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Chapter 15: Problem 14

A journalist is reporting about some research on appropriate amounts of sleep for people 9 to 19 years of age. In that research, a linear regression model is used to describe the relationship between alertness and number of hours of sleep the night before. The researchers reported a \(95 \%\) confidence interval, but newspapers usually report an estimate and a margin of error. Explain how the journalist could determine the margin of error from the reported confidence interval.

Short Answer

Expert verified
To find the margin of error from the reported confidence interval, the journalist should first identify the lower bound (a) and upper bound (b) of the given confidence interval. Then, calculate the midpoint (estimate) as \(\frac{a+b}{2}\), and calculate the margin of error as \(\frac{b - a}{2}\). Finally, report the estimate and margin of error as "Estimate: Midpoint 卤 Margin of Error".

Step by step solution

01

Understand confidence interval and margin of error

A confidence interval is a range of values that estimates a population parameter with a specified level of confidence (in this case, 95%). It is usually represented as `(Lower Bound, Upper Bound)`. The margin of error is the measurement of uncertainty when estimating the true population value. It is calculated as half of the range within the confidence interval.
02

Extract the lower and upper bound from the given confidence interval

Suppose the researchers reported a 95% confidence interval like `(a, b)`, where `a` represents the lower bound, and `b` represents the upper bound.
03

Calculate the midpoint of the confidence interval

To do this, we will need to find the average of the lower and upper bounds. That is: \(Midpoint = \frac{a+b}{2}\)
04

Calculate the margin of error

Now, we need to calculate the margin of error. Since the margin of error is half of the range within the confidence interval, we can calculate it as follows: \(Margin\ of\ Error = \frac{b - a}{2}\)
05

Report the estimate and margin of error

Using the midpoint as the estimate and the margin of error calculated in step 4, the journalist can report the relationship between alertness and the number of hours slept as: "Estimate: Midpoint 卤 Margin of Error"

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Key Concepts

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

Margin of Error
When you're trying to figure out what a survey or study is really saying, it's important to consider the margin of error. This term represents how much you can expect your finding to vary from the true population value. Think of it as the wiggle room in a scientific estimate. To calculate the margin of error, look at the reported confidence interval, often written like this: (Lower Bound, Upper Bound). The margin of error is simply half of the range of this interval:\[Margin\ of\ Error = \frac{b - a}{2}\]This tells us how far off the estimate might be from the true population value, providing a clearer picture of the data's accuracy. If you see a large margin of error, it means there's more uncertainty in the estimate.
Linear Regression
Linear regression is a powerful statistical tool used to understand the relationship between two variables. In this exercise, it helps describe the link between alertness and hours of sleep. Essentially, it tries to "fit a line" through the data points that best explains how these two elements are connected.
  • The independent variable: This is what you think is impacting the other. Here, it's the number of hours of sleep.
  • The dependent variable: This is what you measure to see if it changes because of the independent variable. In this case, that鈥檚 alertness.
By plotting these on a graph and using the line, linear regression predicts the dependent variable based on the independent one. The more closely aligned the data points are to this line, the stronger the relationship.
Population Parameter Estimation
In statistics, when you hear about estimating a population parameter, it means we're trying to make an educated guess about an entire group's characteristic based on a sample. This is crucial in research because it's often impractical to study a whole population directly.
  • Parameters: These are the actual values we're interested in that describe the population, like the average hours of sleep needed.
  • Statistics: Values that describe the sample. For example, the average alertness level from those sampled.
By using confidence intervals around these sample statistics, we can provide an estimate for the population parameter, in which we state the likely range (the confidence interval) where the parameter lies, based on our sample.

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Most popular questions from this chapter

