Question

nd the P-value using a0.05, Is there icient evidonce to conlude that there is a linear.comolntion between budgets and arpss receints? Do the results change if the artual budgets listed are $64 000 000 $93 000 000 S46.000 000 and so on? Budget (x) Gross (y) 46 53 204 609 98 143 64 63 38 90 93 66 5! 43 H1:pu H.pu O D. He: p0 Q c. Ho: p 0 Hi p0 H p 0 Construct a scatterplot. Choose the correct graph below. O D. OA. B. acoT 800- The linear correlation coefficient r is (Round to three decimal places as needed.) The test statistict is (Round to two decimal places needed.) The P-value is (Round to three decimal places as needed.) Because the p-value is than the significance level 0.05, there sufficient evidence to support the claim that there is a linear correlation between between budgets and gross receipts for a significance level of 0.05 .

Do the results change if the actual budgets listed are

64,000,000​,

93,000,000​,

46,000,000​,

and so​ on?

A.

​No, the results do not change because it would result in the same linear correlation coefficient.

B.

​Yes, the results would need to be multiplied by​ 1,000,000.

C.

​Yes, the results would change because it would result in a different linear correlation coefficient.

D.

​No, the results do not change because it would result in a different linear correlation coefficient.

*** What are the null and alternative​ hypotheses?

Answer 1

(x-x)2 (y-v) (x-x)(y-y Y S.No 1 64 63 698.4694 7693.7959 2318.1633 -166.4082 2 93 86 6.6122 4187.9388 1973.8980 46 53 9548.0816 4341.3061 4 38 58 2748.7551 8595.9388 4860.8776 5 204 609 12898.4694 210025.7959 52048,1633 6 98 143 57.3265 59.5102 -58,4082 90 43 0.1837 11602.3673 46.1633 Total 633 1055 18383.7143 251713.4286 63389.8571 90.4286 Mean 150.7143 SSX SSY SXY null hypothesis: Alternate Hypothesis: for 0.05 level,two tailed test and n-2= 5 degree of freedom, critical t 2,5706 reject Ho if absolute value of test statistic t|>2.571 Decision rule: Sy/VS *Sy) correlation coefficient r 0.9319 (V(n-2)/(1-)) test statistic: t= 5.7430 P value 0,0022 X st LO N

option B is correct for hypothesis

option B is correct for scatterplot

correlation coefficient r =0.932

test statistic t =5.74

p value =0.002

because the p value is less than the ,,,,there is sufficient evidence,,,,,,,,,,,,

A.

​No, the results do not change because it would result in the same linear correlation coefficient.