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Concept explainers
Residual Plot: Miles per Gallon Consider the data of Problem 9.
- (a) Make a residual plot for the least-squares model.
- (b) Use the residual plot to comment about the appropriateness of the least-squares model for these data. See Problem 19.
9. Weight of Car: Miles per Gallon Do heavier cars really use more gasoline? Suppose a car is chosen at random. Let x be the weight of the car (in hundreds of pounds), and let y be the miles per gallon (mpg). The following information is based on data taken from Consumer Reports (Vol. 62, No. 4).
Complete parts (a) through (e), given ∑x = 299, ∑y = 167, ∑x2 = 11,887, ∑y2 = 3773, ∑xy = 5814, and r ≈ –0.946.
(f) Suppose a car weighs x = 38 (hundred pounds). What does the least-squares line forecast for y = miles per gallon?
Expand Your Knowledge: Residual Plot The least-squares line usually does not go through all the sample data points (x, y). In fact, for a specified x value from a data pair (x, y), there is usually a difference between the predicted value and the y value paired with x. This difference is called the residual.
The residual is the difference between the y value in a specified data pair (x, y) and the value
One way to assess how well a least-squares line serves as a model for the data is a residual plot. To make a residual plot, we put the x values in order on the horizontal axis and plot the corresponding residuals
- (a) If the least-squares line provides a reasonable model for the data, the pattern of points in the plot will seem random and unstructured about the horizontal line at 0. Is this the case for the residual plot?
- (b) If a point on the residual plot seems far outside the pattern of other points, it might reflect an unusual data point (x, y), called an outlier. Such points may have quite an influence on the least-squares model. Do there appear to be any outliers in the data for the residual plot?
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Chapter 9 Solutions
WebAssign Printed Access Card for Brase/Brase's Understandable Statistics: Concepts and Methods, 12th Edition, Single-Term
- Question 1 The data shown in Table 1 are and R values for 24 samples of size n = 5 taken from a process producing bearings. The measurements are made on the inside diameter of the bearing, with only the last three decimals recorded (i.e., 34.5 should be 0.50345). Table 1: Bearing Diameter Data Sample Number I R Sample Number I R 1 34.5 3 13 35.4 8 2 34.2 4 14 34.0 6 3 31.6 4 15 37.1 5 4 31.5 4 16 34.9 7 5 35.0 5 17 33.5 4 6 34.1 6 18 31.7 3 7 32.6 4 19 34.0 8 8 33.8 3 20 35.1 9 34.8 7 21 33.7 2 10 33.6 8 22 32.8 1 11 31.9 3 23 33.5 3 12 38.6 9 24 34.2 2 (a) Set up and R charts on this process. Does the process seem to be in statistical control? If necessary, revise the trial control limits. [15 pts] (b) If specifications on this diameter are 0.5030±0.0010, find the percentage of nonconforming bearings pro- duced by this process. Assume that diameter is normally distributed. [10 pts] 1arrow_forward4. (5 pts) Conduct a chi-square contingency test (test of independence) to assess whether there is an association between the behavior of the elderly person (did not stop to talk, did stop to talk) and their likelihood of falling. Below, please state your null and alternative hypotheses, calculate your expected values and write them in the table, compute the test statistic, test the null by comparing your test statistic to the critical value in Table A (p. 713-714) of your textbook and/or estimating the P-value, and provide your conclusions in written form. Make sure to show your work. Did not stop walking to talk Stopped walking to talk Suffered a fall 12 11 Totals 23 Did not suffer a fall | 2 Totals 35 37 14 46 60 Tarrow_forwardQuestion 2 Parts manufactured by an injection molding process are subjected to a compressive strength test. Twenty samples of five parts each are collected, and the compressive strengths (in psi) are shown in Table 2. Table 2: Strength Data for Question 2 Sample Number x1 x2 23 x4 x5 R 1 83.0 2 88.6 78.3 78.8 3 85.7 75.8 84.3 81.2 78.7 75.7 77.0 71.0 84.2 81.0 79.1 7.3 80.2 17.6 75.2 80.4 10.4 4 80.8 74.4 82.5 74.1 75.7 77.5 8.4 5 83.4 78.4 82.6 78.2 78.9 80.3 5.2 File Preview 6 75.3 79.9 87.3 89.7 81.8 82.8 14.5 7 74.5 78.0 80.8 73.4 79.7 77.3 7.4 8 79.2 84.4 81.5 86.0 74.5 81.1 11.4 9 80.5 86.2 76.2 64.1 80.2 81.4 9.9 10 75.7 75.2 71.1 82.1 74.3 75.7 10.9 11 80.0 81.5 78.4 73.8 78.1 78.4 7.7 12 80.6 81.8 79.3 73.8 81.7 79.4 8.0 13 82.7 81.3 79.1 82.0 79.5 80.9 3.6 14 79.2 74.9 78.6 77.7 75.3 77.1 4.3 15 85.5 82.1 82.8 73.4 71.7 79.1 13.8 16 78.8 79.6 80.2 79.1 80.8 79.7 2.0 17 82.1 78.2 18 84.5 76.9 75.5 83.5 81.2 19 79.0 77.8 20 84.5 73.1 78.2 82.1 79.2 81.1 7.6 81.2 84.4 81.6 80.8…arrow_forward
