How to calculate this question ? 2.15. The following data give the monthly machi
ID: 3355746 • Letter: H
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How to calculate this question ?
2.15. The following data give the monthly machine maintenance cost (y) in hundreds of dollars and the number of machine hours per month (x), taken over the last 7 months. Cost (y Hours (x) 26 25 20 18 30 40 110 98 121 116 90 84 a. Fit a linear regression. Construct the ANOVA table. Find R and test the hypothesis that A =0 using the F ratio. b. Obtain the standard errors of and A . Using the t distribution, test the hypothesis: (i) A) = 0; (ii) A = 0. Construct a 99% confidence interval for Bi c. Find the fitted value at x = 100 and estimate its standard error. Calculate the 95% confidence interval for Ao + 00. d. Repeat (c), only this time take x = 84. Explain the change in the interval length.Explanation / Answer
a)
The linear regression equation of y on x is estimated as,
Y = 68.4459 + -0.4104x.
R-squared = 0.6767. The ANOVA table of the regression analysis is given below:
Source Df Sum Sq Mean Sq F value Pr(>F)
x 1 217.9 217.90 10.47 0.0231 *
Residuals 5 104.1 20.82
From the above test it is evident that the regression of y on x is significant at 5% level of significance and hence beta_1 (coefficient of x) is significant.
b) The estimates and the standard errors are given below. Also the t-statistic (estimate/s.e.) and the corresponding p-value are also listed in the following table:
Coefficients Estimate Std. Error t value Pr(>|t|)
Beta_0 68.4459 12.9270 5.295 0.00321 **
Beta_1 -0.4104 0.1268 -3.235 0.02308 *
Hence the hypothesis beta_0 = 0 is rejected at 1% level of significance and the hypothesis beta_1=0 is significant at 5% level of significance.
The 99% confidence interval for beta_1 is given by,
[ -0.4104 – 3.25*0.1268 / sqrt(10), -0.4104 + 3.25*0.1268 / sqrt(10) ] = [-0.5407175, -0.2800825 ].
c) The fitted value at x=100 is 68.4459-0.4104*100 = 27.4059.
The standard error of the fitted value is given by,
sqrt(sy*(1+1/n+(x1-mx)^2/((n-1)*sx))), where sy & sx is the variances of y and x respectively, mx is the mean of x and the the value is calculated as 7.362462. The 95% CI for the estimate is,
[ est - 2.26*7.36246/sqrt(10), est + 2.26*7.36246/sqrt(10) ] = [ 22.14413, 32.66767].
d) The estimatedvalue at x=84 is 33.9723 and the standard error is 7.411461.
The 95% CI is given by the interval [ 28.67552, 39.26908 ].
The length of the interval is slightly wider in the second case since mean of x (101) is much nearer to 100 than that of 84.
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