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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

Expert verified
Answer: No, there is not enough evidence to indicate that x and y are linearly related at the 1% significance level since the p-value (0.95) is much greater than the significance level (0.01). The 99% confidence interval for the slope of the line is (-23.52, 24.12).

Step by step solution

01

Organize the Data

First, let's organize the data into two separate lists for x and y values: $$ x = [-2, -1, 0, 1, 2] $$ $$ y = [1, 1, 3, 5, 5] $$
02

Calculate the Sum of the Data

Now calculate the sum of x, sum of y, sum of x times y, sum of x squared, and sum of y squared: $$ \Sigma x = -2 - 1 + 0 + 1 + 2 = 0 $$ $$ \Sigma y = 1 + 1 + 3 + 5 + 5 = 15 $$ $$ \Sigma xy = (-2)(1) + (-1)(1) + (0)(3) + (1)(5) + (2)(5) = 3 $$ $$ \Sigma x^2 = (-2)^2 + (-1)^2 + (0)^2 + (1)^2 + (2)^2 = 10 $$ $$ \Sigma y^2 = (1)^2 + (1)^2 + (3)^2 + (5)^2 + (5)^2 = 51 $$
03

Perform Hypothesis Test with t Statistic

We are going to perform the hypothesis test as follows, where n is the number of data points: - Null Hypothesis (H0): The slope of the regression line is equal to zero (no linear relation between x and y). - Alternative Hypothesis (H1): The slope of the regression line is not equal to zero (there is a linear relation between x and y). First, find the mean of x and y: $$ \bar{x} = \frac{\Sigma x}{n} = \frac{0}{5} = 0 $$ $$ \bar{y} = \frac{\Sigma y}{n} = \frac{15}{5} = 3 $$ Calculate the slope, b: $$ b = \frac{n\Sigma(xy) - (\Sigma x)(\Sigma y)}{n\Sigma x^2 - (\Sigma x)^2} = \frac{5(3) - (0)(15)}{5(10) - (0)^2} = \frac{15}{50} = 0.3 $$ Calculate the intercept, a: $$ a = \bar{y} - b\bar{x} = 3 - 0.3(0) = 3 $$ The linear regression equation is: $$ y = 3 + 0.3x $$ Calculate the standard error: $$ \text{SE} = \sqrt{\frac{\Sigma y^2 - b\Sigma(xy)}{n - 2}} = \sqrt{\frac{51 - 0.3(3)}{5 - 2}} = \sqrt{\frac{50.1}{3}} = 4.08 $$ Calculate the t statistic: $$ t = \frac{b}{\text{SE}} = \frac{0.3}{4.08} = 0.07 $$
04

Calculate the p-value and Determine Significance

Check the p-value associated with the t statistic (df = 3) in a t-table or using software. In this case, the p-value is approximately 0.95. Since the p-value is much greater than the 1% significance level, we fail to reject the null hypothesis. This means that we cannot conclude that there is a linear relationship between x and y at the 1% significance level.
05

Calculate the 99% Confidence Interval for the Slope

Using software or a t-table, find the critical t-value for a 99% confidence interval with 3 degrees of freedom (df = n -2). The critical t-value is approximately 5.84. Calculate the margin of error: $$ \text{Margin of Error} = \text{Critical t-value} \cdot \text{SE} = 5.84 \cdot 4.08 = 23.82 $$ Calculate the 99% confidence interval for the slope: $$ \text{Lower Limit} = 0.3 - 23.82 = -23.52 $$ $$ \text{Upper Limit} = 0.3 + 23.82 = 24.12 $$ The 99% confidence interval for the slope is: (-23.52, 24.12)
06

Explain the Phrase "99% Confident"

The phrase "99% confident" means that if we were to repeat this process of constructing a 99% confidence interval on numerous samples, 99% of the time, the true slope would be contained within the confidence interval. In this case, the 99% confident interval for the slope is so wide (-23.52, 24.12) that it provides little information about the actual slope of the line.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

T-test for Linearity
The t-test for linearity is a statistical tool used to establish whether there's a linear relationship between two variables, generally denoted as x and y. In essence, when we perform a linear regression analysis, we want to know if the independent variable x has a significant effect on the dependent variable y.

The Null Hypothesis (H0) posits there is no linear relation, meaning that the slope (b) of the regression line is zero. The Alternative Hypothesis (H1) claims otherwise—that there is a linear relationship, and the slope is not zero.

During this test, we calculate the t-statistic for the slope of the regression line using the formula:
\[ t = \frac{b}{SE_b}\]
where b is the estimated slope, and SE_b is the standard error of the slope. This statistic is then compared against a critical value from the t-distribution based on our level of significance and the degrees of freedom which, for regression, is typically n - 2 (where n is the number of observations). If the absolute value of the t-statistic is greater than the critical value, we reject the null hypothesis and conclude that there is evidence of linearity.
99% Confidence Interval Calculation
Understanding the 99% confidence interval involves knowing what we mean by being '99% confident' in our statistical analysis. When we're calculating a confidence interval for the slope of our regression line, we're essentially creating a range of values within which we believe the actual slope of our population lies, with a certain degree of certainty.

To calculate a 99% confidence interval, we need a critical value that corresponds to our chosen level of confidence—99% in this case. This value is obtained from the t-distribution table or through statistical software, which considers the degrees of freedom (df = n - 2).

