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Exercises 9.5 to 9.8 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. $$ \begin{array}{lrrrr} \text { The regression equation is } \mathrm{Y}=89.4 & -8.20 \mathrm{X} & \\ \text { Predictor } & \text { Coef } & \text { SE Coef } & \mathrm{T} & \mathrm{P} \\ \text { Constant } & 89.406 & 4.535 & 19.71 & 0.000 \\ \mathrm{X} & -8.1952 & 0.9563 & -8.57 & 0.000 \end{array} $$

Short Answer

Expert verified
The sample slope is \(-8.1952\). The null hypothesis (\(H0: \Beta = 0\)) and the alternative hypothesis (\(H1: \Beta ≠ 0\)). The p-value is \(0.000\). Since the p-value is less than \(0.05\), we reject the null hypothesis and conclude that the model is effective.

Step by step solution

01

Identify the Sample Slope

The sample slope is the coefficient of the predictor variable \(X\) in the regression equation. Looking at the regression results, the estimated sample slope coefficient is \(-8.1952\). This is the average change in \(Y\) for each unit change in \(X\), provided that all other conditions remain constant.
02

State the Null and Alternative Hypotheses

The null hypothesis states that there is no relationship between the predictor variable (in this case \(X\)) and the outcome variable \(Y\). Mathematically, this means the slope of the regression line in the population equals zero. Thus, \(H0: \Beta = 0\). The alternative hypothesis is that there is a relationship between \(X\) and \(Y\), which means the slope of the regression line in the population is not zero. Hence, \(H1: \Beta ≠ 0\).
03

Identify the p-value

The p-value is a probability that measures the evidence against the null hypothesis. In the regression results given, the p-value is shown as 0.000.
04

Decision and Conclusion

Given a \(5\%\) significance level, a p-value less than \(0.05\) would lead to a rejection of the null hypothesis in favor of the alternative. Thus, since the p-value found (0.000) is less than \(0.05\), we conclude that at the \(5 \%\)-level, there is significant evidence to suggest that the slope of the regression line in the population is different from zero. Therefore, the model is effective.

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Most popular questions from this chapter

FIBER IN CEREALS AS A PREDICTOR OF CALORIES In Example 9.10 on page \(592,\) we look at a model to predict the number of calories in a cup of breakfast cereal using the number of grams of sugars. In Exercises 9.64 and 9.65 , we give computer output with two regression intervals and information about a specific amount of sugar. 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 cereals with 10 grams of sugars: \(\begin{array}{rrrrr}\text { Sugars } & \text { Fit } & \text { SE Fit } & & 95 \% \text { Cl } & 95 \% \text { PI } \\ 10 & 132.02 & 4.87 & (122.04,142.01) & (76.60,187.45)\end{array}\)

In Exercises 9.1 to \(9.4,\) 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. $$ \begin{aligned} &\text { The regression equation is } Y=29.3+4.30 \mathrm{X}\\\ &\begin{array}{lrrrr} \text { Predictor } & \text { Coef } & \text { SE Coef } & \text { T } & \text { P } \\ \text { Constant } & 29.266 & 6.324 & 4.63 & 0.000 \\ \text { X } & 4.2969 & 0.6473 & 6.64 & 0.000 \end{array} \end{aligned} $$

Test the correlation, as indicated. Show all details of the test. Test for a negative correlation; \(r=-0.41\); \(n=18\).

In Exercise 9.27 we see that the conditions are met for fitting a linear model to predict life expectancy (LifeExpectancy) from the percentage of government expenditure spent on health care (Health) using the data in SampCountries. Use technology to examine this relationship further, as requested below. (a) Find the correlation between the two variables and give the p-value for a test of the correlation. (b) Find the regression line and give the t-statistic and p-value for testing the slope of the regression line. (c) Find the F-statistic and the p-value from an ANOVA test for the effectiveness of the model. (d) Comment on the effectiveness of this model.

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

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