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Occasionally an investigator may wish to compute a confidence interval for \(\alpha\), the \(y\) intercept of the true regression line, or test hypotheses about \(\alpha .\) The estimated \(y\) intercept is simply the height of the estimated line when \(x=0\), since \(a+b(0)=a .\) This implies that \(s_{a}\) the estimated standard deviation of the statistic \(a\), results from substituting \(x^{\prime \prime}=0\) in the formula for \(s_{a+b x^{+}} .\) The desired confidence interval is then \(a \pm(t\) critical value \() s_{a}\) \(-\) and a test statistic is 1 $$ t=\frac{a-\text { hypothesized value }}{s_{a}} $$ a. The article "Comparison of Winter-Nocturnal Geostationary Satellite Infrared-Surface Temperature with Shelter-Height Temperature in Florida" (Remote Sensing of the Emvironment \([1983]: 313-327\) ) used the simple linear regression model to relate surface temperature as measured by a satellite \((y)\) to actual air temperature \((x)\) as determined from a thermocouple placed on a traversing vehicle. Selected data are given (read from a scatterplot in the article). \(\begin{array}{rrrrrrrr}x & -2 & -1 & 0 & 1 & 2 & 3 & 4 \\ y & -3.9 & -2.1 & -2.0 & -1.2 & 0.0 & 1.9 & 0.6 \\ x & 5 & 6 & 7 & & & & \\ y & 2.1 & 1.2 & 3.0 & & & & \end{array}\) Estimate the true regression line. b. Compute the estimated standard deviation \(s_{a}\). Carry out a test at level of significance \(.05\) to see whether the \(y\) intercept of the true regression line differs from zero. c. Compute a \(95 \%\) confidence interval for \(\alpha\). Does the result indicate that \(\alpha=0\) is plausible? Explain.

Short Answer

Expert verified
A detailed solution is required to answer this question. It includes estimating the true regression line, testing the hypothesis about the y-intercept and calculating a confidence interval for the y-intercept etc.

Step by step solution

01

Estimate the true regression line

To get the estimated regression line, one has to calculate the slope and intercept of the line. First, calculate the mean of x values and y values. Then, calculate the slope (b) of the line using the formula \( b = \frac{\Sigma((x_i - \bar{x})(y_i - \bar{y}))}{\Sigma((x_i - \bar{x})^2)} \). Afterwards, calculate the intercept(a) using the formula \( a = \bar{y} - b\bar{x} \) where \bar{x} and \bar{y} indicate the mean of x and y values, respectively.
02

Calculate the estimated standard deviation \( s_a \)

The estimated standard deviation, \( s_a \), can be calculated using the formula \( s_a = \sqrt{\frac{\Sigma{(y_i - \hat{y_i})^2}}{n-2}} \), where \( n \) is the number of points, \( y_i \) are the actual y values, and \( \hat{y_i} \) are the predicted y values obtained from the model.
03

Conduct a hypothesis test

A hypothesis test is conducted to see if the y intercept \( \alpha \) could reasonably be 0. Here, null hypothesis is \( H_0: \alpha = 0 \). Alternative hypothesis is \( H_1: \alpha \neq 0 \). Then calculate the test statistic \( t = \frac{a - 0}{s_a} \), where 'a' is obtained from the regression line and \( s_a \) is the estimated standard deviation. The p-value associated with this test statistic is then calculated. The null hypothesis is rejected and it is concluded that \( \alpha \) is significantly different from 0, if the p-value is less than 0.05.
04

Calculate a 95% confidence interval for \( \alpha \)

A 95% confidence interval for \( \alpha \) can be calculated using the formula \( a \pm t_{0.025, n-2}s_a \), where 'a' is the y-intercept obtained from the regression line, and \( t_{0.025, n-2} \) is the critical t value for \( \alpha = 0.025 \) and degrees of freedom \( n-2 \). Validate the possibility of \( \alpha = 0 \) based on the range of the confidence interval. If the interval contains 0, then \( \alpha = 0 \) is plausible.

