Question

An aircraft company wanted to predict the number of worker-hours necessary to finish the design of a new plane. Relevant explanatory variables were thought to be the plane’s top speed, its weight, and the number of parts it had in common with other models built by the company. A sample of 27 of the company’s planes was taken, and the following model was estimated: y = b0 + b1x1 + b2x2 + b3x3 + e where y = design effort, in millions of worker-hours x1 = plane’s top speed, in miles per hour x2 = plane’s weight, in tons x3 = percentage of parts in common with other models The estimated regression coefficients were as follows: b1 = 0.661 b2 = 0.065 b3 = -0.018 The estimated standard errors were as follows: sb1 = 0.099 sb2 = 0.032 sb3 = 0.0023 a. Find 90% and 95% confidence intervals for b1. b. Find 95% and 99% confidence intervals for b2. c. Test against a two-sided alternative the null hypothesis that, all else being equal, the plane’s weight has no linear influence on its design effort. d. The error sum of squares for this regression was 0.332. Using the same data, a simple linear regression of design effort on the percentage of common parts was fitted, yielding an error sum of squares of 3.311. Test, at the 1% level, the null hypothesis that, taken together, the variable’s top speed and weight contribute nothing in a linear sense to explaining the changes in the variable, design effort, given that the variable percentage of common parts is also used as an explanatory variable.

Answer 1

The model to be estimated is

$y = \beta_0 + \beta_1 x_1 + \beta_2 x_2 + \beta_3 x_3 + \epsilon$

the number of observations

n=27

The number of independent variables

k=3

The estimated regression coefficients were as follows:

The estimated standard errors were as follows:

a) the level of significance for 90% confidence interval is $\alpha=1=90/100=0.1$

The upper critical value of t is obtained using $P(T>t_{\alpha.2})=\alpha/2=0.1/2=0.05$

the degrees of freedom is n-k-1=27-3-1=23

Using t table for df=23 and right tail area = 0.05 we get $t_{\alpha/2}=1.714$

The 90% confidence interval for b1 is

$\begin{align*} &b_1\pm t_{\alpha/2}s_{b_1}\\ \implies &0.661\pm 1.714\times 0.099\\ \implies &[0.491, 0.831] \end{align*}$

ans: The 90% confidence interval for b1 is [0.491,0.831]

the level of significance for 95% confidence interval is $\alpha=1=95/100=0.05$

The upper critical value of t is obtained using $P(T>t_{\alpha.2})=\alpha/2=0.05/2=0.025$

the degrees of freedom is n-k-1=27-3-1=23

Using t table for df=23 and right tail area = 0.025 we get $t_{\alpha/2}=2.069\\$

The 95% confidence interval for b1 is

$\begin{align*} &b_1\pm t_{\alpha/2}s_{b_1}\\ \implies &0.661\pm 2.069\times 0.099\\ \implies &[0.456,0.866] \end{align*}$

ans: The 95% confidence interval for b1 is [0.456,0.866]

b) The 95% confidence interval for b2 is

$\begin{align*} &b_2\pm t_{\alpha/2}s_{b_2}\\ \implies &0.065\pm 2.069\times 0.032\\ \implies &[-0.001, 0.131] \end{align*}$

ans: The 95% confidence interval for b2 is [-0.001,0.131]

the level of significance for 99% confidence interval is $\alpha=1=99/100=0.01$

The upper critical value of t is obtained using $P(T>t_{\alpha.2})=\alpha/2=0.01/2=0.005$

the degrees of freedom is n-k-1=27-3-1=23

Using t table for df=23 and right tail area = 0.005 we get $t_{\alpha/2}=2.807\\$

The 99% confidence interval for b2 is

$\begin{align*} &b_2\pm t_{\alpha/2}s_{b_2}\\ \implies &0.065\pm 2.807\times 0.032\\ \implies &[-0.025,0.155] \end{align*}$

ans: The 99% confidence interval for b2 is [-0.025,0.155]

c) the plane’s weight (X2) has no linear influence on its design effort (Y) if the coefficient of X2, is zero, that means   $\begin{align*} \beta_2=0 \end{align*}$

We want to test the following hypotheses

$\begin{align*} &H_0:\beta_2=0\leftarrow\text{null hypothesis: the plane weight has no linear influence on design effort}\\ &H_a:\beta_2\ne 0\leftarrow\text{alternative hypothesis: the plane weight has linear influence on design effort}\\ &\alpha=0.05\leftarrow\text{level of significance to test the hypotheses: Assumed}\\ \end{align*}$

The hypothesized value of the slope is coefficient for X2 is $\begin{align*} \beta_{2H_0}=0 \end{align*}$ (from the null hypothesis)

This is a 2 tailed test (The alternative hypothesis has "not equal to"). We can use the 95% confidence interval to test the hypotheses.

The 95% confidence interval for b2 is [-0.001,0.131]. This interval contains the hypothesized value $\begin{align*} \beta_{2H_0}=0 \end{align*}$ . Hence we do not reject the null hypothesis.

ans: We conclude that at 5% level of significance, all else being equal, the plane’s weight has no linear influence on its design effort

d) the full regression model is

$y = \beta_0 + \beta_1 x_1 + \beta_2 x_2 + \beta_3 x_3 + \epsilon$

The error sum of squares for this full regression is

$\begin{align*} SSE_f= 0.332 \end{align*}$

The number of independent variables k=3

The degrees of freedom is $\begin{align*} df_f= n-k-1=27-3-1=23 \end{align*}$

The reduce model is a simple linear regression of design effort (Y) on the percentage of common parts (X3)

$y = \beta_0 + \beta_3 x_3 + \epsilon$

The error sum of squares for this reduced model is

$\begin{align*} SSE_r= 3.311 \end{align*}$

The number of independent variables k=1

The degrees of freedom is $\begin{align*} df_r= n-k-1=27-1-1=25 \end{align*}$

We want to test the following hypotheses

$\begin{align*} &H_0:\beta_1=\beta_2=0\leftarrow\text{null hypothesis: The variables X1,X2 have no linear influence on Y, given that X3 is used}\\ &H_a:\text{at least one of }\beta_1,\beta_2\ne 0\leftarrow\text{alternative hypothesis: At least one of variables X1,X2 have linear influence on Y, given that X3 is used}\\ &\alpha=0.01\leftarrow\text{level of significance to test the hypotheses}\\ \end{align*}$

The test statistics for this partial F test is

$\begin{align*} F=\frac{(SSE_r-SSE_f)(df_r-df_f)}{SSE_f/df_f}=\frac{(3.311-0.332)/(25-23)}{0.332/23}=103.188 \end{align*}$

the numerator degrees of freedom is 25-23=2 and the denominator df is 23

From the F table using alpha=0.01 and the numerator degrees of freedom is 25-23=2 and the denominator df is 23 we get the critical value = 5.66

We will reject the null hypothesis if the test statistics is greater than the critical value.

Here, The test statistics is 103.188 and it is greater than the critical value 5.66. Hence we reject the null hypothesis.

We conclude that at the 1% level, taken together, at least one of the variables top speed and weight contribute in a linear sense to explaining the changes in the variable, design effort, given that the variable percentage of common parts is also used as an explanatory variable.