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Chapter 6: Problem 71

An automobile manufacturer is interested in the fuel efficiency of a proposed new car design. Six nonprofessional drivers were selected, and each one drove a prototype of the new car from Phoenix to Los Angeles. The resulting fuel efficiencies \((x,\) in miles per gallon \()\) are: $$ \begin{array}{llllll} 27.2 & 29.3 & 31.2 & 28.4 & 30.3 & 29.6 \end{array} $$ The normal scores for a sample of size 6 are $$ \begin{array}{llllll} -1.282 & -0.643 & -0.202 & 0.202 & 0.643 & 1.282 \end{array} $$ a. Construct a normal probability plot for the fuel efficiency data. Does the plot look linear? b. Calculate the correlation coefficient for the (normal score, \(x\) ) pairs. Compare this value to the appropriate critical \(r\) value from Table 6.2 to determine if it is reasonable to think that the fuel efficiency distribution is approximately normal.

Short Answer

Expert verified
In summary, the fuel efficiency data for the new car design was analyzed by constructing a normal probability plot and calculating the correlation coefficient. The plot appeared linear and the correlation coefficient was approximately -0.9797, which is greater than the critical r value of 0.811 (in absolute terms) for a sample size of 6 at a 0.05 significance level. Thus, it is reasonable to conclude that the fuel efficiency distribution is approximately normal.

Step by step solution

01

Plot fuel efficiency data against normal scores

First, make a table with two columns; one for the given fuel efficiency data (in ascending order, \(x\)) and the other for the given normal scores (in ascending order, \(z\)). Our table will look like this: $$ \begin{array}{c|c} x & z \\ \hline 27.2 & -1.282 \\ 28.4 & -0.643 \\ 29.3 & -0.202 \\ 29.6 & 0.202 \\ 30.3 & 0.643 \\ 31.2 & 1.282 \end{array} $$ Now, create a scatter plot by plotting fuel efficiency values (x) against normal scores (z). If the plot appears linear, then the fuel efficiency values are likely to follow a normal distribution.
02

Calculate the correlation coefficient

To determine if it is reasonable to believe the fuel efficiency data's distribution is approximately normal, we need to find the correlation coefficient (\(r\)) for the given \((z, x)\) pairs. A high correlation (close to 1) implies a linear relationship and thus, indicate that the data distribution is approximately normal. The formula to calculate the correlation coefficient 'r' is given by: $$ r = \frac{n \sum (zx) - (\sum z)(\sum x)}{\sqrt{[n\sum z^2 - (\sum z)^2][n\sum x^2 - (\sum x)^2]}} $$ Using the provided data, compute the summations needed for the formula: $$ \sum z = -1.282 - 0.643 - 0.202 + 0.202 + 0.643 + 1.282 = 0 $$ $$ \sum x = 27.2 + 28.4 + 29.3 + 29.6 + 30.3 + 31.2 = 176.0 $$ $$ \sum z^2 = (-1.282)^2 + (-0.643)^2 + (-0.202)^2 + (0.202)^2 + (0.643)^2 + (1.282)^2 = 3.000 $$ $$ \sum x^2 = 27.2^2 + 28.4^2 + 29.3^2 + 29.6^2 + 30.3^2 + 31.2^2 = 5{,}240.36 $$ $$ \sum zx = (-1.282)(27.2) + (-0.643)(28.4) + (-0.202)(29.3) + (0.202)(29.6) + (0.643)(30.3) + (1.282)(31.2) = 鈭�8.04 $$ Now, plug the summations into the correlation coefficient formula: $$ r = \frac{6(-8.04) - (0)(176)}{\sqrt{[6(3) - (0)^2][6(5{,}240.36) - (176)^2]}} \approx -0.9797 $$
03

Compare the correlation coefficient with the critical r value

The correlation coefficient we have calculated is \(r \approx -0.9797\). The closer the value is to 1, the more it is likely to follow a normal distribution. Now, to determine if it is reasonable to believe that the fuel efficiency distribution is approximately normal, compare the calculated value with the appropriate critical r value from the table. For a sample size of 6, and at a 0.05 significance level, the critical r value is 0.811. Since \(|-0.9797| > 0.811\), it is reasonable to conclude that the fuel efficiency distribution is approximately normal.

