The data below are the temperatures on randomly chosen days during the summer and the number of employee absences at a local company on those days. Construct a 95% prediction interval for y, the number of days absent,

given x = 95 degrees and y= 0.449x - 30.27

Temperature, x 72 85 91 90 88 98 75 100 80
Number of absences, y 3 7 10 10 8 15 4 15 5

Since the calculated value of t= 18.50249 falls in the critical region we reject the null hypothesis and conclude that there is sufficient reason to support the claim of a linear relationship between the two variables.

Step-by-step explanation:

We set up our hypotheses as

H0: β= 0 the two variable X and Y are not related

Ha: β ≠ 0. the two variables X and Y are related.

The significance level is set at α =0.05

The test statistic if, H0 is true, is t= b/s_b

Where Sb =S_yx/√(∑(X-X`)^2 )

Syx = √((∑(Y-Y`)^2 )/(n-2))

In the given question we have the estimated regression line as y= 0.449x - 30.27

X Y X2 Y2 XY

72 3 5184 9 216

85 7 7225 49 595

91 10 8281 100 910

90 10 8100 100 900

88 8 7744 64 704

98 15 9604 225 1470

75 4 5625 16 300

100 15 10000 225 1500

80 5 6400 25 400

∑779 77 68163 813 6995

Now finding the variances

∑(Y-Y`)^2 = ∑〖Y^2- a〗 ∑Y- b∑XY

= 813 – (- 30.27)77 - 0.449(6995)

= 813+2330.79 – 3140.755

= 3.035

∑(X-X`)^2 = ∑X^2 – (∑〖X)〗^2 /n

= 68163 – (779)2/9

= 736.22

Syx = √((∑(Y-Y`)^2 )/(n-2)) = √(3.035/7) = 0.65846 and

Sb =S_yx/√(∑(X-X`)^2 ) = (0.65846 )/27.13337 = 0.024267

t= b/s_b = 0.449/ 0.024267 = 18.50249

The critical region is t ≥ t(0.025, 7) = 2.365

Since the calculated value of t= 18.50249 falls in the critical region we reject the null hypothesis and conclude that there is sufficient reason to support the claim of a linear relationship between the two variables.