Respuesta :
Answer:
Step-by-step explanation:
From the given information:
x y xy x² y²
4 4 16 16 16
12 10 120 144 100
14 13 182 196 169
20 15 300 400 225
23 15 345 529 225
30 25 750 900 625
40 27 1080 1600 729
48 44 2112 2304 1936
55 38 2090 3025 1444
67 46 3082 4489 2116
72 53 3816 5184 2809
85 71 6035 7225 5041
96 82 7872 9216 6724
112 99 11088 12544 9801
127 101 12827 16129 10201
[tex]\sum _{xi} = 805[/tex] [tex]\sum_{yi} = 643[/tex] [tex]\sum_{x_iy_I}= 51715[/tex] [tex]\sum_{x_i^2}= 63901[/tex] [tex]\sum_{y_i^2} = 42161[/tex]
The least-square regression equation is: [tex]\hat y = b_o+b_1 x[/tex]
[tex]b_1 = \dfrac{n \sum xy - ( \sum _x) ( \sum_y)}{n \sum x^2 - ( \sum x)^2}[/tex]
[tex]b_1 = \dfrac{15(51715) - (803) (643)}{15(63901)-(805)^2}[/tex]
[tex]b_1 = \dfrac{775725-516329}{958515-648025}[/tex]
[tex]b_1 = \dfrac{259396}{310490}[/tex]
b₁ = 0.835440
∴ Slope term, b₁ = 0.835
[tex]SS_{XX} = \sum x^2 - \dfrac{(\sum x)^2}{n}= 63901 - \dfrac{(805)^2}{15}=20699.33[/tex]
[tex]SS_{yy} = \sum y^2 - \dfrac{(\sum y)^2}{n }= 42161 - \dfrac{(643)^2}{15}= 14597.73[/tex]
[tex]SS_{xy} = \sum xy - \dfrac{(\sum x) (\sum y)}{n}= 51715 - \dfrac{(803)(643)}{15}= 17293.067[/tex]
[tex]SST = SS_{yy}= 14597.73[/tex]
[tex]SSR = \dfrac{SS_{xy}^2}{SS_{xx}}= \dfrac{17293.067^2}{20699.33}=14447.34[/tex]
SSE = SST - SSR = 14597.73 - 14447.34 = 150.39
The hypothesis test for the significance of [tex]\beta_1[/tex] is:
[tex]H_o: \beta_1 = 0 \\ \\ H_1: \beta_1 \ne 0[/tex]
Significance level ∝ = 1 - 0.95 = 0.05
The sample slope [tex]b_1[/tex] = 0.835440
[tex]Test \ statistic = t_{observed} = \dfrac{b_1-0}{\sqrt{\dfrac{SSE}{(n-2)SS_{xx}}}}[/tex]
[tex]t_{o} = \dfrac{0.835440-0}{\sqrt{\dfrac{150.39}{(15-2)20699.33}}}[/tex]
[tex]t_{o} = \dfrac{0.835440}{\sqrt{\dfrac{150.39}{269091.29}}}[/tex]
[tex]t_o = 35.339[/tex]
Degree of freedom df = n - 2
df = 15 -2
df = 13
Using the Excel formula to determine the P_value.
[tex]P-value = P(t, \Big|35.339 \Big|)[/tex]
P-value = 2 × t.dist(35.339,13,1)
P-value = 0.0000
P-value = 0
Critical value: [tex]t_{critical} = t_{\alpha/2,df} = t_{0.05/2,13}= 2.160[/tex]
Rejection region: To reject [tex]H_o[/tex]; if [tex]\Big | t_o \Big | > t_c[/tex]
Decision: Since [tex]\Big | t_o \Big | > t_c[/tex]; we reject [tex]H_o[/tex]
Conclusion: There is enough evidence to conclude that the linear relationship between x & y
Thus; we reject [tex]H_o[/tex] & there is a useful linear relationship between x & y.
The 95% C.I for slope is given by the equation:
[tex]=b_1 \pm t_{(\alpha/2,n-2)} \sqrt{\dfrac{SSE}{n-2}}\sqrt{\dfrac{1}{SS_{xx}} }[/tex]
[tex]=0.835440 \pm 2.160 \sqrt{\dfrac{150.39}{15-2}}\sqrt{\dfrac{1}{20699.33} }[/tex]
= 0.835440 ± 2.160 (3.40124)(0.006951)
= 0.835440 ± 0.0511
= (0.835440 - 0.0511, 0.835440 + 0.0511)
= (0.78434, 0.88654)
= (0.784, 0.887) to three decimal places.
∴ 95% C.I of slope = [tex]\mathbf{( 0.784 < \beta_1 < 0.887) \ to \ 3 \ d.p}[/tex]
