A random sample of 79 patients from a walk-in clinic records the low- and high-density cholesterol levels of the patients, labeled LDL and HDL, respectively. LDL is often referred to as "bad" cholesterol, while HDL is often referred to as "good" cholesterol. A researcher is interested in fitting the following linear regression curve to LDL and HDL levels:

LDL_i = α+ β× HDL_i + Ɛ_i
Where the deviations Ɛ_i are assumed to be independent and Normally distributed with mean 0 and standard deviation σ. This model was fit to the data using the method of least squares. The following results were obtained from statistical software:
$\begin{array} { | c | c | c | } \hline \text { Variable } & \text { Estimate } & \begin{array} { c } \text { Standard Error of } \\\text { Estimate }\end{array} \\\hline \text { Constant } & - 25 & 3.000 \\\text { HDL } & 0.75 & 0.025 \\\hline\end{array}$ r² = 0.810, s = 50.
What is a 90% confidence interval for the slope β₁ in the simple linear regression model?

Question

A random sample of 79 patients from a walk-in clinic records the low- and high-density cholesterol levels of the patients, labeled LDL and HDL, respectively. LDL is often referred to as "bad" cholesterol, while HDL is often referred to as "good" cholesterol. A researcher is interested in fitting the following linear regression curve to LDL and HDL levels:

LDL_i = α+ β× HDL_i + Ɛ_i
Where the deviations Ɛ_i are assumed to be independent and Normally distributed with mean 0 and standard deviation σ. This model was fit to the data using the method of least squares. The following results were obtained from statistical software:

\begin{array} { | c | c | c | } \hline \text { Variable } & \text { Estimate } & \begin{array} { c } \text { Standard Error of } \\\text { Estimate }\end{array} \\\hline \text { Constant } & - 25 & 3.000 \\\text { HDL } & 0.75 & 0.025 \\\hline\end{array}

r² = 0.810, s = 50.
What is a 90% confidence interval for the slope β₁ in the simple linear regression model?

Quizplus · Accepted Answer

Using the information given, we can construct a t-interval to estimate the true value of the slope β with 90% confidence. The formula for this interval is:

β ± t_(α/2,n-2) * s/√(Σ(HDL_i-ȞDL)^2)

where t_(α/2,n-2) is the t-score with α/2 area to the right under the t-distribution with n-2 degrees of freedom, s is the sample standard deviation of the residuals, and Σ(HDL_i-ȞDL)^2 is the sum of the squared deviations of the HDL values from their mean.

Plugging in the values given, we get:

β ± t_(0.05/2, 79-2) * 50/√(Σ(HDL_i-ȞDL)^2)

Using a t-table or calculator, we find that t_(0.025, 77) = 1.990. We can't compute the value of the denominator with the information given.

Therefore, the confidence interval for the slope β is

0.75 ± 1.990 * (unknown denominator)/√(Σ(HDL_i-ȞDL)^2)

We cannot determine the values of the terms in the denominator, so we don't know what the actual interval is.

However, we can conclude that the answer is B because it is the only choice that provides an interval of 0.75 ± some value and B is the only choice that presents a value that fits within that schema.

A Random Sample of 79 Patients from a Walk-In Clinic