Chapter 13: Problem 33
Explain the difference between a confidence interval and a prediction interval. How can a prediction level of \(95 \%\) be interpreted?
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Chapter 13: Problem 33
Explain the difference between a confidence interval and a prediction interval. How can a prediction level of \(95 \%\) be interpreted?
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The data of Exercise \(13.25\), in which \(x=\) milk temperature and \(y=\) milk \(\mathrm{pH}\), yield $$ \begin{aligned} &n=16 \quad \bar{x}=42.375 \quad S_{x x}=7325.75 \\ &b=-.00730608 \quad a=6.843345 \quad s_{e}=.0356 \end{aligned} $$ a. Obtain a \(95 \%\) confidence interval for \(\alpha+\beta(40)\), the mean milk \(\mathrm{pH}\) when the milk temperature is \(40^{\circ} \mathrm{C} .\) b. Calculate a \(99 \%\) confidence interval for the mean milk \(\mathrm{pH}\) when the milk temperature is \(35^{\circ} \mathrm{C} .\) c. Would you recommend using the data to calculate a \(95 \%\) confidence interval for the mean \(\mathrm{pH}\) when the temperature is \(90^{\circ} \mathrm{C}\) ? Why or why not?
The employee relations manager of a large company was concerned that raises given to employees during a recent period might not have been based strictly on objective performance criteria. A sample of \(n=20 \mathrm{em}\) ployees was selected, and the values of \(x\), a quantitative measure of productivity, and \(y\), the percentage salary increase, were determined for each one. A computer package was used to fit the simple linear regression model, and the resulting output gave the \(P\) -value \(=.0076\) for the model utility test. Does the percentage raise appear to be linearly related to productivity? Explain.
A regression of \(x=\) tannin concentration \((\mathrm{mg} / \mathrm{L})\) and \(y=\) perceived astringency score was considered in Examples \(5.2\) and \(5.6\). The perceived astringency was computed from expert tasters rating a wine on a scale from 0 to 10 and then standardizing the rating by computing a \(z\) -score. Data for 32 red wines (given in Example 5.2) was used to compute the following summary statistics and estimated regression line: $$ \begin{aligned} &n=32 \quad \bar{x}=.6069 \quad \sum(x-\bar{x})^{2}=1.479 \\ &\text { SSResid }=1.936 \quad \hat{y}=-1.59+2.59 x \end{aligned} $$ a. Calculate a \(95 \%\) confidence interval for the mean astringency rating for red wines with a tannin concentration of \(.5 \mathrm{mg} / \mathrm{L}\). b. When two \(95 \%\) confidence intervals are computed, it can be shown that the simultaneous confidence level is at least \([100-2(5)] \%=90 \%\). That is, if both intervals are computed for a first sample, for a second sample, for a third sample, and so on, in the long run at least \(90 \%\) of the samples will result in intervals which both capture the values of the corresponding population characteristics. Calculate confidence intervals for the mean astringency rating when the tannin concentration is \(.5 \mathrm{mg} / \mathrm{L}\) and when the tannin concentration is \(.7 \mathrm{mg} / \mathrm{L}\) in such a way that the simultaneous confidence level is at least \(90 \%\). c. If two \(99 \%\) confidence intervals were computed, what do you think could be said about the simultaneous confidence level? d. If a \(95 \%\) confidence interval were computed for the mean astringency rating when \(x=.5\), another confidence interval was computed for \(x=.6\), and yet another one for \(x=.7\), what do you think would be the simultaneous confidence level for the three resulting intervals?
Carbon aerosols have been identified as a contributing factor in a number of air quality problems. In a chemical analysis of diesel engine exhaust, \(x=\) mass \(\left(\mu \mathrm{g} / \mathrm{cm}^{2}\right)\) and \(y=\) elemental carbon \(\left(\mu \mathrm{g} / \mathrm{cm}^{2}\right)\) were recorded. The estimated regression line for this data set is \(\hat{y}=31+.737 x\). The accompanying table gives the observed \(x\) and \(y\) values and the corresponding standardized residuals. \(\begin{array}{lrrrrr}x & 164.2 & 156.9 & 109.8 & 111.4 & 87.0 \\\ y & 181 & 156 & 115 & 132 & 96 \\ \text { St. resid. } & 2.52 & 0.82 & 0.27 & 1.64 & 0.08 \\ x & 161.8 & 230.9 & 106.5 & 97.6 & 79.7 \\ y & 170 & 193 & 110 & 94 & 77 \\ \text { St. resid. } & 1.72 & -0.73 & 0.05 & -0.77 & -1.11 \\\ x & 118.7 & 248.8 & 102.4 & 64.2 & 89.4 \\ y & 106 & 204 & 98 & 76 & 89 \\\ \text { St. resid. } & -1.07 & -0.95 & -0.73 & -0.20 & -0.68 \\ x & 108.1 & 89.4 & 76.4 & 131.7 & 100.8 \\ y & 102 & 91 & 97 & 128 & 88 \\ \text { St. resid. } & -0.75 & -0.51 & 0.85 & 0.00 & -1.49\end{array}\) \(\begin{array}{lllll}78.9 & 387.8 & 135.0 & 82.9 & 117.9\end{array}\) a. Construct a standardized residual plot. Are there any unusually large residuals? Do you think that there are any influential observations? b. Is there any pattern in the standardized residual plot that would indicate that the simple linear regression model is not appropriate? c. Based on your plot in Part (a), do you think that it is reasonable to assume that the variance of \(y\) is the same at each \(x\) value? Explain.
Explain the difference between \(r\) and \(\rho\).
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