14-2 The least-squares regression model is given by where y i is the value of the response variable for the i th individual 0 and 1 are the parameters to be estimated based on sample data x i is the value of the explanatory variable for the i th individual i is a random error term with mean 0 and variance, the error terms are independent and normally distributed. i=1,…,n, where n is the sample size (number of ordered pairs in the data set)
Formulas for the slope and intercept estimates. For the estimated regression equation given by the formula: The slope b 1 is calculated by: And the intercept b 0 can be found with :
14-4 The standard error of the estimate, s e, is found using the formula
14-5 Parallel Example 2: Compute the Standard Error Compute the standard error of the estimate for the drilling data which is presented on the next slide.
14-7 Solution Step 1: Using technology (i.e. Minitab), we find the least squares regression line to be Step 2, 3: The predicted values as well as the residuals for the 12 observations are given in the table on the next slide
Conclusion: We have insufficient evidence at the 5% level of significance to support the claim that the residual errors from this model are not normally distributed.
14-14 Hypothesis Test Regarding the Slope Coefficient, 1 To test whether two quantitative variables are linearly related, we use the following steps provided that 1.the sample is obtained using random sampling. 2.the residuals are normally distributed with constant error variance.
14-15 Step 1: Determine the null and alternative hypotheses. The hypotheses can be structured in one of three ways: Step 2: Select a level of significance, , depending on the seriousness of making a Type I error. Two-tailedLeft-TailedRight-Tailed H 0 : 1 = 0 H 1 : 1 0H 1 : 1 < 0H 1 : 1 > 0
14-16 Step 3: Compute the test statistic which follows Student’s t-distribution with n-2 degrees of freedom. Remember, when computing the test statistic, we assume the null hypothesis to be true. So, we assume that 1 =0.
14-17 Step 4: Use Table VI to estimate the P-value using n-2 degrees of freedom. P-Value Approach
14-23 CAUTION! Before testing H 0 : 1 = 0, be sure to draw a residual plot to verify that a linear model is appropriate.
14-24 Parallel Example 5: Testing for a Linear Relation Test the claim that there is a linear relation between drill depth and drill time at the = 0.05 level of significance using the drilling data.
14-25 Solution Verify the requirements: We assume that the experiment was randomized so that the data can be assumed to represent a random sample. In Parallel Example 4 we confirmed that the residuals were normally distributed by constructing a normal probability plot. To verify the requirement of constant error variance, we plot the residuals against the explanatory variable, drill depth.
14-27 Solution Step 1: We want to determine whether a linear relation exists between drill depth and drill time without regard to the sign of the slope. This is a two-tailed test with H 0 : 1 = 0 versus H 1 : 1 0 Step 2: The level of significance is = 0.05. Step 3: Using technology, we obtained an estimate of 1 in Parallel Example 2, b 1 =0.0116. To determine the standard deviation of b 1, we compute. The calculations are on the next slide.
14-29 Solution Step 3, cont’d: We have The test statistic is
14-30 Solution: P-Value Approach Step 4: Since this is a two-tailed test, the P-value is the sum of the area under the t-distribution with 12- 2=10 degrees of freedom to the left of -t 0 = -3.867 and to the right of t 0 = 3.867. Using Table VI we find that with 10 degrees of freedom, the value 3.867 is between 3.581 and 4.144 corresponding to right-tail areas of 0.0025 and 0.001, respectively. Thus, the P-value is between 0.002 and 0.005. Step 5: Since the P-value is less than the level of significance, 0.05, we reject the null hypothesis.
14-31 Solution Step 6: There is sufficient evidence at the = 0.05 level of significance to conclude that a linear relation exists between drill depth and drill time.
14-32 Confidence Intervals for the Slope of the Regression Line A (1- ) 100% confidence interval for the slope of the true regression line, 1, is given by the following formulas: Lower bound: Upper bound: Here, t /2 is computed using n-2 degrees of freedom.
14-33 Note: The confidence interval formula for 1 can be computed only if the data are randomly obtained, the residuals are normally distributed, and there is constant error variance.
14-34 Parallel Example 7: Constructing a Confidence Interval for the Slope of the True Regression Line Construct a 95% confidence interval for the slope of the least-squares regression line for the drilling example.
14-35 Solution The requirements for the usage of the confidence interval formula were verified in previous examples. We also determined b 1 = 0.0116 in previous examples.
14-36 Solution Since t 0.025 =2.228 for 10 degrees of freedom, we have Lower bound = 0.0116-2.228 0.003=0.0049 Upper bound = 0.0116+2.228 0.003=0.0183. We are 95% confident that the mean increase in the time it takes to drill 5 feet for each additional foot of depth at which the drilling begins is between 0.005 and 0.018 minutes.
The Coefficient of Determination The Coefficient of Determination is the proportion of the variability in the response variable that can be attributed to the least squares regression model.
How to calculate R 2 Using the sum of squares technique: But for the SLR models we can simplify the calculation slightly and, where r is the correlation between the response and predictor variables.
14-39 Parallel Example 8: Calculating the Coefficient of Determination Using technology for our drilling example we can calculate the correlation between the response and predictor to be 0.772822. Using the simplified calculation for the coefficient of determination that means:
Interpretation: Our model using the depth at which drilling begins as a predictor is able to explain 59.73% of the natural variability in the time it takes to drill 5 feet.