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

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.

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

XYX2         Y2      XY

72351849    216

857722549    595

91108281100      910

90108100100      900

888774464       704

98159604225      1470

754562516       300

1001510000225       1500

805640025        400        

∑77977681638136995

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.

step-by-step explanation:

compare

v = 1/3 a^2h

step-by-step explanation:

if the height is 57.7cm and the height of the slant edge is 18.2cm, then the volume is 1/3 a^2h.