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Chapter 13: Problem 2

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 true regression line \(y=-0.12+0.095 x\). a. What is the true average flow rate for a pressure drop of 10 in.? A drop of 15 in.? b. What is the true average change in flow rate associated with a 1 -in. increase in pressure drop? Explain. c. What is the average change in flow rate when pressure drop decreases by 5 in.?

Short Answer

Expert verified
a. The average flow rate for a pressure drop of 10 in. is \(0.83\) and for 15 in. it's \(1.285\). b. The average change in flow rate associated with a 1-in. increase in pressure drop is \(0.095\). c. The average change in flow rate for a 5-in. decrease in pressure drop is \(-0.475\).

Step by step solution

01

Calculate the average flow rate for a pressure drop of 10 in.

To calculate the average flow rate for a pressure drop of 10 in., substitute \(x = 10\) into the equation. Thus, the calculation will be \(y = -0.12 + 0.095*10\).
02

Calculate the average flow rate for a pressure drop of 15 in.

Similarly, for an average flow rate at a pressure drop of 15 in., substitute \(x = 15\) into the equation and calculate \(y = -0.12 + 0.095*15\).
03

Calculate average change in flow rate for 1-in. increase in pressure drop

The true average change in flow rate associated with a 1-in. increase in pressure drop is represented by the coefficient in the equation, which is 0.095.
04

Calculate average change in flow rate for 5-in. decrease in pressure drop

When the pressure drop decreases by 5 in., compute the change in flow rate by multiplying the coefficient (0.095) by the change in pressure (5), i.e., \(-5 * 0.095\). The negative sign is used to signify the decrease in pressure.

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Key Concepts

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

Flow Rate Calculation
Flow rate calculation is crucial in understanding the relationship between pressure drop and the flow rate in a device. In our scenario, we are using a simple linear regression model to explore this relationship. The model is expressed as follows:\[ y = -0.12 + 0.095x \]Where:
  • \(y\) represents the flow rate (in cubic feet per minute or another appropriate unit).
  • \(x\) is the pressure drop measured in inches of water.
To find the flow rate for any given pressure drop, we substitute the value of \(x\) into the equation. For instance, if the pressure drop \(x = 10\), then substituting would give\[ y = -0.12 + 0.095 \times 10 \]Calculating, we get that the flow rate \(y\) for a 10-inch pressure drop is approximately 0.83. Similarly, for a pressure drop of 15 inches:\[ y = -0.12 + 0.095 \times 15 \]This results in a flow rate of about 1.305. These calculations allow us to interpret how the flow rate behaves under different conditions of pressure.
Pressure Drop
Pressure drop is an essential factor in assessing the performance of devices that measure air quality by indicating how much resistance is present as air passes through the filter. In this model, pressure drop is used as the independent variable \(x\) in the linear regression equation to predict how it affects flow rate.Understanding pressure drop helps determine operational efficiency. A high-pressure drop often implies that the filter is becoming clogged and may need replacement. It indicates how pressure differences influence air movement through the system.By consistently monitoring pressure drop, operators can proactively manage maintenance schedules and optimize device functionality. The direct relationship established by the linear regression model informs us that if the pressure increases by one unit, the flow rate changes by the coefficient of the pressure in the regression equation, specifically 0.095 as described above.
Average Change in Flow Rate
The average change in flow rate associated with changes in pressure drop is determined by the slope of the regression line. In our exercise, the slope is 0.095, which describes how much the flow rate changes per unit increase in pressure drop.

1-inch Increase

For a 1-inch increase in pressure drop, the flow rate increases by the value of the slope, which is 0.095.

5-inch Decrease

When considering a 5-inch decrease in pressure drop, first note the negative sign associated with a decrease. Multiply the slope by this pressure change:\[ ext{Change in flow rate} = -5 \times 0.095 \]This results in a decrease of 0.475 in the flow rate. By interpreting this data, we can predict and adjust for changes in device performance, ensuring consistent results.

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Most popular questions from this chapter

The authors of the article "Age, Spacing and Growth Rate of Tamarix as an Indication of Lake Boundary Fluctuations at Sebkhet Kelbia, Tunisia" (Journal of Arid Environments [1982]: 43-51) used a simple linear regression model to describe the relationship between \(y=\) vigor (average width in centimeters of the last two annual rings) and \(x=\) stem density (stems/m \(^{2}\) ). The estimated model was based on the following data. Also given are the standardized residuals. \(\begin{array}{lrrrrr}x & 4 & 5 & 6 & 9 & 14 \\ y & 0.75 & 1.20 & 0.55 & 0.60 & 0.65 \\ \text { St. resid. } & -0.28 & 1.92 & -0.90 & -0.28 & 0.54\end{array}\) $$ \begin{array}{lrrrrr} x & 15 & 15 & 19 & 21 & 22 \\ y & 0.55 & 0.00 & 0.35 & 0.45 & 0.40 \\ \text { St. resid. } & 0.24 & -2.05 & -0.12 & 0.60 & 0.52 \end{array} $$ a. What assumptions are required for the simple linear regression model to be appropriate? b. Construct a normal probability plot of the standardized residuals. Does the assumption that the random deviation distribution is normal appear to be reasonable? Explain. c. Construct a standardized residual plot. Are there any unusually large residuals? d. Is there anything about the standardized residual plot that would cause you to question the use of the simple linear regression model to describe the relationship between \(x\) and \(y ?\)

