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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.)

Short Answer

Expert verified
Yes, based on the given F statistic (8.94) and p-value (0.005), the linear model is effective at predicting the number of Facebook friends. Because the p-value is less than 0.05, we conclude there's a statistically significant relationship between grey matter density and the number of friends a person has on Facebook.

Step by step solution

01

Understanding the ANOVA Table

The table provided is from an ANOVA for a regression model predicting the number of Facebook friends based on a person's grey matter density in the brain. The columns represent Source of variation, Degrees of Freedom (DF), Sum of Squares (SS), Mean Square (MS), F, and P.
02

Reading the F statistic and P-Value

The F statistic is found at the bottom of the 'F' column, under 'Regression'. Here, the table shows an F statistic of 8.94. The associated p-value is found in the 'P' column, also under 'Regression'. The table shows a p-value of 0.005.
03

Interpreting the F-statistic and P-value

The F-statistic tests whether the overall regression model is a good fit for the data. A larger F-statistic indicates a more significant statistical model. Here, an F statistic of 8.94 can be considered relatively large, suggesting that the model is a good fit for the data. Next, consider the p-value. A p-value below 0.05 indicates that the model is statistically significantly better at predicting the outcome than would be the case if the predictors had no effect. Here, the p-value of 0.005 is less than 0.05, which indicates that the number of Facebook friends can be predicted by the grey matter density in the brain to a statistically significant degree.

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Key Concepts

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

F-statistic interpretation
The F-statistic in ANOVA for regression plays a crucial role in understanding the overall significance of the regression model. Simply put, the F-statistic is used to compare the model that includes the variable in question, in this case, grey matter density, against a model that does not. It tests the null hypothesis that the variable has no linear predictive value for the response variable, which would be the number of Facebook friends in this analysis.

The F-statistic is calculated by the ratio of Mean Square of the Regression (MSR) to the Mean Square of the Residual (MSE). In our exercise, the F-statistic is 8.94, which is substantial enough to suggest that our regression model provides a better fit for the data than a model without the predictor variable. In simpler terms, it shows that grey matter density does have an impact on the number of Facebook friends a person has. The larger this statistic, the more evidence there is against the null hypothesis, leading us to believe that the predictor variable is relevant.
p-value significance
The p-value is a vital component in validating the strength of the results obtained from the F-statistic. It represents the probability of observing the results purely by chance, given that the null hypothesis is true. A low p-value suggests that the observed results are unlikely to have occurred due to random variation alone, and thus, one can infer that the model has statistical significance.

In the context of our study, the p-value of 0.005 is well below the commonly accepted significance level of 0.05. This means there's a very small chance that the relationship between grey matter density and the number of Facebook friends occurred by chance. Thus, we confidently reject the null hypothesis and accept that there is a positive relationship between grey matter density and the number of Facebook friends. This underscores the importance of understanding the p-value: it quantifies the evidence against the null hypothesis and guides us towards a statistical conclusion that is derived from the data.
Regression model analysis
Regression model analysis involves evaluating the performance and adequacy of a regression model in describing the relationship between variables. One of the key outputs of this analysis is the ANOVA table, which we have already discussed in terms of its F-statistic and p-value. However, to grasp the quality of the regression, we should look into other components like the coefficient estimates, R-squared value, residual analysis, and potential outlier impact.

The regression equation provided, 'FBfriends = 367 + 82.4 GMdensity', gives us specific information about the influence of grey matter density on the number of Facebook friends. The coefficient of GMdensity, 82.4, tells us how much the predicted number of Facebook friends changes for a one-unit increase in grey matter density. Furthermore, while the F-statistic and p-value give us insights into whether the model is statistically significant, the R-squared value (not provided in the exercise but often found in regression outputs) would indicate the proportion of variance in the response variable that's explained by the predictor.

Ultimately, regression model analysis informs us not just whether a variable matters, but how much it matters, the direction of its effect, and the proportion of the overall effect it explains. Our exercise underlines the significance of grey matter density on the number of Facebook friends, a conclusion drawn from thorough analysis of the regression model.

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

Hantavirus is carried by wild rodents and causes severe lung disease in humans. A study \(^{5}\) on the California Channel Islands found that increased prevalence of the virus was linked with greater precipitation. The study adds "Precipitation accounted for \(79 \%\) of the variation in prevalence." (a) What notation or terminology do we use for the value \(79 \%\) in this context? (b) What is the response variable? What is the explanatory variable? (c) What is the correlation between the two variables?

in Fish In Exercise 9.21 , we see that the conditions are met for using the \(\mathrm{pH}\) of a lake in Florida to predict the mercury level of fish in the lake. The data are given in FloridaLakes. Computer output is shown for the linear model with several values missing: The regression equation is AvgMercury \(=1.53-0.152 \mathrm{pH}\) Predictor Constant \(\mathrm{pH}\) \(\begin{array}{rrrr}\text { Coef } & \text { SE Coef } & \text { T } & \text { P } \\ 1.5309 & 0.2035 & 7.52 & 0.000 \\ -0.15230 & \text { * (c) }^{* *} & -5.02 & 0.000\end{array}\) stant \(S=* *(b)^{* *} \quad R-S q=* *(a)^{* x}\) Analysis of Variance Source Regression Residual Erri Total \(\begin{array}{rrrrrr} & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { P } \\ \text { ion } & 1 & 2.0024 & 2.0024 & 25.24 & 0.000 \\ \text { Error } & 51 & 4.0455 & 0.0793 & & \\ & 52 & 6.0479 & & & \end{array}\) (a) Use the information in the ANOVA table to compute and interpret the value of \(R^{2}\). (b) Show how to estimate the standard deviation of the error term, \(s_{\epsilon}\). (c) Use the result from part (b) and the summary statistics below to compute the standard error of the slope, \(S E,\) for this model: $$ \begin{array}{lrrrrr} \text { Variable } & \text { N } & \text { Mean } & \text { StDev } & \text { Minimum } & \text { Maximum } \\ \text { pH } & 53 & 6.591 & 1.288 & 3.600 & 9.100 \\ \text { AvgMercury } & 53 & 0.5272 & 0.3410 & 0.0400 & 1.3300 \end{array} $$

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

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

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