Question

An emergency service wishes to see whether a relationship exists between the outside temperature and the number of emergency calls it receives for a 7-hour period. The data are shown. Emergency Calls and Temperatures Temperature x 68 74 82 88 93 99 101 No. of calls y 7 4 8 10 11 9 13

a. Describe the linear relationship between the temperature and the number of calls.

b. Calculate the correlation coefficient, r.

c. Is r statistically significant at the 0.05 level. Explain.

d. Determine the equation of the line of best fit.

e. Calculate and interpret the coefficient of determination, 2 r .

f. Predict the number of calls when the temperature is 80degrees.

g. Predict the temperature outside when the number of calls is 6.

h. Predict the number of calls when the temperature is 59 degrees

Answer 1

Data as below

Temperature (x)	No. of calls (y)
68	7
74	4
82	8
88	10
93	11
99	9
101	13

1. Linear relationship between variables

The equation has the form Y=a+bX, where Y is the dependent variable (that’s the variable that goes on the Y axis), X is the independent variable (i.e. it is plotted on the X axis), b is the slope of the line and a is the y-intercept

From the table we can calculate

Temperature (x)	No. of calls (y)		xy	x^2	y^2
68	7		476	4624	49
74	4		296	5476	16
82	8		656	6724	64
88	10		880	7744	100
93	11		1023	8649	121
99	9		891	9801	81
101	13		1313	10201	169
605	62		5535	53219	600

We can say that

Σx	605
Σy	62
Σxy	5535
Σx^2	53219
Σy^2	600

using the formula as below to calculate a and b

In our case n = 7 (sample size)

a = [(62)(53219) - (605)(5535)] / [(7)(53219) - (605)^2]

Σy * Σx^2	3299578
Σx * Σxy	3348675
n* Σx^2	372533
Σx ^ 2	366025
a	-7.5

b = [(7)(5535) - (605)(62)] / [(7)(53219) - (605)^2]

n* Σxy	38745
Σx * Σy	37510
n* Σx^2	372533
Σx ^ 2	366025
b	0.19

Hence the regression equation is

y = a + b x

y = -7.5 + 0.19 x

Hence we can say that

No. of calls = -7.5 + 0.19 Temperature

2. To calculate r

Formula is

Putting the values in the formula

n* Σxy	38745
Σx * Σy	37510
n* Σx^2	372533
Σx ^ 2	366025
n*Σy^2	4200
Σy ^2	3844
r	0.811357
R^2	65.8%

Calculating ANOVA Table

Using below formula for ANOVA table

Source of Variation	DF	SS	MS	F
Regression	1	SSR=∑ni=1(y^i−y¯)2	MSR=SSR/1	F∗=MSR/MSE
Residual error	n-2	SSE=∑ni=1(yi−y^i)2	MSE=SSE/n−2
Total	n-1	SSTO=∑ni=1(yi−y¯)2

Looking for p-value using F value

Analysis of Variance

Source	DF	SS	MS	F	P
Regression	1	33.4802	33.4802	9.63	0.027
Error	5	17.3769	3.4754
Total	6	50.8571

P-value = 0.027

Since P value is less than the level of significance (0.05)

A low p-value (< 0.05) indicates that, a predictor that has a low p-value is likely to be a meaningful addition to your model because changes in the predictor's value are related to changes in the response variable. Conversely, a larger (insignificant) p-value suggests that changes in the predictor are not associated with changes in the response

Hence we can say that the regression model is significant