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weeeeeb [17]
3 years ago
6

Triangle ABC with vertices A(-4, 2), B(4,7), and

Mathematics
1 answer:
alex41 [277]3 years ago
4 0

Step-by-step explanation:

A(-4,2)=A'(-4,-2)

B(4,7)=B'(4,-7)

C(5,1)=C'(5,-1)

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The data below are the temperatures on randomly chosen days during the summer and the number of employee absences at a local com
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Answer:

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

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

<u>80 5 6400 25        400         </u>

<u>∑779 77 68163 813 6995</u>

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.

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6 0
3 years ago
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