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Key Concepts

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

Normal Probability Plot
The normal probability plot is a graphical technique used to help assess whether or not a dataset is approximately normally distributed. This involves plotting each data point in the dataset against a corresponding theoretical normal score.
The process begins by arranging the data in ascending order and matching each observation with its respective normal score, which you have when you know the sample size. For instance, in our exercise with the fuel efficiency data and normal scores of size 6, each data point was paired with one of these scores.
When you present these pairs on a scatter plot, a nearly straight line indicates that the data are likely following a normal distribution. The more the points deviate from this line, the less the data resemble a normal distribution, making this plot a simple visual indicator in statistical analyses.
Correlation Coefficient
The correlation coefficient, denoted as \(r\), quantitatively measures the strength and direction of a linear relationship between two variables. In the context of the normal probability plot, the correlation coefficient helps further establish whether the data points conform to a normal distribution when plotted against theoretical normal scores.
For our task of evaluating fuel efficiency data for normality, the correlation coefficient was calculated using the formula:
  • \( r = \frac{n \sum zx - (\sum z)(\sum x)}{\sqrt{[n\sum z^2 - (\sum z)^2][n\sum x^2 - (\sum x)^2]}} \)
The resulting coefficient, close to 1 in magnitude, signifies a strong linear relationship, implying the data's distribution is approximately normal. Values nearer to 0 suggest weak or no linear relation. Hence, in our case, a high correlation coefficient of approximately \(-0.9797\) suggested a near-normal distribution of the fuel efficiency data.
Critical Value
A critical value in statistics is a point on the scale of the test statistic beyond which we reject the null hypothesis, and it helps decide whether results from statistical tests are significant. When evaluating a distribution's normality, we compare our computed correlation coefficient to a predefined critical value.
For this exercise, the critical value comes from a statistical table, developed based on sample size and the chosen significance level (commonly 0.05) for normality tests. With six samples, the critical value was noted at 0.811. This particular threshold serves as a benchmark: the calculated correlation coefficient must have an absolute value beyond this critical limit to substantiate normality.
In simpler terms, if the absolute value of the correlation is greater than the critical value, we accept that the data distribution could reasonably be normal, as was the case here.
Normal Scores
Normal scores are critical when constructing a normal probability plot as they represent the expected positions of data points in a theoretical normal distribution. These scores are typically based on the inverse standard normal distribution, meaning they mirror the quantiles which the sample sizes would align with in a truly normal dataset.
For six data points, each score reflects its rank within a standard normal distribution, such as
  • \(-1.282\)
  • \(-0.643\)
  • \(-0.202\)
  • 0.202
  • 0.643
  • 1.282
These are arranged symmetrically around zero, with negative values for lower ranks and positive ones for higher ranks. Notably, these values correspond with increasing data points in the observed dataset. Correct alignment of these scores with the empirical data in plotting helps determine the data's adherence to a normal distribution through visual representation.

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Most popular questions from this chapter

6.9 Two six-sided dice, one red and one white, will be rolled. List the possible values for each of the following random variables. a. \(x=\) sum of the two numbers showing b. \(y=\) difference between the number on the red die and the number on the white die (red - white) c. \(w=\) largest number showing

Consider the variable \(x=\) time required for a college student to complete a standardized exam. Suppose that for the population of students at a particular university, the distribution of \(x\) is well approximated by a normal curve with mean 45 minutes and standard deviation 5 minutes. a. If 50 minutes is allowed for the exam, what proportion of students at this university would be unable to finish in the allotted time? b. How much time should be allowed for the exam if you wanted \(90 \%\) of the students taking the test to be able to finish in the allotted time? c. How much time is required for the fastest \(25 \%\) of all students to complete the exam?

Let \(x\) denote the duration of a randomly selected pregnancy (the time elapsed between conception and birth). Accepted values for the mean value and standard deviation of \(x\) are 266 days and 16 days, respectively. Suppose that the probability distribution of \(x\) is (approximately) normal. a. What is the probability that the duration of a randomly selected pregnancy is between 250 and 300 days? b. What is the probability that the duration is at most 240 days? c. What is the probability that the duration is within 16 days of the mean duration? d. A "Dear Abby" newspaper column dated January 20, 1973 , contained a letter from a woman who stated that the duration of her pregnancy was exactly 310 days. (She wrote that the last visit with her husband, who was in the navy, occurred 310 days before the birth of her child.) What is the probability that the duration of pregnancy is at least 310 days? Does this probability make you skeptical of the claim? e. Some insurance companies will pay the medical expenses associated with childbirth only if the insurance has been in effect for more than 9 months ( 275 days). This restriction is designed to ensure that benefits are only paid if conception occurred during coverage. Suppose that conception occurred 2 weeks after coverage began. What is the probability that the insurance company will refuse to pay benefits because of the 275 -day requirement?

A pizza company advertises that it puts 0.5 pound of real mozzarella cheese on its medium-sized pizzas. In fact, the amount of cheese on a randomly selected medium pizza is normally distributed with a mean value of 0.5 pound and a standard deviation of 0.025 pound. a. What is the probability that the amount of cheese on a medium pizza is between 0.525 and 0.550 pounds? b. What is the probability that the amount of cheese on a medium pizza exceeds the mean value by more than 2 standard deviations? c. What is the probability that three randomly selected medium pizzas each have at least 0.475 pounds of cheese?

The time that it takes a randomly selected job applicant to perform a certain task has a distribution that can be approximated by a normal distribution with a mean of 120 seconds and a standard deviation of 20 seconds. The fastest \(10 \%\) are to be given advanced training. What task times qualify individuals for such training?
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