Regression Chapter 8

What would you expect for other heights? Height (in.) Weight (lbs.) What would you expect for other heights? 140 ( is called the mean response) How much would an adult female weigh if she were 5 feet tall? 130 This distribution is normally distributed. (we hope) She could weigh varying amounts – in other words, there is a distribution of weights for adult females who are 5 feet tall. 120 110 Where would you expect the TRUE LSRL (Least Squares Regression Line) to be? What about the standard deviations of all these normal distributions? We want the standard deviations of all these normal distributions to be the same. 100

Regression Model The mean response my has a straight-line relationship with x: Where: the slope b and the y-intercept a are unknown parameters For any fixed value of x, the response y varies according to a normal distribution. Repeated responses of y are independent of each other. The standard deviation of y (sy) is the same for all values of x. (sy is also an unknown parameter)

What distribution does their weight have? Person # Ht Wt 1 64 130 10 175 15 150 19 125 21 145 40 186 47 121 60 137 63 143 68 120 70 112 78 108 83 160 Suppose we look at part of a population of adult women. These women are all 64 inches tall. What distribution does their weight have?

Because there are two unknowns a & b We use to estimate The slope b of the LSRL is an unbiased estimator of the true slope b. The y-intercept a of the LSRL is an unbiased estimator of the true intercept a. The standard error s is an unbiased estimator of the true standard deviation of y (sy). Where y is the observed value and is the expected value. Note: df = n-2 Because there are two unknowns a & b

Height Weight Suppose you took many samples of the same size from this population & calculated the LSRL for each. Using the slope from each of these LSRLs – we can create a sampling distribution for the slope of the true LSRL. What is the standard deviation of the sampling distribution? What is the mean of the sampling distribution equal? What shape will this distribution have? b b b b b b b mb = b

Let’s review the regression model! x & y have a linear relationship with the true LSRL going through the my (mean response) sy is the same for each x-value. For a given x-value, the responses (y) are normally distributed

Assumptions for inference on slope The observations are independent Check that you have an SRS The true relationship is linear Check the scatter plot & residual plot The standard deviation of the response is constant. The responses vary normally about the true regression line. Check a histogram or normal probability plot of residuals

What is the slope of a horizontal line? Height Weight Suppose the LSRL has a horizontal line –would height be useful in predicting weight? A slope of zero – means that there is NO relationship between x & y!

Hypotheses Be sure to define b! H0: b = 0 1 Ha: b > 0 Ha: b < 0 This implies that there is no relationship between x & y Or that x should not be used to predict y What would the slope equal if there were a perfect relationship between x & y? H0: b = 0 Ha: b > 0 Ha: b < 0 Ha: b ≠ 0 1 Be sure to define b!

Because there are two unknowns a & b Formulas: Confidence Interval: Hypothesis test: df = n -2 Because there are two unknowns a & b

Body fat = -27.376 + 0.250 weight r = 0.697 r2 = 0.485 Example: It is difficult to accurately determine a person’s body fat percentage without immersing him or her in water. Researchers hoping to find ways to make a good estimate immersed 20 male subjects, and then measured their weights. Find the LSRL, correlation coefficient (r), and coefficient of determination (r2). Body fat = -27.376 + 0.250 weight r = 0.697 r2 = 0.485

b) Explain the meaning of slope in the context of the problem. There is approximately .25% increase in body fat for every pound increase in weight. c) Explain the meaning of the coefficient of determination in context. Approximately 48.5% of the variation in body fat can be explained by the regression of body fat on weight.

a = -27.376 b = 0.25 s = 7.049 d) Estimate a, b, and s. e) Create a scatter plot and residual plot for the data. Weight Body fat Weight Residuals

f) Is there sufficient evidence that weight can be used to predict body fat? Assumptions: Have an SRS of 20 male subjects Since the residual plot is randomly scattered, weight & body fat are linear Since the points are evenly spaced across the LSRL on the scatterplot, sy is approximately equal for all values of weight Since the normal probability plot of the residuals is approximately linear, the responses are approximately normally distributed. H0: b = 0 Where b is the true slope of the LSRL of weight Ha: b ≠ 0 & body fat Since the p-value < a, I reject H0. There is sufficient evidence to suggest that weight can be used to predict body fat.

Be sure to show all graphs! g) Give a 95% confidence interval for the true slope of the LSRL. Assumptions: SAME AS T-TEST!!!! Have an SRS of 20 male subjects Since the residual plot is randomly scattered, weight & body fat are linear Since the points are evenly spaced across the LSRL on the scatterplot, sy is approximately equal for all values of weight Since the normal probability plot of the residuals is approximately linear, the responses are approximately normally distributed. We are 95% confident that the true slope of the LSRL of weight & body fat is between 0.12 and 0.38. Be sure to show all graphs!

What does “s” represent (in context)? h) Here is the computer-generated result from the data: Sample size: 20 R-square = 48.53% s = 7.0491323 df? What does “s” represent (in context)? Parameter Estimate Std. Err. Intercept -27.376263 11.547428 Weight 0.24987414 0.060653996 Correlation coeficient? Be sure to write as decimal first! What does this number represent? What do these numbers represent?