For a sample of people, three measurements are taken on each person (skin, thigh, arm) in an attempt to obtain variables useful in predicting their body fat (also measured). Output of SAS regression of fat on skin, thigh, and arm variables and regression of fat on skin are given on the following pages.

1. Given a formula(s) to construct a confidence interval for 0+501+202+303. Make it clear that you could calculate the lower and upper bounds if asked. Include the Student's t value for a 95% confidence level.

You said that you only needed help with 2 out of the 3 questions. Consider the next two my official answers. I think I can help you with this one, but I'm not 100% on it.

The estimate of 0+501+202+303 can be calculated by plugging in the estimates of 0-3 (from the top of page D), then multiplying and adding:

(-67.35) + 50(2.04) + 20(-1.35) + 30(0.60) = -67.35 + 102 - 27 + 18 = 25.65

To make a confidence interval, we need to know the standard deviations. We know the standard errors (again top of page D). First, we can find the variances of 0-3, then we can multiply and add them together. Then we will turn the variance into the standard deviation.

(1) find the variances

(standard error)2 = variance / N. For each of the variables, N = 10 because there were 10 original observations (I think. This is where I'm least sure about the answer. You might use N = 2 because there were originally t-tests with df = 1.).

(91.10)2 = variance / 10 var = 82992.1
(2.37)2 = variance / 10 var = 56.169
(2.78)2 = variance / 10 var = 77.284
(1.45)2 = variance / 10 var = 21.025

(2) combine the variances

(82992.1) + 50(56.169) + 20(77.284) + 30(21.025) = 82992.1 + 2808.45 + 1545.68 + 630.75 = 87976.98

(3) find the standard deviation

sd = √var/√N

sd = √87976.98/√40 = 46.898

Now we can calculate the confidence interval. It will look like

estimate ± t*(sd)

Our estimate is 25.65, the sd is 46.898, and t* is the critical value of t at the 95% level. t* = 2.023 using 39 degrees of freedom (this is based on N, and I'm not sure about it).

The 95% confidence interval for 0+501+202+303 is:

25.65 ± (2.023)(46.898)

2. The regression coefficient associated with skin differs in the full and simple linear regression models. Explain the difference.

The simple linear regression model: fat = 0 + 1(skin). Fat is the dependent (y) variable and skin is the independent (x) variable. 0 is estimated as -0.54 and 1 is estimated as 0.79 (bottom of page D).

The full linear regression model includes more independent variables. It is fat = 0 + 1(thigh) + 2(skin) + 3(arm). 0 is estimated as -67.35, 1 is estimated as 2.04, 2 is estimated as -1.35, and 3 is estimated as 0.60 (top of page D).

The regression coefficient, r2, for the full model is r2 = 0.9413 and r2 for the simple model is r2 = 0.7106. That means that the full model explains 94.13% of the variation in fat, and the simple model explains 71.06% of the variation in fat. Therefore the model with more independent variables is able to explain more of the variation in the dependent variable.

3. There are 10 confidence intervals in the middle of page D. What confidence level can be associated with the claim that all 10 are correct inferences?

Each of the confidence intervals in the middle of page D is a 99% confidence interval, so for each confidence interval, there is a 1% chance that the observation in question does not fall in that interval and a 99% chance that it does.

Remember that when you are calculating the probability of events A and B happening simultaneously, you multiply the individual probabilities together. Here, the individual probabilities are 0.99 and they occur 10 times.

0.9910 = 0.9044

Therefore, there is a 90.44% chance that all 10 inferences are correct, and the confidence level is 90.44%.

PFA