Question

(a) For each of 12 regions the following table relates the divorce rate (number of divorces per 100 married couples over a year) to an affluence measure (average house value). Note: Summary statistics for these data are shown below the data table. House price (£000s), x 100 130 180 210 220 250 Divorce rate, y 1.50 2.14 1.98 2.80 2.43 3.01 House price (£000s), X Divorce rate, y 250 253 290 4.02 2.50 3.80 295 350 360 4.30 4.15 4.40 Summary statistics for these data: x = 2888, 9 r2 = 764034, y = 37.03, y2 = 125.4779, X xy = 9704.2 i. Using the graph paper provided, draw a scatter diagram for these data. ii. Calculate the correlation coefficient. Is this a high value? iii. Comment on the relationship between 'House price' and 'Divorce rate'. iv. Compute the best fitting straight line for these data and draw this line on your scatter diagram. Label it carefully. v. What would you expect the divorce rate to be in a region with a house price of £150,000?

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Answer 1

Solution:The given data is as follows:

House price(x)	Divorce rate (y)
100	1.5
130	2.14
180	1.98
210	2.8
220	2.43
250	3.01
250	4.02
253	2.5
290	3.8
295	4.3
350	4.15
360	4.4

i)The scatter diagram for the above data is as below.

Let us consider house price on x-axis and divorce rate on y-axis.

Scatter plot 5 4.5 4. 3.5 3 Divorce rate 2.5 2 1.5 1 0.5 0 0 50 100 150 200 250 300 350 400 House price(£000s)

ii)Given:$\sum$x=2888,$\sum$x²=764034,$\sum$y=37.03,$\sum$y²=125.4779,$\sum$xy=9704.2 and n=12

The correlation coefficient formula is given below:

Γ = n(Σxy) - (Σx)(Σy) ν[nΣx? - (Σx)?][ηΣy? - (Σy)2]

=[(12*9704.2)-(2888*37.03)] / $\sqrt{}$ [(12*764034)-2888²] [(12*125.4779)-37.03²]

=9507.76/10552.69

=0.90098

The correlation coefficient is 0.90098.

iii)Since r=0.90098, we can say that there is a strong positive correlation between the two variables house price and divorce rate.

iv)Let us assume y=a+bx be the linear equation.

where y=dependent variable, a=intercept,b=slope and x=independent variable.

Let us find the value of 'a' and 'b' using the below formula.

Σν) Σx25- Σx)Σχy) n (Σ.x2) - (Σκ)2 n(Σχν) (Σχ)(Σ) n (Σx2) - (Σκ)?

a=[(37.03*764034)-(2888*9704.2)] / [(12*764034)-2888²]

=266449.42/827864

=0.3219

b=[(12*9704.2)-(2888*37.03)]/ [(12*764034)-2888²]

=9507.76/827864

=0.0115

So the best fit line is given by y'=0.3219+0.0115x

The best fit line on the scatterplot is as shown below.

Scatter plot 5 4.5 y=0.0115x + 0.3219 4. 3.5 3 Divorce rate 2.5 2 1.5 1 0.5 0 0 50 100 150 200 250 300 350 400 House price(£000s)

v)Given:x=150

We need to predict the value of y' using the regression equation y'=0.3219+0.0115x

Substituting x=150, we have y'=0.3219+(0.0115*150)=2.0469$\approx$2.05

2.05 is the expected divorce rate in a region with a house price of $\pounds$ 150,000.