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A random sample of 50 countries is stored in the dataset SampCountries. Two variables in the dataset are life expectancy (LifeExpectancy) and percentage of government expenditure spent on health care (Health) for each country. We are interested in whether or not the percent spent on health care can be used to effectively predict life expectancy. (a) What are the cases in this model? (b) Create a scatterplot with regression line and use it to determine whether we should have any serious concerns about the conditions being met for using a linear model with these data. (c) Run the simple linear regression, and report and interpret the slope. (d) Find and interpret a \(95 \%\) confidence interval for the slope. (e) Is the percentage of government expenditure on health care a significant predictor of life expectancy? (f) The population slope (for all countries) is 0.467 . Is this captured in your \(95 \%\) CI from part (d)? (g) Find and interpret \(R^{2}\) for this linear model.

Short Answer

Expert verified
Among the 50 countries studied, each individual country represents a case. Based on the scatter plot and regression line plotted, preliminary investigation of whether the data fits the conditions for a linear model would have been made. After running a simple linear regression, an interpretation of the slope in the context of the variables would have been made, which then followed by an interpretation of the \(95\%\) confidence interval for the slope as well as significance of Health as a predictor of Life Expectancy. Comparison of the population slope with the observed confidence interval would give insight about the precision and reliability of the sample result. Lastly, \(R^{2}\) would be calculated and interpreted to understand the model-fit for the data.

Step by step solution

01

Identifying Cases

Cases in this model would be the individual countries used in the random sample of the dataset 'SampCountries'.
02

Creating Scatterplot and Regression Line

Create a scatter plot using a statistical software or a graphing calculator with 'LifeExpectancy' as the dependent variable on the vertical axis and 'Health' as the independent variable on the horizontal axis. Use the software to fit a regression line to the scatter plot. Check for any obvious deviations from linearity or any outliers in the plot which would signify the data may not meet the conditions for a linear model.
03

Running Linear Regression and Interpreting Slope

Use statistical software to run a simple linear regression. The slope of the regression line will indicate the estimated increase in Life Expectancy for each one-percent increase in the Health expenditure (all other factors being equal).
04

Finding and Interpreting the Confidence Interval for Slope

Find the \(95\%\) confidence interval for the slope using statistical software. This interval provides a range of values which are likely to contain the true population slope.
05

Determining Significance

Check if the percentage of government expenditure on health care is a significant predictor of life expectancy. This can be done by looking at the p-value for the Health variable in the regression output. If the p-value is less than the significance level (typically 0.05), then the predictor is statistically significant.
06

Comparing Population Slope to Confidence Interval

Compare the given population slope (0.467) with the \(95\%\) confidence interval obtained in step 4. If the given population slope lies within this confidence interval, it would indicate that the sample evidence supports this value for the population slope.
07

Finding and Interpreting \(R^{2}\)

Calculate \(R^{2}\) (the coefficient of determination) for this linear model. This statistic measures the proportion of the variance in the dependent variable (Life Expectancy) that is predictable from the independent variable (Health). The closer \(R^{2}\) is to 1, the better the model fits the data.

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Most popular questions from this chapter

The dataset OttawaSenators contains information on the number of points and the number of penalty minutes for 24 Ottawa Senators NHL hockey players. Computer output is shown for predicting the number of points from the number of penalty minutes: The regression equation is Points \(=29.53-0.113\) PenMins \(\begin{array}{lrrrr}\text { Predictor } & \text { Coef } & \text { SE Coef } & \text { T } & \text { P } \\ \text { Constant } & 29.53 & 7.06 & 4.18 & 0.000 \\ \text { PenMins } & -0.113 & 0.163 & -0.70 & 0.494\end{array}\) \(\mathrm{S}=21.2985 \quad \mathrm{R}-\mathrm{Sq}=2.15 \% \quad \mathrm{R}-\mathrm{Sq}(\mathrm{adj})=0.00 \%\) Analysis of Variance Source Regression Residual Error Total 2 \(\begin{array}{rrrrr}\text { DF } & \text { SS } & \text { MS } & \text { F } & \text { P } \\ 1 & 219.5 & 219.5 & 0.48 & 0.494 \\ 22 & 9979.8 & 453.6 & & \\ 23 & 10199.3 & & & \end{array}\) (a) Write down the equation of the least squares line and use it to predict the number of points for a player with 20 penalty minutes and for a player with 150 penalty minutes. (b) Interpret the slope of the regression equation in context. (c) Give the hypotheses, t-statistic, p-value, and conclusion of the t-test of the slope to determine whether penalty minutes is an effective predictor of number of points. (d) Give the hypotheses, F-statistic, p-value, and conclusion of the ANOVA test to determine whether the regression model is effective at predicting number of points. (e) How do the two p-values from parts (c) and (d) compare? (f) Interpret \(R^{2}\) for this model.

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. $$ \begin{array}{l} \text { Analysis of Variance } \\ \begin{array}{lrrrr} \text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { P } \\ \text { Regression } & 1 & 303.7 & 303.7 & 1.75 & 0.187 \\ \text { Residual Error } & 174 & 30146.8 & 173.3 & & \\ \text { Total } & 175 & 30450.5 & & & \end{array} \end{array} $$

In Exercises 9.11 to \(9.14,\) test the correlation, as indicated. Show all details of the test. Test for a positive correlation; \(r=0.35 ; n=30\).

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 16 grams of sugars: Sugars 95 \(\mathrm{P}\) \(\begin{array}{rrr}\text { rs Fit } & \text { SE Fit } \\\ 6 & 157.88 & 7.10 & \text { (143.3 }\end{array}\) \(95 \% \mathrm{Cl}\) 35,172.42) \(9 \%\) \(\begin{array}{lllll}16 & 15788 & 7.10 & (143.35,172.42) & (101.46\end{array}\) 214.31)

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: \(\begin{array}{rrrrr}\text { Sugars } & \text { Fit } & \text { SE Fit } & & 95 \% \text { Cl } & 95 \% \text { PI } \\ 10 & 132.02 & 4.87 & (122.04,142.01) & (76.60,187.45)\end{array}\)

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