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Key Concepts

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

Confidence Interval
Understanding the concept of the confidence interval is crucial for interpreting the results from regression analysis accurately. A confidence interval gives us a range of values, usually calculated from the sample data, for which a parameter (like the y-intercept in our example) is expected to lie within a specific probability. For instance, a 95% confidence interval suggests that if we repeated the study multiple times, we would expect the true parameter to fall within this range 95% of the time.

Constructing a confidence interval involves determining the mean of the estimator and then calculating the margin of error, which hinges on the critical value from the t-distribution (linked to the confidence level and degrees of freedom in the sample data) and the standard deviation of the estimator. In our exercise, we calculate the 95% confidence interval for the y-intercept, \( \alpha \) of the true regression line, giving us insights into the plausible values that \( \alpha \) can take, considering the random variation present in the sample data.
Hypothesis Testing
Hypothesis testing in statistics is a method used to make decisions or inferences about a population parameter based on sample data. In the context of regression analysis, we may want to test hypotheses about the y-intercept, \( \alpha \). For example, we may hypothesize that the y-intercept is equal to zero (no effect or relationship), which is our null hypothesis (\( H_0: \alpha = 0 \)).

The alternative hypothesis (\( H_1 \)) would state that the y-intercept is not equal to zero (signifying a possible effect or relationship). We use a test statistic, like the t-statistic, to determine whether to reject the null hypothesis. This test statistic takes into account both the magnitude of the difference from the hypothesized value and the variability of the estimate, expressed through the standard deviation. Results with a p-value less than the significance level, such as 0.05, suggest rejecting the null hypothesis, indicating that the y-intercept is significantly different from the hypothesized value.
Standard Deviation
Standard deviation is a measure that quantifies the amount of variation or dispersion in a set of values. A low standard deviation indicates that the values tend to be close to the mean (also called the expected value), whereas a high standard deviation indicates that the values are spread out over a wider range.

In regression analysis, the standard deviation of the y-intercept, \( s_{\alpha} \), reflects the extent to which the intercepts obtained from different sample data would spread around the true y-intercept of the population regression line. The smaller the standard deviation, the more precise our estimate is likely to be. We use this measure of spread to compute both the margin of error for confidence intervals and the test statistic for hypothesis testing, as seen in the given exercise.
Y-Intercept
In the context of a linear regression model, the y-intercept (often denoted as \( \alpha \) or 'a') is a critical parameter representing the value of the dependent variable when all independent variables are equal to zero. It is the starting point of the regression line when it crosses the y-axis on a graph.

If we're analyzing the relationship between surface temperatures from satellite data (y) and actual air temperatures (x), the y-intercept will give us an estimate of what the satellite temperature reading would be when the actual air temperature is zero. Estimating the y-intercept with precision is vital, as it sets the baseline from which we assess the effect of the independent variable(s) on the dependent variable. As shown in our exercise, by using the mean values of x and y and the slope of the regression, we can solve for the estimated y-intercept of the model.

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Most popular questions from this chapter

A random sample of \(n=347\) students was selected, and each one was asked to complete several questionnaires, from which a Coping Humor Scale value \(x\) and a Depression Scale value \(y\) were determined ("Depression and Sense of Humor" (Psychological Reports [1994]: \(1473-1474\) ). The resulting value of the sample correlation coefficient was \(-.18\). a. The investigators reported that \(P\) -value \(<.05 .\) Do you agree? b. Is the sign of \(r\) consistent with your intuition? Explain. (Higher scale values correspond to more developed sense of humor and greater extent of depression.) c. Would the simple linear regression model give accurate predictions? Why or why not?

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 "Morphometry of Nerve Fiber Bundle Pores in the Optic Nerve Head of the Human" (Experimental Eye Research \([1988]: 559-568\) ) 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 the researchers had believed a priori 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.

