Assignment 12 Solutions

# Assignment 12 Solutions - CE93 – Engineering Data...

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Unformatted text preview: CE93 – Engineering Data Analysis Prof. Joan Walker, Spring 2011 Assignment #12 – Final Exam Review: “Due” May 4 th , 2011 (This assignment will not be graded. Solutions will be posted on due date) Note: This final exam review can be treated as a practice test – it is to your benefit to attempt this assignment in the conditions of the actual final exam (i.e. no MATLAB, crib sheet only, etc.) Problem #1 The structural soundness of aircraft is determined using alignment checks, which involve measuring distances between different points on the aircraft and comparing them with specifications. When a measurement is outside of a prescribed tolerance, it is termed an “alignment error.” The number of alignment errors is related to the number of missing rivets on the aircraft. Data for four aircraft is presented below. Number of Missing Rivets, x Number of Alignment Errors, y 13 7 15 7 10 6 – changed value since initial posting 22 12 From the above data, the following quantities were calculated: ? = 15 ? ¡ = 8 ? ? = 5.1 ? ? = 2.7 ¢ ? ? 2 4 ? =1 = 978 ¢ ? ? 2 4 ? =1 = 278 ¢ ( ? ? − ? 4 ? =1 ) 2 = 78 = ££ ¢ ( ? ? − ? ¡ 4 ? =1 ) 2 = 22 = ?? ¢ ( ? ? − ? 4 ? =1 ) ∗ ¤? ? − ? ¡¥ = 40 = £? a) What are the OLS estimates of the slope ( ¦ ) and intercept ( § ) of the regression line relating x and y? Show your calculations. ¦ ¨ = ? ?? ? ?? = 0.5128 § © = ? ¡ − ¦ ¨ ? ¡ = 8 − 0.5128 ∗ 15 = 0.308 b) Estimate the variance of the residual of this model. (Variance of residual = Variance of ? ? ª since ? ? ª = § © + ¦ ¨ ? ? + 2 « = 2 « = ¬¬ ­ ® − 2 Where: CE93 – Engineering Data Analysis Prof. Joan Walker, Spring 2011 ?? ? = ? ?? ? ?? − ? ?? 2 ¡ ? ?? = 78 ∗ 22 − 40 2 78 = 1.487 Thus: 2 ¢ = 1.487 £ − 2 = 0.7435 c) How would you measure the “fit” of the OLS equation to your data? Perform the calculation. Measure “fit” using the coefficient of determination , R 2 (Note that this is different from R – the correlation coefficient) ? 2 = ? ?? − ?? ? ¡ ? ?? = 1 − 1.487 22 = 0.9324 d) What is the 95% confidence interval for the slope? ¤ ¥ −? ¦ 2 , £− 2 ≤ § ¨ − § © ?? ? £ − 2 ¡ ∗ ? ?? ≤ ? ¦ 2 , £− 2 ª « = 1 − ¦ ¤ ¥ − 4.303 ≤ 0.5128 − §¡ © 1.487 2 ∗ 78 ≤ 4.303 ª « = 0.95 0.5128 − 0.4201 ≤ § ≤ 0.5128 + 0.4201 ¡ = 0.95 0.0927 ≤ § ≤ 0.9329 ¡ = 0.95 Thus, the 95% CI for = [0.0.0927,0.9329] e) Test the hypothesis that § = 0.5 against the alternative hypothesis that § ≠ 0.5 , using a significance level of 0.05. Since § = 0.5 is inside the 95% confidence interval (therefore the 5% significance… ), we fail to reject the null hypothesis, cannot accept the alternate hypothesis....
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Assignment 12 Solutions - CE93 – Engineering Data...

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