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Chapter 9: Problem 6

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. $$ \begin{array}{lrrrr} \text { The regression equation is } \mathrm{Y}=82.3-0.0241 \mathrm{X} & \\ \text { Predictor } & \text { Coef } & \text { SE Coef } & \mathrm{T} & \mathrm{P} \\ \text { Constant } & 82.29 & 11.80 & 6.97 & 0.000 \\ \mathrm{X} & -0.02413 & 0.02018 & -1.20 & 0.245 \end{array} $$

Short Answer

Expert verified
The sample slope is \(-0.0241\). The null hypothesis is that the population slope is zero, while the alternative hypothesis is that it's non-zero. The p-value is \(0.245\). Given the p-value is higher than the 5% significance level, we cannot reject the null hypothesis. Hence, there's not enough evidence to establish a relationship between \(X\) and \(Y\).

Step by step solution

01

Understand the Regression Equation

The regression equation is \(Y = 82.3 - 0.0241X\). This equation represents a straight line where \(Y\) is the dependent variable we are trying to predict or explain, \(X\) is the independent variable we are using to do the prediction and \(82.3\) is the y-intercept鈥攊.e., the predicted value of \(Y\) when \(X\) equals 0. The slope of this line (i.e., the coefficient of \(X\)) is \(-0.0241\), meaning that each unit increase in \(X\) decreases \(Y\) by \(0.0241\) units.
02

State the Null and Alternative Hypotheses

The test here is to see whether there is a significant relationship between \(X\) and \(Y\). The null hypothesis (\(H0\)) is that the slope is not different from zero in the population (i.e., there is no relationship). The alternative hypothesis (\(HA\)) is that the slope is different from zero (i.e., there is a relationship). Thus, \(H0: \beta = 0\) vs. \(HA: \beta \neq 0\), where \(\beta\) denotes the population slope.
03

Identify the P-Value

The p-value associated with the coefficient of \(X\) is \(0.245\), which is much greater than the 5% significance level. This value provides a measure of evidence against \(H0\). The lower the p-value, the stronger the evidence against \(H0\).
04

Make a Conclusion

Given that the p-value is \(0.245\), which is above the 5% significance level, we do not reject the null hypothesis. This means there is not enough evidence to conclude that there is a relationship between \(X\) and \(Y\) in the population.

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Key Concepts

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

Null Hypothesis
In the world of statistics, the null hypothesis (\(H_0\)) plays a critical role in hypothesis testing. It is essentially a statement that claims "no effect" or "no difference." In the context of simple linear regression, the null hypothesis typically states that there is no relationship between the independent variable and the dependent variable in the population.

For the regression equation given in the exercise, \(Y = 82.3 - 0.0241X\), the null hypothesis is based on the slope of the line. The null hypothesis is:
  • \(H_0: \beta = 0\), where \(\beta\) is the population slope.
This implies that changes in \(X\) do not lead to changes in \(Y\). If we fail to reject this hypothesis, it suggests that the independent variable, \(X\), is not a predictor of changes in the dependent variable, \(Y\).

Understanding the null hypothesis allows researchers to set a baseline expectation for what they might observe if there really is no effect. It's crucial because it provides a way to test whether observed effects might merely be due to chance.
Alternative Hypothesis
The alternative hypothesis (\(H_A\)) is what researchers hope to prove or find evidence for. It represents the situation where there is an actual effect or relationship between the variables being studied. In simple linear regression, the alternative hypothesis challenges the null hypothesis by proposing that the slope is not zero, indicating a relationship.

In our exercise, the regression equation is \(Y = 82.3 - 0.0241X\). The alternative hypothesis suggests that the slope \(\beta\) is different from zero:
  • \(H_A: \beta eq 0\)
This means that there is some relationship between \(X\) and \(Y\). If we were to reject the null hypothesis, it would provide evidence supporting the alternative hypothesis, suggesting \(X\) does impact \(Y\).

Understanding the alternative hypothesis is key in research because it represents what the scientist aims to demonstrate through their experiment or study. It sets the stage for potential discovery or confirmation of new theories.
P-Value
The p-value is a statistical measure that helps determine the significance of results in hypothesis testing. It gives us an idea of how extreme the observed results are under the assumption that the null hypothesis is true. Simply put, the p-value tells us about the strength of evidence against \(H_0\).

In hypothesis testing, a common threshold used is a 5% significance level. If the p-value is less than or equal to 0.05, it suggests there is strong evidence against the null hypothesis, leading us to reject it. In the context of our regression problem, the p-value associated with the coefficient of \(X\) is 0.245.

