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Chapter 9: Problem 8

Show some computer output for fitting simple linear models. State the value of the sample slope for each model and give the null and alternative hypotheses for testing if the slope in the population is different from zero. Identify the p-value and use it (and a \(5 \%\) significance level) to make a clear conclusion about the effectiveness of the model.$$ \begin{array}{lrrrr} \text { Coefficients: } & \text { Estimate } & \text { Std.Error } & \mathrm{t} \text { value } & \operatorname{Pr}(>|\mathrm{t}|) \\ \text { (Intercept) } & 807.79 & 87.78 & 9.30 & 0.000 \\ \mathrm{~A} & -3.659 & 1.199 & -3.05 & 0.006 \end{array} $$

Short Answer

Expert verified
The sample slope of the linear model is -3.659. The null hypothesis \(H_0: \beta = 0\) is rejected in favour of the alternative \(H_1: \beta \neq 0\) as per the p-value 0.006. Thus, the model is considered effective.

Step by step solution

01

Identifying the Sample Slope

The sample slope can be found in the coefficient column of the A variable. In this case it is -3.659, which suggests that for every one unit increase in variable A, the dependent variable decreases by 3.659 units on average, holding everything else constant.
02

Null and Alternative Hypotheses

The null hypothesis for testing the effectiveness of the slope is that the slope coefficient is equal to zero in the population. The alternative hypothesis is that the slope coefficient is different from zero. Mathematically, the hypotheses can be written as follows: Null Hypothesis \(H_0: \beta = 0\) and Alternative Hypothesis \(H_1: \beta \neq 0\).
03

Identifying the P-value and Interpretation

The p-value of A variable is 0.006. A p-value less than the significance level (0.05 in this case) indicates a strong evidence against the null hypothesis. Thus, concluding that the slope is different from zero in the population, and variable A has a significant impact.
04

Conclusion about the Model's Effectiveness

Since the p-value is less than the significance level, there is evidence to reject the null hypothesis in favour of the alternative. Therefore, it can be concluded that the model is effective and variable A significantly affects the dependent variable.

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Key Concepts

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

Hypothesis Testing
Hypothesis testing in statistics is a crucial method used to make inferences about populations based on sample data. It allows us to determine if there is enough statistical evidence to support a specific hypothesis about a population parameter. In the context of simple linear regression, hypothesis testing helps us decide if the relationship between two variables is significant.

Here's how it typically works:
  • **Null Hypothesis (\( H_0 \)):** This implies that there is no effect or no relationship. In the case of linear regression, the null hypothesis asserts that the slope of the regression line is zero (\( \beta = 0 \)), indicating no effect of the independent variable on the dependent variable.
  • **Alternative Hypothesis (\( H_1 \)):** This states that there is a relationship, or an effect exists. For regression, it suggests the slope is not zero (\( \beta eq 0 \)).
  • **Decision Making:** We use sample data to test these hypotheses, often with the help of statistical software that provides outputs including p-values. The analysis will lead to either the rejection of the null hypothesis or failing to reject it, thereby supporting or refuting the claim of an effect.
Understanding hypothesis testing is fundamental when analyzing linear regression models in any empirical research because it frames the way we interpret the results of regression analysis.
P-Value Interpretation
The p-value is a key concept in hypothesis testing which aids in making informed decisions. It represents the probability that the observed data would occur if the null hypothesis were true.

Here's what you need to know:
  • **Significance Level:** Commonly set at 0.05, the significance level defines the threshold for determining if a result is significant. It represents a 5% risk of concluding that a difference exists when there is none (Type I error).
  • **P-value Analysis:** In our model, the p-value for the slope of variable A is 0.006. This is significantly below 0.05, indicating strong evidence against the null hypothesis. Hence, we infer that variable A has a statistically significant impact on the dependent variable.
  • **Conclusion:** If the p-value is less than or equal to the significance level, we reject the null hypothesis. If it is greater, we fail to reject it. A small p-value (typically 鈮� 0.05) indicates strong evidence against the null hypothesis, so you reject the null hypothesis.
By interpreting p-values correctly, researchers can make strong inferences about their data and trust the implications of their statistical models.
Regression Analysis
Regression analysis is a powerful statistical method used to examine the relationship between two or more variables. In simple linear regression, we focus on the relationship between an independent variable, denoted here as A, and a dependent variable.

