I need help with question 61 60. A professor of statistics in Southern Universit
ID: 3317978 • Letter: I
Question
I need help with question 61
60. A professor of statistics in Southern University wants to determine whether the average starting salaries amoog graduates of the 15 universities in Louisiana are equal. A sample of 25 recent graduutes from each university was randomly taken. The appropriate critical value for the ANOV A test is obtained from the F-distribution with degrees of freedom is equal to A). 15 and 25 B). 14 and 360 C). 360 and 14 D). 15 and 360 E. 25 and 10 61. A randomized block design with 4 treatments without replicates and 5 blocks produced the following surn of squares values: Total SS 1951 , SST 349, SSE 88, The value of F for treatments mest be A). 16.3 B). 1.34 C). 115 D)- 8.34 E·13 62. In a regression problem the folowing pairs of (x y) are given: (1, 1) (-L.-1).(0,0). (2, 2) and (3, 3) That indicates thar: A). the corrclation coefficient is -1 B) the corelation coefticiens-is 0 C)·the correlation coefficient is 1 D), the coefficient of deternimalion is between-2 and 2 E). pope of these 63. Given the least squares regressioa line y 463+ L38r, and a cotrelation of determination of 0.90, the coefficient deterinination mast be: A. 095 B) +095 D). 0.81 E. -0.81 4. In a regnession model involving 40 observations, the following estimated megnessine model was obtained:13.5+41, + 6 2, +5 Ett Fortis mde the following statistics are given: SSR-501 and SSE-99. The, the cvefficient of determination is: A) 0.198 B L00 0.835 Di 167 E). 0.914 65. In a chi-square arodress of- fit test with 5 degrees of freedom and a significaence level of D.I.5, the chisquare value from the tables 17es, which of the fallovwing computed values of the chl-square test statistic wil lead torejection of the null hypothesis?Explanation / Answer
61)
for above df for treatments =4-1=3
hence MS for treatments =SSTr/df =349/3=116.33
for df for error =(4-1)*(5-1)=12
hence MSE =SSE/df=188/12=15.667
herefore test statistic for treatment F =MSTr/MSE =116.33/15.667 =7.43
option B is correct