A. [pic] B. = 0.44, there is a weak positive correlation. C. df = n-2 = 25-2 = 23 α = 0.05 one-tailed critical value T = 1.319 Test statistic T = r*Sqrt[(n-2)/(1-r²)] = 0.44*Sqrt[(25-2)/(1-0.44²)] = 2.35 Conclusion: Since 2.35 > 1.319 we conclude that there is a positive relationship between the variables D. r² = 0.44² = 0.1936, so approximately 19% of the variation in revenue is explained by the variation in the number of rooms occupied. Chap. 14 Ex. 17 A. Sales = 14 - 1 * Number of competitors + 30 * Population + 0.20 * Advertising expense The estimated value is Sales = 14 -1 * 4 + 30 * 0.4 + 0.20 * 30 =$28000 B. R^2 = SSR/SST =3050/5250 = 0.58095 C. Standard error of estimate = [pic]=[pic] = 9.1989 D. ANOVA (F test) is used to determine if any of the regression coefficients are not 0. F = 1016.67/84.62 = 12.0145 Critical value of F with (3, 26) = 2.975 Since 12.0145 > 2.975, we reject the null hypothesis that all variables are insignificant E. Critical value =2.056, X1 = number of competitors in the region.  -1.43 Fails to reject The significance test suggests that the variable X1 = number of competitors in the region has an insignificant regression coefficient in the model. This variable can be removed from the model. Ex. 18 A. Income has the largest correlation with sales, 0.964, yes; this is called multi-co linearity. B. SSR/SST =1593.81/1602.89 = 0.9943% C. F -test= 318.76/2.27=140.42. Critical value= F (0.05, 5, 4) =6.26 Since F statistic is larger than critical value, we reject the null hypothesis. Conclusion: at least one of the coefficients is non-zero. D. critical value=+/-2.78, the reject region is greater than 2.78 or less than -2.78, the t-ratio of all variables except age, outlets and bosses are in reject region. Therefore, we cannot reject the hypothesis that coefficient of variables age, outlets and bosses is zero. We can eliminate outlets and bosses variables. E. The change of R2 = 1593.81/1602.89 – 1593.66/1602.89 = 0.9943 – 0.9942 = 0.0001. The small change due to the two models differ only two variables which are not significant different from 0. F. Since the histogram and stem and leaf plot are symmetric about 0, the normality assumption appear reasonable. Also the residual plot seems randomly, so is not violated. G. There is a slight arc to the residuals. The values in the middle tend to be higher than at either end, which might indicate that a linear regression model is not appropriate. However, this trend is very slight. We’ve already seen that the residuals are fairly normal. This graph gives us no reason to think that they are not independent or that the display non-constant variance. The regression assumptions do not appear to be violated in this case. Chap. 17 Ex. 22 F(0)=300*0.135=40.6 F(1)=300*0.271=81.2 F(2)=300*0.271=81.2 F(3)=300*0.18=54.1 F(4)=300*0.09=27.1 F(5)=300*0.053=15.8 H0: population distribution is Poisson with a mean of 2.0; Ha: population distribution is not Poisson with a mean of 2.0 chi-square statistics=(50-40.6)^2/40.6+(77-81.2)^/81.2+...+(13-15.8)^2/15.8=4.16 while chi-square critical with 6-1=5 degree at 0.05 level is 11.07 since 4.16