Let's consider why this is important:
  • A p-value of 0.245 is greater than 0.05, indicating weak evidence against \(H_0\). As a result, we do not reject the null hypothesis.
  • This high p-value implies that the changes in \(X\) do not have a statistically significant effect on \(Y\).
  • Thus, according to this result, there isn't adequate proof to claim a relationship exists between the independent variable and the dependent variable.
Understanding the p-value and significance level helps guide decision-making in research, allowing us to interpret results and ensure our conclusions align with the evidence at hand.

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Most popular questions from this chapter

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.$$ \begin{array}{lrrrr} \text { Coefficients: } & \text { Estimate } & \text { Std.Error } & \mathrm{t} \text { value } & \operatorname{Pr}(>|\mathrm{t}|) \\ \text { (Intercept) } & 7.277 & 1.167 & 6.24 & 0.000 \\ \text { Dose } & -0.3560 & 0.2007 & -1.77 & 0.087 \end{array} $$

Use the computer output (from different computer packages) to estimate the intercept \(\beta_{0},\) the slope \(\beta_{1},\) and to give the equation for the least squares line for the sample. Assume the response variable is \(Y\) in each case. $$ \begin{array}{lrrrr} \text { Coefficients: } & \text { Estimate } & \text { Std.Error } & \mathrm{t} \text { value } & \mathrm{Pr}(>|\mathrm{t}|) \\ \text { (Intercept) } & 7.277 & 1.167 & 6.24 & 0.000 \\ \text { Dose } & -0.3560 & 0.2007 & -1.77 & 0.087 \end{array} $$

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. $$ \begin{aligned} &\text { Response: }\\\ &\begin{array}{lrrrrr} & \text { Df } & \text { Sum Sq } & \text { Mean Sq } & \text { F value } & \operatorname{Pr}(>\text { F) } \\ \text { ModelB } & 1 & 10.380 & 10.380 & 2.18 & 0.141 \\ \text { Residuals } & 342 & 1630.570 & 4.768 & & \\ \text { Total } & 343 & 1640.951 & & & \end{array} \end{aligned} $$

Data 9.1 on page 577 introduces the dataset InkjetPrinters, which includes information on all-in-one printers. Two of the variables are Price (the price of the printer in dollars) and CostColor (average cost per page in cents for printing in color). Computer output for predicting the price from the cost of printing in color is shown: $$ \begin{aligned} &\text { The regression equation is Price }=378-18.6 \text { CostColor }\\\ &\begin{array}{lrrrrr} \text { Analysis of Variance } & & & & \\ \text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { P } \\ \text { Regression } & 1 & 57604 & 57604 & 13.19 & 0.002 \\ \text { Residual Error } & 18 & 78633 & 4369 & & \\ \text { Total } & 19 & 136237 & & & \end{array} \end{aligned} $$ (a) What is the predicted price of a printer that costs 10 cents a page for color printing? (b) According to the model, does it tend to cost more or less (per page) to do color printing on a cheaper printer? (c) Use the information in the ANOVA table to determine the number of printers included in the dataset. (d) Use the information in the ANOVA table to compute and interpret \(R^{2}\). (e) Is the linear model effective at predicting the price of a printer? Use information from the computer output and state the conclusion in context.

Golf Scores In a professional golf tournament the players participate in four rounds of golf and the player with the lowest score after all four rounds is the champion. How well does a player's performance in the first round of the tournament predict the final score? Table 9.6 shows the first round score and final score for a random sample of 20 golfers who made the cut in a recent Masters tournament. The data are also stored in MastersGolf. Computer output for a regression model to predict the final score from the first-round score is shown. Use values from this output to calculate and interpret the following. Show your work. (a) Find a \(95 \%\) interval to predict the average final score of all golfers whoshoot a 0 on the first round at the Masters. (b) Find a \(95 \%\) interval to predict the final score of a golfer who shoots a -5 in the first round at the Masters. (c) Find a \(95 \%\) interval to predict the average final score of all golfers who shoot a +3 in the first round at the Masters. The regression equation is Final \(=0.162+1.48\) First \(\begin{array}{lrrrr}\text { Predictor } & \text { Coef } & \text { SE Coef } & \text { T } & \text { P } \\ \text { Constant } & 0.1617 & 0.8173 & 0.20 & 0.845 \\ \text { First } & 1.4758 & 0.2618 & 5.64 & 0.000 \\ S=3.59805 & R-S q=63.8 \% & \text { R-Sq }(a d j) & =61.8 \%\end{array}\) Analysis of Variance Source Regression Residual Error Total \(\begin{array}{rrrrr}\text { DF } & \text { SS } & \text { MS } & \text { F } & \text { P } \\ 1 & 411.52 & 411.52 & 31.79 & 0.000 \\ 18 & 233.03 & 12.95 & & \\ 19 & 644.55 & & & \end{array}\)
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