Key aspects include:
  • **Linear Relationship:** Assumes a straight-line relationship between the variables, which is captured by a linear equation (\( Y = eta_0 + eta_1X + ext{error} \)), where \( eta_1 \) represents the slope.
  • **Estimation and Interpretation:** The coefficient for variable A is -3.659. This suggests that an increase of one unit in A results in an average decrease of -3.659 units in the dependent variable. Negative values indicate an inverse relationship.
  • **Effectiveness of the Model:** The effectiveness of a regression model can be evaluated using statistical tests to determine the significance of the coefficients. As observed, a significant p-value (like our 0.006) reflects meaningful implications of the model's predictions.
Regression analysis turns complex data relationships into useful insights that can guide decision-making and improve predictions in various fields, ranging from economics to biology.

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Most popular questions from this chapter

In Exercises 9.62 and 9.63 , we give computer output with two regression intervals and information about the percent of calories eaten during the day. Interpret each of the intervals in the context of this data situation. (a) The \(95 \%\) confidence interval for the mean response (b) The \(95 \%\) prediction interval for the response The intervals given are for mice that eat \(50 \%\) of their calories during the day: \(\begin{array}{rrrrr}\text { DayPct } & \text { Fit } & \text { SE Fit } & 95 \% \mathrm{Cl} & 95 \% \mathrm{PI} \\ 50.0 & 7.476 & 0.457 & (6.535,8.417) & (2.786,12.166)\end{array}\)

We show an ANOVA table for regression. State the hypotheses of the test, give the F-statistic and the p-value, and state the conclusion of the test. $$ \begin{aligned} &\text { Response: } Y\\\ &\begin{array}{lrrrrr} & \text { Df } & \text { Sum Sq } & \text { Mean Sq } & \mathrm{F} \text { value } & \operatorname{Pr}(>\mathrm{F}) \\ \text { ModelA } & 1 & 352.97 & 352.97 & 11.01 & 0.001 * * \\ \text { Residuals } & 359 & 11511.22 & 32.06 & & \\ \text { Total } & 360 & 11864.20 & & & \end{array} \end{aligned} $$

Exercise 9.19 on page 588 introduces a study examining the relationship between the number of friends an individual has on Facebook and grey matter density in the areas of the brain associated with social perception and associative memory. The data are available in the dataset FacebookFriends and the relevant variables are GMdensity (normalized \(z\) -scores of grey matter density in the brain) and \(F B\) friends (the number of friends on Facebook). The study included 40 students at City University London. Computer output for ANOVA for regression to predict the number of Facebook friends from the normalized brain density score is shown below. The regression equation is FBfriends \(=367+82.4\) GMdensity Analysis of Variance \(\begin{array}{lrrrrr}\text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { P } \\ \text { Regression } & 1 & 245400 & 245400 & 8.94 & 0.005 \\ \text { Residual Error } & 38 & 1043545 & 27462 & & \\ \text { Total } & 39 & 1288946 & & & \end{array}\) Is the linear model effective at predicting the number of Facebook friends? Give the F-statistic from the ANOVA table, the p-value, and state the conclusion in context. (We see in Exercise 9.19 that the conditions are met for fitting a linear model in this situation.)

Predicting Re-Election Margin Data 2.9 on page 106 introduces data on the approval rating of an incumbent US president and the margin of victory or defeat in the subsequent election (where negative numbers indicate the margin by which the incumbent president lost the re-election campaign). The data are reproduced in Table 9.5 and are available in ElectionMargin. Computer output for summary statistics for the two variables and for a regression model to predict the margin of victory or defeat from the approval rating is shown: The regression equation is Margin \(=-36.76+0.839\) Approval \(\begin{array}{lrrrr}\text { Term } & \text { Coef } & \text { SE Coef } & \text { T-Value } & \text { P-Value } \\ \text { Constant } & -36.76 & 8.34 & -4.41 & 0.001 \\ \text { Approval } & 0.839 & 0.155 & 5.43 & 0.000 \\\ S=5.66054 & R-S q=74.64 \% & R-S q(a d j) & =72.10 \%\end{array}\) \(\begin{array}{lrrrrr}\text { Analysis of Variance } & & & & & \\ \text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { P } \\ \text { Regression } & 1 & 943.0 & 943.04 & 29.43 & 0.000 \\ \text { Residual Error } & 10 & 320.4 & 32.04 & & \\ \text { Total } & 11 & 1263.5 & & & \end{array}\) Use values from this output to calculate and interpret the following. Show your work. (a) A \(95 \%\) confidence interval for the mean margin of victory for all presidents with an approval rating of \(50 \%\) (b) A \(95 \%\) prediction interval for the margin of victory for a president with an approval rating of \(50 \%\) (c) A \(95 \%\) confidence interval for the mean margin of victory if we have no information about the approval rating. (Hint: This is just an ordinary confidence interval for a mean based only on the single sample of Margin values.)

Test the correlation, as indicated. Show all details of the test. Test for evidence of a linear association; \(r=0.28 ; n=100\).
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