Question

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Answer 1

Answer to 1^st Question:
The correct scatter – plot of the data is given in Option A

Answer to 2^nd Question:
Let Overhead weight be represented by x and Weight by y

The following table shows all the calculations –

	Overhead Width, (x)	Weight (y)	x^2	y^2	xy
	7.2	115	51.84	13225	828
	7.6	187	57.76	34969	1421.2
	9.6	230	92.16	52900	2208
	9.3	201	86.49	40401	1869.3
	8.8	202	77.44	40804	1777.6
	8.2	183	67.24	33489	1500.6
Total	50.7	1118	432.93	215788	9604.7

Total number of observations, n = 6

Mean of x, $\bar{x}$ = 50.7/6 = 8.45

Mean of y, $\bar{y}$ = 1118/6 = 186.3333

Standard Deviation of x, Sx = {($\sum$(x^2)/n) - ($\bar{x}$^2)}^0.5 = {(432.93/6) – (8.45^2)}^0.5 = 0.8675

Standard Deviation of y, Sy = {($\sum$(y^2)/n) - ($\bar{y}$^2)}^0.5 = {(215788/6) – (186.3333^2)}^0.5 = 35.2784

Covariance between x and y, Cov.(x, y) = ($\sum$xy / n) - ($\bar{x}$$\bar{y}$) = (9604.7/6) – (8.45 x 186.3333) = 26.2669

Correlation Coefficient, r = Cov.(x, y) / (Sx. Sy) = 26.2669 / (0.8675 x 35.2784) = 0.858

Answer to 3^rd Question:
To test for significance of the linear correlation, we can perform a hypothesis test,

The null hypothesis being the two variables have 0 correlation, H0: $\rho$ = 0

the alternative hypothesis the two variables are correlated, Ha: $\rho$ ≠ 0

(where $\rho$ - Population Correlation Coefficient)

This is a two – sided test

Assuming that the data has been taken from a Bivariate Normal Distribution,

The test statistic for this test would be t = [r x (n - 2) ^0.5]/ (1 – (r^2)) ^0.5, the test statistic t follows the t – distribution with (n - 2) degrees of freedom under the null hypothesis

Substituting all values, we get t = 3.341

The level of significance, $\alpha$ = 1% = 0.01

To find the critical values, we need to consider ($\alpha$/2) = 0.005

The Critical Value, t(0.005, 4) = 4.604

Now, if the value of the test statistic is greater than the critical value, we reject the null hypothesis else we fail to reject the null hypothesis

Here, value of the test statistic < Critical Value

Hence, we fail to reject the null

Because the absolute value of the test statistic is lesser than the positive critical value, there is no sufficient evidence to support the claim that there is linear correlation between the Overhead Width of Seals from photographs and the Weights of the Seals at significance level $\alpha$ = 0.01