15.19 Acrylamide is a chemical that is sometimes found in cooked starchy foods and which is thought to increase the risk of certain kinds of cancer. The paper "A Statistical Regression Model for the Estimation of Acrylamide Concentrations in French Fries for Excess Lifetime Cancer Risk Assessment" (Food and Chemical Toxicology [2012]: \(3867-3876\) ) describes a study to investigate the effect of frying time (in seconds) and acrylamide concentration (in micrograms per kilogram) in french fries. The data in the accompanying table are approximate values read from a graph that appeared in the paper. \begin{tabular}{|cc|} \hline Frying Time & Acrylamide Concentration \\ \hline 150 & 155 \\ 240 & 120 \\ 240 & 190 \\ 270 & 185 \\ 300 & 140 \\ 300 & 270 \\ \hline \end{tabular} a. For these data, the estimated regression line for predicting \(y=\) acrylamide concentration based on \(x=\) frying time is \(y=87+0.359 x\). What is an estimate of the average change in acrylamide concentration associated with a 1-second increase in frying time? b. What would you predict for acrylamide concentration for a frying time of 250 seconds? c. Use the given Minitab output to decide if there is convincing evidence of a useful linear relationship between acrylamide concentration and frying time. You may assume that the necessary conditions have been met. R-sq \(\begin{array}{cc}\text { R-sq(adj) } & \text { R-sq(pred) } \\ 0.00 \% & 0.00 \%\end{array}\) \(\mathrm{S}\) 3 \(\mathrm{q}\) \(8 \%\) Coefficients \(\mathrm{K}-\mathrm{Sq}\) \(14.38 \%\) \(\begin{array}{lccccc}\text { Term } & \text { Coef } & \text { SE Coef } & \text { T-Value } & \text { P-Value } & \text { VIF } \\ \text { Constant } & 87 & 112 & 0.78 & 0.480 & \\ x & 0.359 & 0.438 & 0.82 & 0.459 & 1.00\end{array}\) Regression Equation \(y=87+0.359 x\)

Consider the accompanying data on \(x=\) advertising share and \(y=\) market share for a particular brand of soft drink during 10 randomly selected years. \(\begin{array}{llllllllll}x & 0.103 & 0.072 & 0.071 & 0.077 & 0.086 & 0.047 & 0.060 & 0.050 & 0.070 & 0.052\end{array}\) \(\begin{array}{llllllllll}\underline{y} & 0.135 & 0.125 & 0.120 & 0.086 & 0.079 & 0.076 & 0.065 & 0.059 & 0.051 & 0.039\end{array}\) a. Construct a scatterplot for these data. Do you think the simple linear regression model would be appropriate for describing the relationship between \(x\) and \(y ?\) b. Calculate the equation of the estimated regression line and use it to obtain the predicted market share when the advertising share is 0.090 c. Compute \(r^{2}\). How would you interpret this value? d. Calculate a point estimate of \(\sigma_{e^{-}}\) How many degrees of freedom is associated with this estimate?

The flow rate in a device used for air quality measurement depends on the pressure drop \(x\) (inches of water) across the device's filter. Suppose that for \(x\) values between 5 and \(20,\) these two variables are related according to the simple linear regression model with population regression line \(y=-0.12+0.095 x\) a. What is the mean flow rate for a pressure drop of 10 inches? A drop of 15 inches? b. What is the average change in flow rate associated with a 1 inch increase in pressure drop? Explain.

The paper "Predicting Yolk Height, Yolk Width, Albumen Length, Eggshell Weight, Egg Shape Index, Eggshell Thickness, Egg Surface Area of Japanese Quails Using Various Egg Traits as Regressors" (International journal of Poultry Science [2008]: \(85-88\) ) suggests that the simple linear regression model is reasonable for describing the relationship between \(y=\) eggshell thickness (in micrometers) and \(x=\) egg length (mm) for quail eggs. Suppose that the population regression line is \(y=0.135+0.003 x\) and that \(\sigma=0.005 .\) Then, for a fixed \(x\) value, \(y\) has a normal distribution with mean \(0.135+0.003 x\) and standard deviation 0.005 . a. What is the mean eggshell thickness for quail eggs that are \(15 \mathrm{~mm}\) in length? For quail eggs that are \(17 \mathrm{~mm}\) in length? b. What is the probability that a quail egg with a length of \(15 \mathrm{~mm}\) will have a shell thickness that is greater than \(0.18 \mu \mathrm{m} ?\) c. Approximately what proportion of quail eggs of length \(14 \mathrm{~mm}\) have a shell thickness of greater than \(0.175 ?\) Less than \(0.178 ?\)

Do taller adults make more money? The authors of the paper "Stature and Status: Height, Ability, and Labor Market Outcomes" (Journal of Political Economics [2008]: 499-532) investigated the association between height and earnings. They used the simple linear regression model to describe the relationship between \(x=\) height (in inches) and \(y=\) log(weekly gross earnings in dollars) in a very large sample of men. The logarithm of weekly gross earnings was used because this transformation resulted in a relationship that was approximately linear. The paper reported that the slope of the estimated regression line was \(b=0.023\) and the standard deviation of \(b\) was \(s_{b}=0.004\). Carry out a hypothesis test to decide if there is convincing evidence of a useful linear relationship between height and the logarithm of weekly earnings. You can assume that the basic assumptions of the simple linear regression model are met.
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