Question

... Use the given data set to complete parts (a) through (C) below. (Use a = 0.05.) X 6 7 10 7.45 8 6.76 13 12.74 9 7.11 11 7.81 14 8.84 4 5.39 12 8.14 5 5.72 y 6.08 6.41 Click here to view a table of critical values for the correlation coefficient. a. Construct a scatterplot. Choose the correct graph below. OA. OB. OC. OD 16- AY 16- AY 167 AY 16 12- 12 12 a 12- 8- 8- 8- 8- 4- 4- 4- 4- X 0- 0- 0 0- 4 8 12 16 12 16 4 12 16 . -00 12 16 b. Find the linear correlation coefficient, r, then determine whether there is sufficient evidence to support the claim of a linear correlation between the two variables. The linear correlation coefficient is r=0 (Round to three decimal places as needed.) Using the linear correlation coefficient found in the previous step, determine whether there is sufficient evidence to support the claim of a linear correlation between the two variables. Choose the correct answer below. O A. There is insufficient evidence to support the claim of a nonlinear correlation between the two variables.

Answer 1

(a)

Following is the scatter plot of data:

14 12 10 8 Y 6 4 N 0 0 2 4 6 8 10 12 14 16 X

Correct option is A.

(b)

Following table shows the calculations:

	X	Y	X^2	Y^2	XY
	10	7.45	100	55.5025	74.5
	8	6.76	64	45.6976	54.08
	13	12.74	169	162.3076	165.62
	9	7.11	81	50.5521	63.99
	11	7.81	121	60.9961	85.91
	14	8.84	196	78.1456	123.76
	6	6.08	36	36.9664	36.48
	4	5.39	16	29.0521	21.56
	12	8.14	144	66.2596	97.68
	7	6.41	49	41.0881	44.87
	5	5.72	25	32.7184	28.6
Total	99	82.45	1001	659.2861	797.05

Sample size: n = 11 From table we have: 2 SS=Sxx = -ΣΑ 1001- ((99)^2/11) = 110 2 SST = Syy y2 n 659.2861- (82.45)^2/11) = 41.285873 * SP=S7 = xy n = 797.05- (82.45 * 99) /11) = 55 The correlation coefficient is: S 1 = = 55/ sqrt( 110 * 41.285873) S.xxSyy 1' = 0.816

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Following is the test for correlation:

Here we have 1 = 0.816, n = 11 Hypotheses are: H: P = 0 Ha:Po Level of significance: a =0.05 Test is two tailed. Degree of freedom: df =n-2=9 Test Statistics: 1 = 0.816/sqrt ((1 -(0.816)^2)/(11-2)) (1–-2)/(1-2) = 4.235 Critical value: Rejection Region: 2.262 Ift<-2.262 or t > 2.262, Reject HO Decision: Since test statistics lies in rejection region so we reject the null hypothesis P-value: P-value=0.0022 Decision: Since p-value is less than level of significance so we reject the null hypothesis Excel function for critical value: ETINV(0.05,9) Excel function for p-value: =TDIST(4.235,9,2)

Correct option is D.

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c:

Correct option is C.