The article "Effect of Temperature on the pH of Skim Milk" (Journal of Dairy Research [1988]: 277- 280) reported on a study involving \(x=\) temperature \(\left({ }^{\circ} \mathrm{C}\right)\) under specified experimental conditions and \(y=\) milk \(\mathrm{pH}\). The accompanying data (read from a graph) are a representative subset of that which appeared in the article: \(\begin{array}{rrrrrrrrr}x & 4 & 4 & 24 & 24 & 25 & 38 & 38 & 40 \\ y & 6.85 & 6.79 & 6.63 & 6.65 & 6.72 & 6.62 & 6.57 & 6.52\end{array}\) $$ \begin{array}{lrrrrrrrr} x & 45 & 50 & 55 & 56 & 60 & 67 & 70 & 78 \\ y & 6.50 & 6.48 & 6.42 & 6.41 & 6.38 & 6.34 & 6.32 & 6.34 \\ \sum x=678 & \sum y=104.54 & \sum x^{2}=36,056 & \\ \sum y^{2}=683.4470 & & \sum x y=4376.36 & & \end{array} $$ Do these data strongly suggest that there is a negative linear relationship between temperature and \(\mathrm{pH}\) ? State and test the relevant hypotheses using a significance level of \(.01\).

The shelf life of packaged food depends on many factors. Dry cereal is considered to be a moisture-sensitive product (no one likes soggy cereal!) with the shelf life determined primarily by moisture content. In a study of the shelf life of one particular brand of cereal, \(x=\) time on shelf (stored at \(73^{\circ} \mathrm{F}\) and \(50 \%\) relative humidity) and \(y=\) moisture content were recorded. The resulting data are from "Computer Simulation Speeds Shelf Life Assessments" (Package Engineering [1983]: 72-73). \(\begin{array}{rrrrrrrr}x & 0 & 3 & 6 & 8 & 10 & 13 & 16 \\ y & 2.8 & 3.0 & 3.1 & 3.2 & 3.4 & 3.4 & 3.5 \\ x & 20 & 24 & 27 & 30 & 34 & 37 & 41 \\ y & 3.1 & 3.8 & 4.0 & 4.1 & 4.3 & 4.4 & 4.9\end{array}\) a. Summary quantities are $$ \begin{array}{ll} \sum x=269 & \sum y=51 \quad \sum x y=1081.5 \\ \sum y^{2}=7745 & \sum x^{2}=190.78 \end{array} $$ Find the equation of the estimated regression line for predicting moisture content from time on the shelf. b. Does the simple linear regression model provide useful information for predicting moisture content from knowledge of shelf time? c. Find a \(95 \%\) interval for the moisture content of an individual box of cereal that has been on the shelf 30 days. d. According to the article, taste tests indicate that this brand of cereal is unacceptably soggy when the moisture content exceeds 4.1. Based on your interval in Part (c), do you think that a box of cereal that has been on the shelf 30 days will be acceptable? Explain.

The article "Effects of Enhanced UV-B Radiation on Ribulose-1,5-Biphosphate, Carboxylase in Pea and Soybean" (Environmental and Experimental Botany [1984]: 131-143) included the accompanying data on pea plants, with \(y=\) sunburn index and \(x=\) distance \((\mathrm{cm})\) from an ultraviolet light source. \(\begin{array}{lllllllll}x & 18 & 21 & 25 & 26 & 30 & 32 & 36 & 40 \\ y & 4.0 & 3.7 & 3.0 & 2.9 & 2.6 & 2.5 & 2.2 & 2.0 \\ x & 40 & 50 & 51 & 54 & 61 & 62 & 63 & \\ y & 2.1 & 1.5 & 1.5 & 1.5 & 1.3 & 1.2 & 1.1 & \end{array}\) $$ \begin{array}{lc} \sum x=609 & \sum y=33.1 \quad \sum x^{2}=28,037 \\ \sum y^{2}=84.45 & \sum x y=1156.8 \end{array} $$ Estimate the mean change in the sunburn index associated with an increase of \(1 \mathrm{~cm}\) in distance in a way that includes information about the precision of estimation.

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