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Chapter 13: Problem 15

Exercise \(13.10\) presented \(y=\) hardness of molded plastic and \(x=\) time elapsed since the molding was completed. Summary quantities included \(n=15 \quad b=2.50 \quad\) SSResid \(=1235.470\) \(\sum(x-\bar{x})^{2}=4024.20\) a. Calculate the estimated standard deviation of the statistic \(b\). b. Obtain a \(95 \%\) confidence interval for \(\beta\), the slope of the population regression line. c. Does the interval in Part (b) suggest that \(\beta\) has been precisely estimated? Explain.

Short Answer

Expert verified
The short answer will be formed after the numerical calculations are performed with the given inputs and subsequent interpretation of percentile estimates. However, a general perspective is that the precision of \(\beta\) would be high if the calculated interval excludes zero, indicating a significant relation between variables.

Step by step solution

01

Calculating the estimated standard deviation of b

The estimated standard deviation of \(b\), denoted as \(se_b\), can be calculated using formula: \(se_{b}= \sqrt{SSResid / (n-2)} / \sqrt{\sum{(x-\bar{x})^2}}\). Using given values, \(SSResid =1235.470, \quad n = 15, \quad \sum{(x-\bar{x})^2} = 4024.20\). Substitute these into the formula to get the answer.
02

Obtaining a 95% confidence interval for 尾

A 95% confidence interval for the population slope (\(\beta\)) can be calculated using the formula: \((b - t_{\alpha/2, n-2}*se_{b}, b + t_{\alpha/2, n-2}*se_{b})\). where \(t_{\alpha/2, n-2}\) is the t value for a two-tailed test with alpha level equalling 0.05 and \(n-2\) degrees of freedom. The value of \(t_{0.025, 13}\) can be found in t-distribution table. After finding \(t_{0.025, 13}\), substitute the calculated \(se_{b}\) and given \(b = 2.5\) into the formula to get the confidence interval.
03

Analyzing the precision of 尾 estimation

If the confidence interval includes zero, it suggests that the estimate of \(\beta\) is not precise as it indicates that there is not enough evidence to conclude that\(\beta\) is different from zero. However, if the confidence interval does not contain zero, then the estimate of \(\beta\) can be considered relatively precise, as there's statistical evidence showing a significant relationship between time elapsed and hardness of molded plastic.

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Key Concepts

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

Confidence Interval
A confidence interval provides a range of values that likely include the true value of an unknown parameter. In regression analysis, when we talk about the confidence interval for the population slope \( \beta \), it helps us estimate how much change in the dependent variable (e.g., hardness of plastic) can be expected with a one-unit change in the predictor (e.g., time elapsed). A 95% confidence interval, for instance, says that if we were to take 100 different random samples and compute an interval estimate for each sample, we would expect about 95 of the 100 intervals to cover the true population slope \( \beta \). This confidence level is based on the idea of repeated sampling. Calculating a confidence interval involves:
  • Estimating the standard deviation of the slope, \( se_b \).
  • Using a \( t \)-distribution to find \( t_{\alpha/2, n-2} \), which considers the variability and sample size.
  • Applying the formula: \( (b - t_{\alpha/2, n-2} \cdot se_b, \, b + t_{\alpha/2, n-2} \cdot se_b) \).
This interval allows us to assess whether the estimated slope is significantly different from zero, suggesting a meaningful relationship between the predictor and outcome variables.
Standard Deviation
Standard deviation measures how spread out numbers are in a data set. In the context of regression analysis, the standard deviation of the slope estimate \( se_b \) provides insight into the precision of our slope estimation. The formula for calculating \( se_b \) is: \[se_b = \sqrt{\frac{SSResid}{n-2}} \/ \sqrt{\sum{(x-\bar{x})^2}}\]where:
  • \( SSResid \) is the sum of squared residuals, indicating the variation not explained by the regression model.
  • \( n \) is the sample size, telling how many observations are under consideration.
  • \( \sum{(x-\bar{x})^2} \) represents variability in the predictor variable.
By understanding \( se_b \), you can judge the reliability of the \( b \) value. A smaller \( se_b \) means your slope estimate is more precise, implying less variability in the prediction of the dependent variable based on changes in the predictor.
Population Slope
Understanding the population slope \( \beta \) is central in regression analysis. It represents the average change in the dependent variable for each one-unit change in the independent variable. In practical terms, if \( \beta \) is significantly different from zero, it suggests a real relationship between the variables. When estimating \( \beta \), researchers typically rely on sample data, as calculating \( \beta \) for an entire population is often impractical.The precision of the estimated \( \beta \) can be gauged from:
  • The width of its confidence interval: A narrow interval means a more precise estimation.
  • The standard deviation of the slope estimate: A smaller \( se_b \) indicates less dispersion and more confidence in the estimated slope value.
A precisely estimated \( \beta \) allows for reliable predictions and inferences to be made about the relationship between variables, making it a powerful tool in many areas of research and decision-making.

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Most popular questions from this chapter

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?

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Give a brief answer, comment, or explanation for each of the following. a. What is the difference between \(e_{1}, e_{2}, \ldots, e_{n}\) and the \(n\) residuals? b. The simple linear regression model states that \(y=\alpha+\beta x\). c. Does it make sense to test hypotheses about \(b\) ? d. SSResid is always positive. e. A student reported that a data set consisting of \(n=6\) observations yielded residuals \(2,0,5,3,0\), and 1 from the least-squares line. f. A research report included the following summary quantities obtained from a simple linear regression analysis: \(\sum(y-\bar{y})^{2}=615 \quad \sum(y-\hat{y})^{2}=731\)

Reduced visual performance with increasing age has been a much-studied phenomenon in recent years. This decline is due partly to changes in optical properties of the eye itself and partly to neural degeneration throughout the visual system. As one aspect of this problem, the article presented the accompanying data on \(x=\) age and \(y=\) percentage of the cribriform area of the lamina scleralis occupied by pores. \(\begin{array}{llllllllll}x & 22 & 25 & 27 & 39 & 42 & 43 & 44 & 46 & 46 \\ y & 75 & 62 & 50 & 49 & 54 & 49 & 59 & 47 & 54 \\ x & 48 & 50 & 57 & 58 & 63 & 63 & 74 & 74 & \\ y & 52 & 58 & 49 & 52 & 49 & 31 & 42 & 41 & \end{array}\) a. Suppose that prior to this study the researchers had believed that the average decrease in percentage area associated with a 1 -year age increase was \(.5 \%\). Do the data contradict this prior belief? State and test the appropriate hypotheses using a. 10 significance level. b. Estimate true average percentage area covered by pores for all 50 -year- olds in the population in a way that conveys information about the precision of estimation.
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