The formula for the confidence interval is:
\[\text{Confidence Interval} = b \pm \text{Critical t-value} \times \text{SE}_b\]
Here, b represents the sample slope and SE_b represents the standard error of b. The resulting confidence interval tells us that we can be 99% confident the interval captures the true population slope. In practice, wider intervals indicate less precision, while narrower intervals suggest our estimate is more precise.
Hypothesis Testing in Statistics
Hypothesis testing is a foundational concept in statistics used to make inferences about populations based on sample data. It is a method through which we can test claims or hypotheses about a parameter, like the mean or proportion, in a quantitative manner.

It starts with two competing hypotheses: the null hypothesis (H0), which represents the status quo or a statement to be tested, and the alternative hypothesis (H1), which represents what we're trying to provide evidence for. The hypothesis test uses sample data to determine whether to reject H0, thereby supporting H1, or to not reject H0 based on a pre-determined level of significance (typically 5%, 1%, or 0.1%).

The process involves calculating a p-value, which is the probability of observing test results as extreme as those recorded, assuming the null hypothesis is true. If the p-value is less than our significance level, we reject H0. Otherwise, we do not have enough evidence to support the alternative hypothesis. This method, applied across scientific research and data analysis, isn't about proving the null hypothesis true but rather about assessing the strength of evidence against it.

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Most popular questions from this chapter

An informal experiment was conducted at McNair Academic High School in Jersey City, New Jersey. Twenty freshman algebra students were given a survey at the beginning of the semester, measuring his or her skill level. They were then allowed to use laptop computers both at school and at home. At the end of the semester, their scores on the same survey were recorded \((x)\) along with their score on the final examination \((y) .^{9}\) The data and the MINITAB printout are shown here. $$ \begin{array}{ccc} \hline \text { Student } & \text { End-of-Semester Survey } & \text { Final Exam } \\ \hline 1 & 100 & 98 \\ 2 & 96 & 97 \\ 3 & 88 & 88 \\ 4 & 100 & 100 \\ 5 & 100 & 100 \\ 6 & 96 & 78 \\ 7 & 80 & 68 \\ 8 & 68 & 47 \\ 9 & 92 & 90 \\ 10 & 96 & 94 \\ 11 & 88 & 84 \\ 12 & 92 & 93 \\ 13 & 68 & 57 \\ 14 & 84 & 84 \\ 15 & 84 & 81 \\ 16 & 88 & 83 \\ 17 & 72 & 84 \\ 18 & 88 & 93 \\ 19 & 72 & 57 \\ 20 & 88 & 83 \\ \hline \end{array} $$ $$ \begin{aligned} &\text { Analysis of Variance }\\\ &\begin{array}{lrrrrr} \text { Source } & \text { DF } & \text { Adj SS } & \text { AdjMS } & \text { F-Value } & \text { P-Value } \\ \hline \text { Regression } & 1 & 3254.03 & 3254.03 & 56.05 & 0.000 \\ \text { Error } & 18 & 1044.92 & 58.05 & & \\ \text { Total } & 19 & 4298.95 & & & \end{array} \end{aligned} $$ $$ \begin{aligned} &\text { Model Summary }\\\ &\begin{array}{ccc} \mathrm{S} & \mathrm{R}-\mathrm{sq} & \mathrm{R}-\mathrm{sq}(\mathrm{adj}) \\ \hline 7.61912 & 75.69 \% & 74.34 \% \end{array} \end{aligned} $$ $$ \begin{aligned} &\text { Coefficients }\\\ &\begin{array}{lrrrr} \text { Term } & \text { Coef } & \text { SE Coef } & \text { T-Value } & \text { P-Value } \\ \hline \text { Constant } & -26.8 & 14.8 & -1.82 & 0.086 \\ \mathrm{x} & 1.262 & 0.169 & 7.49 & 0.000 \end{array} \end{aligned} $$ Regression Equation $$ y=-26.8+1.262 x $$ a. Construct a scatterplot for the data. Does the assumption of linearity appear to be reasonable? b. What is the equation of the regression line used for predicting final exam score as a function of the endof-semester survey score? c. Do the data present sufficient evidence to indicate that final exam score is linearly related to the end-ofsemester survey score? Use \(\alpha=.01\). d. Find a \(99 \%\) confidence interval for the slope of the regression line. e. Use the MINITAB printout to find the value of the coefficient of determination, \(r^{2}\). Show that \(r^{2}=\) SSR/Total SS. f. What percentage reduction in the total variation is achieved by using the linear regression model?

What value does \(r\) assume if all the data points fall on the same straight line in these cases? a. The line has positive slope. b. The line has negative slope.

Calculate the sums of squares and cross-products, \(S_{x x}\) and \(S_{x x}\) $$(3,6) \quad(5,8) \quad(2,6) \quad(1,4) \quad(4,7) \quad(4,6)$$

Exercises \(6-7\) were formed by reversing the slope of the lines in Exercises 4 - 5. Plot the points on graph paper and calculater and \(r^{2}\). Notice the change in the sign of \(r\) and the relationship between the values of \(r^{2}\) compared to Exercises \(4-5 .\) By what percentage was the sum of squares of deviations reduced by using the least-squares predictor \(\hat{y}=a+b x\) rather than \(\bar{y}\) as a predictor of \(y\) ? $$\begin{array}{l|rrrrr}x & -2 & -1 & 0 & 1 & 2 \\\\\hline y & 4 & 4 & 3 & 2 & 2\end{array}$$

Use the information given (reproduced below) to find a prediction interval for a particular value of \(y\) when \(x=x_{0} .\) Is the interval wider than the corresponding confidence interval from Exercises \(3-4 ?\) $$ \begin{array}{l} n=6, s=.639, \Sigma x_{i}=19, \Sigma x_{i}^{2}=71, \\ \hat{y}=3.58+.82 x, x_{0}=2,99 \% \text { prediction interval } \end{array} $$

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