- Name: Lab Time: Quiz 7 & 8 (Take Home) - due Wednesday, Feb. 26 Contingency Analysis (Ch. 9) In lab 5, part 3, you will create a mosaic plot and conducted a chi-square contingency test to evaluate whether elderly patients who did not stop walking to talk (vs. those who did stop) were more likely to suffer a fall in the next six months. I have tabulated the data below. Answer the questions below. Please show your calculations on this or a separate sheet. Did not stop walking to talk Stopped walking to talk Totals Suffered a fall Did not suffer a fall Totals 12 11 23 2 35 37 14 14 46 60 Quiz 7: 1. (2 pts) Compute the odds of falling for each group. Compute the odds ratio for those who did not stop walking vs. those who did stop walking. Interpret your result verbally.arrow_forwardSolve please and thank you!arrow_forward7. In a 2011 article, M. Radelet and G. Pierce reported a logistic prediction equation for the death penalty verdicts in North Carolina. Let Y denote whether a subject convicted of murder received the death penalty (1=yes), for the defendant's race h (h1, black; h = 2, white), victim's race i (i = 1, black; i = 2, white), and number of additional factors j (j = 0, 1, 2). For the model logit[P(Y = 1)] = a + ß₁₂ + By + B²², they reported = -5.26, D â BD = 0, BD = 0.17, BY = 0, BY = 0.91, B = 0, B = 2.02, B = 3.98. (a) Estimate the probability of receiving the death penalty for the group most likely to receive it. [4 pts] (b) If, instead, parameters used constraints 3D = BY = 35 = 0, report the esti- mates. [3 pts] h (c) If, instead, parameters used constraints Σ₁ = Σ₁ BY = Σ; B = 0, report the estimates. [3 pts] Hint the probabilities, odds and odds ratios do not change with constraints.arrow_forward
- Solve please and thank you!arrow_forwardSolve please and thank you!arrow_forwardQuestion 1:We want to evaluate the impact on the monetary economy for a company of two types of strategy (competitive strategy, cooperative strategy) adopted by buyers.Competitive strategy: strategy characterized by firm behavior aimed at obtaining concessions from the buyer.Cooperative strategy: a strategy based on a problem-solving negotiating attitude, with a high level of trust and cooperation.A random sample of 17 buyers took part in a negotiation experiment in which 9 buyers adopted the competitive strategy, and the other 8 the cooperative strategy. The savings obtained for each group of buyers are presented in the pdf that i sent: For this problem, we assume that the samples are random and come from two normal populations of unknown but equal variances.According to the theory, the average saving of buyers adopting a competitive strategy will be lower than that of buyers adopting a cooperative strategy.a) Specify the population identifications and the hypotheses H0 and H1…arrow_forward
- You assume that the annual incomes for certain workers are normal with a mean of $28,500 and a standard deviation of $2,400. What’s the chance that a randomly selected employee makes more than $30,000?What’s the chance that 36 randomly selected employees make more than $30,000, on average?arrow_forwardWhat’s the chance that a fair coin comes up heads more than 60 times when you toss it 100 times?arrow_forwardSuppose that you have a normal population of quiz scores with mean 40 and standard deviation 10. Select a random sample of 40. What’s the chance that the mean of the quiz scores won’t exceed 45?Select one individual from the population. What’s the chance that his/her quiz score won’t exceed 45?arrow_forward
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