The article "Effects of Gamma Radiation on Juvenile and Mature Cuttings of Quaking Aspen" (Forest Science \([1967]: 240-245\) ) reported the following data on \(x=\) exposure to radiation \((\mathrm{kR} / 6\) hours) and \(y=\mathrm{dry}\) weight of roots (mg): 8 $$ \begin{array}{lrrrrr} x & 0 & 2 & 4 & 6 & 8 \\ y & 110 & 123 & 119 & 86 & 62 \end{array} $$ a. Construct a scatterplot for these data. Does the plot suggest that the simple linear regression model might be appropriate? b. The estimated regression line for these data is \(\hat{y}=\) \(127-6.65 x\) and the standardized residuals are as given. \(\begin{array}{rrrrrr} & 0 & 2 & 4 & 6 & 8 \\ \text { resid. } & -1.55 & 0.68 & 1.25 & -0.05 & -1.06\end{array}\) \(x\) St. re Construct a standardized residual plot. What does the plot suggest about the adequacy of the simple linear regression model?

The article "Technology, Productivity, and Industry Structure" (Technological Forecasting and Social Change [1983]: 1-13) included the accompanying data on \(x=\) research and development expenditure and \(y=\) growth rate for eight different industries. $$ \begin{array}{rrrrrrrrr} x & 2024 & 5038 & 905 & 3572 & 1157 & 327 & 378 & 191 \\ y & 1.90 & 3.96 & 2.44 & 0.88 & 0.37 & -0.90 & 0.49 & 1.01 \end{array} $$ a. Would a simple linear regression model provide useful information for predicting growth rate from research and development expenditure? Use a \(.05\) level of significance. b. Use a \(90 \%\) confidence interval to estimate the average change in growth rate associated with a 1 -unit increase in expenditure. Interpret the resulting interval.

Exercise \(13.8\) gave data on \(x=\) treadmill run time to exhaustion and \(y=20-\mathrm{km}\) ski time for a sample of 11 biathletes. Use the accompanying MINITAB output to answer the following questions. The regression equation is ski \(=-88.8-2.33\) tread \(\begin{array}{lrrrr}\text { Predictor } & \text { Coef } & \text { Stdev } & \text { t-ratio } & p \\\ \text { Constant } & 88.796 & 5.750 & 15.44 & 0.000 \\ \text { tread } & 2.3335 & 0.5911 & 3.95 & 0.003 \\ s=2.188 & \text { R-sq }=63.4 \% & \text { R-sq }(a d j)=59.3 \% & \end{array}\) Analysis of Variance \(\begin{array}{lrrrrr}\text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { p } \\ \text { Regression } & 1 & 74.630 & 74.630 & 15.58 & 0.003 \\ \text { Error } & 9 & 43.097 & 4.789 & & \\ \text { Total } & 10 & 117.727 & & & \end{array}\) a. Carry out a test at significance level \(.01\) to decide whether the simple linear regression model is useful. b. Estimate the average change in ski time associated with a 1 -minute increase in treadmill time, and do so in a way that conveys information about the precision of estimation. c. MINITAB reported that \(s_{a+b(10)}=.689\). Predict ski time for a single biathlete whose treadmill time is \(10 \mathrm{~min}\), and do so in a way that conveys information about the precision of prediction. d. MINITAB also reported that \(s_{a+b(11)}=1.029 .\) Why is this larger than \(s_{a+b(10) ?}\)

A regression of \(y=\) sunburn index for a pea plant on \(x=\) distance from an ultraviolet light source was considered in Exercise 13.22. The data and summary statistics presented there give $$ \begin{aligned} &n=15 \quad \bar{x}=40.60 \quad \sum(x-\bar{x})^{2}=3311.60 \\ &b=-.0565 \quad a=4.500 \quad \text { SSResid }=.8430 \end{aligned} $$ a. Calculate a \(95 \%\) confidence interval for the true average sunburn index when the distance from the light source is \(35 \mathrm{~cm}\). 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, yet again for a third, and so on, in the long run at least \(90 \%\) of the samples will result in intervals both of which capture the values of the corresponding population characteristics. Calculate confidence intervals for the true mean sunburn index when the distance is \(35 \mathrm{~cm}\) and when the distance is \(45 \mathrm{~cm}\) in such a way that the simultaneous confidence level is at least \(90 \%\). c. If two \(99 \%\) 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 true mean index when \(x=35\), another \(95 \%\) confidence interval were computed when \(x=40\), and yet another one when \(x=45\), what do you think would be the simultaneous confidence level for the three resulting intervals? e. Retum to Part (d) and answer the question posed there if the individual confidence level for each interval were \(99 \%\).
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