x |
54 |
54 |
61 |
61 |
68 |
68 |
75 |
75 |
75 |
y |
16.47 |
18.69 |
14.3 |
15.12 |
13.51 |
11.64 |
11.17 |
12.53 |
11.22 |
In order to solve this question I used R software.
R codes and output:
> x=c(54,54,61,61,68,68,75,75,75)
> y=c(16.47,18.69,14.3,15.12,13.51,11.64,11.17,12.53,11.22)
> fit=lm(y~x)
> summary(fit)
Call:
lm(formula = y ~ x)
Residuals:
Min 1Q Median 3Q Max
-1.56361 -0.61194 -0.04444 0.30639 1.60806
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 32.04123 2.88431 11.109 1.07e-05 ***
x -0.27702 0.04359 -6.355 0.000384 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 1.057 on 7 degrees of freedom
Multiple R-squared: 0.8523, Adjusted R-squared: 0.8312
F-statistic: 40.38 on 1 and 7 DF, p-value: 0.0003836
a.
Estimated linear regression equation:
Y = 32.04123 - 0.27707 X
b.
Hypothesis:
c.
Test statistic,
Degrees of freedom = n-1 = 8-1 = 7
Critical value = -1.895
Rejection region = Region below -1.895
Since calculated value fall in the rejection region, we reject null hypothesis . And conclude that slope coefficient is significantly negative.
d.
Hypothesis:
95 % confidence interval:
Since above confidence contain values less than -0.1 also, hence we accept null hypothesis and conclude that slope β is significantly lesser than -0.1.
e.
Hypothesis:
95 % confidence interval:
Since above confidence contain values less -0.1, hence we reject null hypothesis and conclude that slope β is significantly lesser than -0.1.
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