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Obtain as much information as you can about the \(P\) -value for the \(F\) test for model utility in each of the following situations: a. \(k=2, n=21\), calculated \(F=2.47\) b. \(k=8, n=25\), calculated \(F=5.98\) c. \(k=5, n=26\), calculated \(F=3.00\) d. The full quadratic model based on \(x_{1}\) and \(x_{2}\) is fit, \(n=20\), and calculated \(F=8.25\). \mathrm{\\{} e . ~ \(k=5, n=100\), calculated \(F=2.33\)

Short Answer

Expert verified
To calculate the actual P-values of these \(F\) tests, one would need to use an \(F\) distribution table or software. The specifics of these calculations exceed the scope of this exercise. However, one can interpret that larger \(F\) values with larger numerator (greater \(k\)) and smaller denominator (smaller \(n-k-1\)) degrees of freedom are likely to generate smaller \(P\)-values, suggesting a strong evidence against the null hypotheses. A full detailed interpretation with P-values calculation requires a statistical software or \(F\) distribution table.

Step by step solution

01

Understand the F Test

The \(F\) test is used to determine if the variances between two populations are equal. It calculates an \(F\) statistic which follows an \(F\) distribution. Here, 'model utility' refers to how useful the model is in explaining the data variances. It's done by comparing the variance explained by the model with the total variance.
02

Note Down The Parameters For Each Scenario

Let's note down the \(k\) (number of predictors), \(n\) (number of observations) and the calculated \(F\) statistic for each scenario to comprehend the variations.
03

Identify The Degrees of Freedom

Degrees of freedom for an \(F\) test depend on the values of \(n\) and \(k\). The denominator degrees of freedom is \(n-k-1\), while the numerator degrees of freedom is \(k\). Identify these for each scenario.
04

Interpreting The \(F\) Values

Generally, larger \(F\) values indicate that model is explaining more variance. However, the significance of this depends on the \(P\)-value. Without actual calculations or table lookups, we can generally say that larger \(F\) values with larger numerator and smaller denominator degrees of freedom are likely to generate smaller \(P\)-values, suggesting a strong evidence against the null hypotheses (hypotheses of no difference).
05

Comparing Different Scenarios

With the information from steps 3 and 4, compare the different scenarios. Make estimations about which model is likely more 'useful' (i.e., explaining more variance).

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Key Concepts

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

P-value interpretation
The P-value, or probability value, is a crucial statistic in hypothesis testing. It tells you the probability of observing your test results, or something more extreme, if the null hypothesis were true. In simple terms, the P-value quantifies the evidence against the null hypothesis. A low P-value indicates that it is unlikely the observed result was due to chance, suggesting that your model is capturing a real effect.

For instance, in an F test for model utility, if you get a P-value of 0.03, this means, assuming no real relationship, there's only a 3% chance that you'd observe the collected data or something more extreme due to random fluctuations alone. Conventional thresholds for P-values are 0.05 or 0.01, with values below these levels considered statistically significant, providing stronger evidence against the null hypothesis.
Variance analysis
Variance analysis in the context of the F test is about evaluating the differences in variability between groups in your data. When you perform an F test, you are essentially comparing the variance explained by your model, which is based on the hypothesized relationships, against the variance found in the data not explained by the model.

This process is central to determining the utility of your model. If your model explains a significant portion of the variance compared to the unexplained variance, it shows that your model has utility. In the scenarios provided, calculating the F statistic is part of this variance analysis process, which compares the model variance to the error variance.
Degrees of freedom
Degrees of freedom (df) are an essential part of variance analysis because they take into account the number of independent pieces of information in your data that go into estimating parameters. In an F test, there are two sets of degrees of freedom to consider: the numerator df, which is related to the number of predictors or groups being compared, and the denominator df, which is related to the number of observations.

For the problems at hand, the numerator df equals k and the denominator df equals n - k - 1. With df, the distributions of your test statistics are defined, allowing you to calculate the P-value and determine the statistical significance of your results. Degrees of freedom are also fundamental when using tables or software to find critical values for the F distribution.
F distribution
The F distribution is the theoretical distribution used for hypothesis testing when comparing variances. It is a ratio of two scaled chi-square distributions and hence is always non-negative and skewed positively. The shape of the F distribution changes based on the degrees of freedom in the numerator and the denominator.

When using F distribution for model utility, higher F values indicate that the model explains a significant amount of the variance in the data, while lower F values suggest the model may not be useful. Since not all F values are created equal, comparing them to a critical value from the F distribution allows us to judge the statistical significance of our test results.
Null hypothesis in statistics
In statistics, the null hypothesis is a general statement or default position that there is no relationship between two measured phenomena or no association among groups. In model utility testing using an F test, the null hypothesis typically asserts that any differences in variances are due to chance. This means that, under the null hypothesis, the model is assumed not to have utility.

The alternative hypothesis, on the contrary, is that the model does provide a better explanation than chance alone. Rejecting the null hypothesis suggests the model has utility and the predictor variables are indeed influencing the response variable. In your scenarios, finding significant F values could lead to the rejection of the null hypothesis, indicating the potential utility of the model in explaining the data.

