Question

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

Answer 1

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

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

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