M9 Problem Set

M9: Problem Set Due No due date Points 5 Questions 5 Time Limit None Attempt History Attempt Time Score LATEST Attempt 1 12,784 minutes 5 out of 5 Score for this quiz: 9 out of 5 Submitted Nov 12 at 7:02am This attempt took 12,784 minutes. Question 1 0/0 pts Suppose you have 17 data points and you calculate the sample correlation coefficient and find that r=.54. Can you be 95% confident that a linear relation exists between the variables? If so, is the relation positive or negative? Justify your answer. Your Answer: n=17,r=0.54 Critical value = 0.48215 The absolute value of r (|r|]) = 0.54, which is greater than the critical value, indicating a positive linear relationship. Note that for n=17 and 95%, we find a Critical Value Correlation Coefficient of .48215. The absolute of r is |r|=.54, which is more than .48215. So a linear relation exists and since the original r is positive, it is a positive linear relation.

Question 2 0/0 pts Suppose you have 45 data points and you calculate the sample correlation coefficient and find that r=-.23. Can you be 95% confident that a linear relation exists between the variables? If so, is the relation positive or negative? Justify your answer. Your Answer: n=45, r=-.23 cri"cal value= .29396 the absolute value of r=.23, which is less than the cri"cal value, meaning there is no linear rela"on Note that for n=45 and 95%, we find a Critical Value Correlation Coefficient of .29396. The absolute of ris |r|=.23, which is less than .29296. So no linear relation exists. Question 3 0/0 pts By hand compute the sample correlation coefficient for the following data: X 1 3 4 9 10 Can you be 95% confident that a linear relation exists between the variables? If so, is the relation positive or negative? Justify you answer. Your Answer: —_— 2ar;l - 2?/1 ‘CB — Y=

There are five data points, so n=5. - 1+3+4+9+10 X = =54 4 o 20+18+10+8+4 5 , L—%? (1-54)"+(3—54)*+(4—54)* +(9—54)* + (10— 5.4) Sx — — = 15.3 n-1 5-1 S, = V15.3 = 3.911 3 Yy-9? (20-12)*+(18—12)* + (10— 12)* + (8 — 12)* + (4 — 12)® _ e ¥ n-1 5—1 S, = V46 = 6.78 We now calculate the sample linear correlation coefficient: (i)Y N n—-1 e I e e b ) )+ )+ ) ) =-3.6958. X; — X Vi — ¥ %( > x)( 3 —) —3.6958 = y — = —.92 n—-1 5—-1 U Note that for n=5 and 95%, we find a Critical Value Correlation Coefficient of .87834. The absolute of r is [r|=.92, which is more ' than .87834. So a linear relation exists and since the original r is | negative, it is a negative linear relation. Question 4 0/0 pts Find the best fit line for the following data:

x 1 3 4 9 10 y 20 18 10 8 4 (Notice that this is the same data as the previous problem. You may use any calculations that you did for that problem to help you with this problem.) Your Answer: X= 5.4 y=12 Sx =3.911 Sy =6.78 r=-.92 Slope: m=r(Sy/Sx) m=.92(6.78/3.911=-1.595 Y intercept: b=y-mX b=12-(-1.595)(5.4)=20.613 When we sub these into the equation for the line: y= mx+b y=-1.595x+20.613