Mathematics, 10.12.2020 16:10 manarsadi6

Consider the following data on x = rainfall volume (m^3) and y = runoff volume (m^3) for a particular location. x 4 12 14 20 23 30 40 48 55 67 72 85 96 112 127
y 4 10 13 15 15 25 27 44 38 46 53 71 82 99 101

Required:
Calculate a 95% confidence interval for the true average change in runoff volume associated with a 1m^3 increase in rainfall volume. (Round your answers to three decimal places.)

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Answer from: 4804053670

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

$\sum _{xi} = 805$ $\sum_{yi} = 643$ $\sum_{x_iy_I}= 51715$ $\sum_{x_i^2}= 63901$ $\sum_{y_i^2} = 42161$

The least-square regression equation is: $\hat y = b_o+b_1 x$

$b_1 = \dfrac{n \sum xy - ( \sum _x) ( \sum_y)}{n \sum x^2 - ( \sum x)^2}$

$b_1 = \dfrac{15(51715) - (803) (643)}{15(63901)-(805)^2}$

$b_1 = \dfrac{775725-516329}{958515-648025}$

$b_1 = \dfrac{259396}{310490}$

b₁ = 0.835440

∴ Slope term, b₁ = 0.835

$SS_{XX} = \sum x^2 - \dfrac{(\sum x)^2}{n}= 63901 - \dfrac{(805)^2}{15}=20699.33$

$SS_{yy} = \sum y^2 - \dfrac{(\sum y)^2}{n }= 42161 - \dfrac{(643)^2}{15}= 14597.73$

$SS_{xy} = \sum xy - \dfrac{(\sum x) (\sum y)}{n}= 51715 - \dfrac{(803)(643)}{15}= 17293.067$

$SST = SS_{yy}= 14597.73$

$SSR = \dfrac{SS_{xy}^2}{SS_{xx}}= \dfrac{17293.067^2}{20699.33}=14447.34$

SSE = SST - SSR = 14597.73 - 14447.34 = 150.39

The hypothesis test for the significance of $\beta_1$ is:

$H_o: \beta_1 = 0 \\ \\ H_1: \beta_1 \ne 0$

Significance level ∝ = 1 - 0.95 = 0.05

The sample slope b_1 = 0.835440

$Test \ statistic = t_{observed} = \dfrac{b_1-0}{\sqrt{\dfrac{SSE}{(n-2)SS_{xx}}}}$

$t_{o} = \dfrac{0.835440-0}{\sqrt{\dfrac{150.39}{(15-2)20699.33}}}$

$t_{o} = \dfrac{0.835440}{\sqrt{\dfrac{150.39}{269091.29}}}$

t_o = 35.339

Degree of freedom df = n - 2

df = 15 -2

df = 13

Using the Excel formula to determine the P_value.

$P-value = P(t, \Big|35.339 \Big|)$

P-value = 2 × t.dist(35.339,13,1)

P-value = 0.0000

P-value = 0

Critical value: $t_{critical} = t_{\alpha/2,df} = t_{0.05/2,13}= 2.160$

Rejection region: To reject H_o ; if $\Big | t_o \Big | t_c$

Decision: Since $\Big | t_o \Big | t_c$ ; we reject H_o

Conclusion: There is enough evidence to conclude that the linear relationship between x & y

Thus; we reject H_o & there is a useful linear relationship between x & y.

The 95% C.I for slope is given by the equation:

$=b_1 \pm t_{(\alpha/2,n-2)} \sqrt{\dfrac{SSE}{n-2}}\sqrt{\dfrac{1}{SS_{xx}} }$

$=0.835440 \pm 2.160 \sqrt{\dfrac{150.39}{15-2}}\sqrt{\dfrac{1}{20699.33} }$

= 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 = $\mathbf{( 0.784 < \beta_1 < 0.887) \ to \ 3 \ d.p}$

Answer from: Quest

The linear equation is: y=ax+b

Answer from: Quest

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