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Most popular questions from this chapter

The accompanying MINITAB output results from fitting the model described in Exercise \(14.12\) to data. $$ \begin{array}{lrrr} \text { Predictor } & \text { Coef } & \text { Stdev } & \text { t-ratio } \\ \text { Constant } & 86.85 & 85.39 & 1.02 \\ \mathrm{X} 1 & -0.12297 & 0.03276 & -3.75 \\ \mathrm{X} 2 & 5.090 & 1.969 & 2.58 \\ \mathrm{X} 3 & -0.07092 & 0.01799 & -3.94 \\ \mathrm{X} 4 & 0.0015380 & 0.0005560 & 2.77 \\ \mathrm{~S}=4.784 & \mathrm{R}-\mathrm{sq}=90.8 \% & \mathrm{R}-\mathrm{s} q(\mathrm{adj})=89.4 \% \end{array} $$ $$ \begin{array}{lrrr} \text { Analysis of Variance } & & & \\ & \text { DF } & \text { SS } & \text { MS } \\ \text { Regression } & 4 & 5896.6 & 1474.2 \\ \text { Error } & 26 & 595.1 & 22.9 \\ \text { Total } & 30 & 6491.7 & \end{array} $$ a. What is the estimated regression equation? b. Using a \(.01\) significance level, perform the model utility test. c. Interpret the values of \(R^{2}\) and \(s_{e}\) given in the output.

Suppose that the variables \(y, x_{1}\), and \(x_{2}\) are related by the regression model $$ y=1.8+.1 x_{1}+.8 x_{2}+e $$ a. Construct a graph (similar to that of Figure \(14.5)\) showing the relationship between mean \(y\) and \(x_{2}\) for fixed values 10,20 , and 30 of \(x_{1}\). b. Construct a graph depicting the relationship between mean \(y\) and \(x_{1}\) for fixed values 50,55, and 60 of \(x_{2}\). c. What aspect of the graphs in Parts (a) and (b) can be attributed to the lack of an interaction between \(x_{1}\) and \(x_{2}\) ? d. Suppose the interaction term \(.03 x_{3}\) where \(x_{3}=x_{1} x_{2}\) is added to the regression model equation. Using this new model, construct the graphs described in Parts (a) and (b). How do they differ from those obtained in Parts (a) and (b)?

Obtain as much information as you can about the \(P\) -value for an upper-tailed \(F\) test in each of the following situations: a. \(\mathrm{df}_{1}=3, \mathrm{df}_{2}=15\), calculated \(F=4.23\) b. \(\mathrm{df}_{1}=4, \mathrm{df}_{2}=18\), calculated \(F=1.95\) c. \(\mathrm{df}_{1}=5, \mathrm{df}_{2}=20\), calculated \(F=4.10\) d. \(\mathrm{df}_{1}=4, \mathrm{df}_{2}=35\), calculated \(F=4.58\)

When coastal power stations take in large quantities of cooling water, it is inevitable that a number of fish are drawn in with the water. Various methods have been designed to screen out the fish. The article "Multiple \(\mathrm{Re}-\) gression Analysis for Forecasting Critical Fish Influxes at Power Station Intakes" (Journal of Applied Ecology [1983]: 33-42) examined intake fish catch at an English power plant and several other variables thought to affect fish intake: $$ \begin{aligned} y &=\text { fish intake (number of fish) } \\ x_{1} &=\text { water temperature }\left({ }^{\circ} \mathrm{C}\right) \\ x_{2} &=\text { number of pumps running } \\ x_{3} &=\text { sea state }(\text { values } 0,1,2, \text { or } 3) \\ x_{4} &=\text { speed }(\text { knots }) \end{aligned} $$ Part of the data given in the article were used to obtain the estimated regression equation $$ \hat{y}=92-2.18 x_{1}-19.20 x_{2}-9.38 x_{3}+2.32 x_{4} $$ (based on \(n=26\) ). SSRegr \(=1486.9\) and SSResid = \(2230.2\) were also calculated. a. Interpret the values of \(b_{1}\) and \(b_{4}\). b. What proportion of observed variation in fish intake can be explained by the model relationship? c. Estimate the value of \(\sigma\). d. Calculate adjusted \(R^{2}\). How does it compare to \(R^{2}\) itself?

The article "Readability of Liquid Crystal Displays: A Response Surface" (Human Factors \([1983]: 185-190\) ) used a multiple regression model with four independent variables, where \(y=\) error percentage for subjects reading a four-digit liquid crystal display $$ \begin{aligned} &\left.x_{1}=\text { level of backlight (from } 0 \text { to } 122 \mathrm{~cd} / \mathrm{m}\right) \\ &x_{2}=\text { character subtense }\left(\text { from } .025^{\circ} \text { to } 1.34^{\circ}\right) \end{aligned} $$ \(x_{3}=\) viewing angle \(\left(\right.\) from \(0^{\circ}\) to \(60^{\circ}\) ) \(x_{4}=\) level of ambient light (from 20 to \(1500 \mathrm{~lx}\) ) The model equation suggested in the article is $$ y=1.52+.02 x_{1}-1.40 x_{2}+.02 x_{3}-.0006 x_{4}+e $$ a. Assume that this is the correct equation. What is the mean value of \(y\) when \(x_{1}=10, x_{2}=.5, x_{3}=50\), and \(x_{4}=100 ?\) b. What mean error percentage is associated with a backlight level of 20 , character subtense of \(.5\), viewing angle of 10, and ambient light level of 30 ? c. Interpret the values of \(\beta_{2}\) and \(\beta_{3}\)

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