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antoniya [11.8K]

2 years ago

9

Solve for X. show your work. (x+2)(x+8)=0

1 answer:

valentinak56 [21]2 years ago

3 0

Step-by-step explanation:

x = -2

x = -8

please follow me

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Consider the following data on x = rainfall volume (m^3) and y = runoff volume (m^3) for a particular location.

nydimaria [60]

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

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

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

6 0

3 years ago