Question

Use the data given in Exercises 5-6 (Exercises 17-18, Section 12.1). Do the data provide sufficient evidence to indicate that \(y\) and \(x\) are linearly related? Test using the \(t\) statistic at the 1\% level of significance. Construct a \(99 \%\) confidence interval for the slope of the line. What does the phrase "99\% confident" mean? $$ \begin{array}{r|rrrrr} x & -2 & -1 & 0 & 1 & 2 \\ \hline y & 1 & 1 & 3 & 5 & 5 \end{array} $$

Short answer

Based on the t-test at a 1% level of significance, there isn't sufficient evidence to conclude that variables \(x\) and \(y\) are linearly related. Additionally, a 99% confidence interval for the slope was constructed and found to lie between -2.5783 and 4.5783. Being "99% confident" means that if we repeated this procedure many times, 99% of the intervals we construct would contain the true slope of the line. The wide range of possible slope values suggests that we cannot be confident in a linear relationship between \(x\) and \(y\).

Step-by-step solution

Calculate the means of \(x\) and \(y\).

First, we need to calculate the mean of each variable, using the formula \(\bar{x}=\frac{1}{n}\sum_{i=1}^n x_i\) and \(\bar{y}=\frac{1}{n}\sum_{i=1}^n y_i\), where \(n\) is the number of observations. In this case, \(n=5\). For \(x\), we have \(\bar{x}=\frac{(-2)+(-1)+0+1+2}{5} = 0\). For \(y\), we have \(\bar{y}=\frac{1+1+3+5+5}{5} = 3\).

Calculate the sums of squares and cross-product.

To run the t-test and calculate the slope, we need the sum of squared deviations for both variables and the sum of cross-product of deviations. We'll use the following formulas: - Sum of squared deviations for \(x\) (SSD_x): \(\sum_{i=1}^n (x_i - \bar{x})^2\) - Sum of squared deviations for \(y\) (SSD_y): \(\sum_{i=1}^n (y_i - \bar{y})^2\) - Sum of cross-product of deviations (CP): \(\sum_{i=1}^n (x_i - \bar{x})(y_i - \bar{y})\) We'll calculate these sums for our given data: $$ SSD_x = (-2-0)^2 + (-1-0)^2 + (0-0)^2 + (1-0)^2 + (2-0)^2 = 10 \newline $$ $$ SSD_y = (1-3)^2 + (1-3)^2 + (3-3)^2 + (5-3)^2 + (5-3)^2 = 20 $$ $$ CP = (-2-0)(1-3) + (-1-0)(1-3) + (0-0)(3-3) + (1-0)(5-3) + (2-0)(5-3) = 10 $$

Compute the test statistic.

With these values calculated, we can now calculate the slope \(b_1\) and the test statistic \(t\): $$ b_1 = \frac{CP}{SSD_x} = \frac{10}{10} = 1 $$ $$ t = \frac{b_1}{\sqrt{\frac{SSD_y - b_1^2\cdot{}SSD_x}{(n-2) \cdot{} SSD_x}}} = \frac{1}{\sqrt{\frac{20 - 1^2\cdot{}10}{3\cdot{}10}}} = 1.632993 $$

Determine the critical values and make a decision.

To test the hypothesis at a 1% significance level, we need the critical values for a two-tailed t-test with \(n-2=3\) degrees of freedom. From the t-distribution table, we find these values to be \(-4.5407\) and \(4.5407\). Since our test statistic \(t=1.632993\) lies between these critical values, we fail to reject the null hypothesis. There isn't sufficient evidence to conclude that \(x\) and \(y\) are linearly related.

Construct a 99% confidence interval.

To construct a 99% confidence interval for the slope, we'll use the following formula: $$ CI = (b_1 - t_{\alpha/2}\cdot{} SE, b_1 + t_{\alpha/2}\cdot{} SE) $$ where \(SE=\sqrt{\frac{SSD_y - b_1^2\cdot{}SSD_x}{(n-2) \cdot{} SSD_x}} = \sqrt{\frac{20 - 1^2\cdot{}10}{3\cdot{}10}} = 0.612372\) and \(t_{\alpha/2}\) is the critical value corresponding to a 99% confidence level with \(n-2=3\) degrees of freedom, which is \(5.8409\). Hence, the 99% confidence interval for the slope is: $$ CI = (1-5.8409\cdot{}0.612372, 1+5.8409\cdot{}0.612372) = (-2.5783, 4.5783) $$

Interpret the results.

Being "99% confident" means that if we repeated this procedure many times, 99% of the intervals we construct would contain the true slope of the line. In this case, our 99% confidence interval for the slope is \((-2.5783, 4.5783)\), suggesting that there's a wide range of possible values for the slope and we cannot be confident in a linear relationship between \(x\) and \(y\).

Key concepts

Understanding Linear Relationship Through t-Tests

When dealing with statistical data, establishing if a linear relationship exists between two variables is often crucial. A linear relationship suggests that as one variable changes, the other variable changes in a directly proportional manner. To determine this, a hypothesis test such as the t-test can be conducted.

In the context of the exercise provided, we are concerned with whether variable y is linearly related to variable x. A t-test, particularly the test for the slope of the regression line, assesses this relationship. If the slope is significantly different from zero (based on the p-value or confidence interval), it suggests evidence of a linear relationship. For the exercise in question, the calculated t statistic does not exceed the critical t value at the 1% significance level, implying no sufficient evidence to claim a linear relationship between the variables.

Confidence Intervals: Interpreting What Being '99% Confident' Means

A confidence interval is a range of values that is likely to contain a population parameter with a certain level of confidence. In the solution example, a 99% confidence interval for the slope of the regression line between x and y is constructed. This interval provides a range within which the true slope is expected to lie with 99% certainty.

The phrase '99% confident' implies that if we were to take multiple samples and calculate a 99% confidence interval for each, approximately 99 out of 100 of those intervals would contain the true population slope. However, the wide interval from the exercise (-2.5783, 4.5783) indicates substantial uncertainty about the slope, leading to the conclusion that any linear relationship claim is weak.

Degrees of Freedom and Their Importance in Hypothesis Testing

The concept of degrees of freedom is pivotal in the context of statistical analysis, particularly in hypothesis testing and construction of confidence intervals. Degrees of freedom usually represent the number of independent values or quantities that can vary in an analysis without breaking any constraints. In terms of a t-test, degrees of freedom are used to determine the critical t values from the t-distribution, which vary according to the sample size.

In the example exercise, we calculate our test statistic with n-2 degrees of freedom, where n is the number of paired observations. The subtraction of 2 accounts for the estimation of two parameters: the slope and the y-intercept of the regression line. As degrees of freedom increase, the t-distribution approaches a normal distribution, which affects the critical values used for hypothesis testing, ultimately impacting the conclusion drawn from statistical tests.