/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 33 The fuel economy information on ... [FREE SOLUTION] | 91Ó°ÊÓ

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The fuel economy information on a new SUV's window sticker indicates that its new owner can expect \(16 \mathrm{mpg}\) (miles per gallon) in city driving and \(20 \mathrm{mpg}\) for highway driving and \(18 \mathrm{mpg}\) overall. Accurate gasoline records for one such vehicle were kept, and a random sample of mileage per tank of gasoline was collected: $$\begin{array}{llllllllll}\hline 17.6 & 17.7 & 18.1 & 22.0 & 17.0 & 19.4 & 18.9 & 17.4 & 21.0 & 19.2 \\\18.3 & 19.1 & 20.7 & 16.7 & 19.4 & 18.2 & 18.4 & 17.1 & 17.4 & 15.8 \\\17.9 & 18.0 & 16.3 & 17.5 & 17.3 & 20.4 & 19.1 & 21.0 & 18.1 & 19.0 \\\19.6 & 18.9 & 16.8 & 18.2 & 17.6 & 19.1 & 18.0 & 16.8 & 20.9 & 17.9 \\\17.7 & 20.3 & 18.6 & 19.0 & 16.5 & 19.4 & 18.6 & 18.6 & 17.3 & 18.7 \\\\\hline\end{array}$$ a. Determine whether an assumption of normality is reasonable. Explain. b. Construct a \(95 \%\) confidence interval for the estimate of the mean mileage per gallon. c. What does the confidence interval suggest about SUV fuel economy expectations as expressed on the window sticker?

Short Answer

Expert verified
To determine if normality is reasonable, use the Central Limit Theorem since the sample is large. Then calculate the sample mean and standard deviation. Use these values to construct a 95% confidence interval for the mean mileage per gallon. Interpret this interval in the context of the number given on the SUV's window sticker.

Step by step solution

01

Evaluate Normality

Given the large data set, one could plot the data as a histogram or a normal Q-Q plot to evaluate if normality is a reasonable assumption. However, lacking such tools, we will infer normality based on the Central Limit Theorem which states that if you have a population with mean \( \mu \) and standard deviation \( \sigma \) and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. Given the size of our sample (n=50) is large, we can say that an assumption of normality is reasonable.
02

Calculate Mean and Standard Deviation

First, calculate the mean (average) of the data set by summing all the data points and dividing by the number of data points. Second, calculate the standard deviation, which is a measure of how spread out the numbers are. Use the following formula for the standard deviation: \( \sigma = \sqrt{\frac{\sum_{i=1}^{n}(x_i - \mu)^2}{n-1}} \); where \( x_i \) represents each data point, \( \mu \) represents the mean, and n is the number of data points.
03

Construct a 95% Confidence Interval for the Mean Mileage Per Gallon

A confidence interval gives an estimated range of values which is likely to include an unknown population parameter. For a 95% accuracy, the z-score is 1.96. The formula for confidence interval is: \( CI = \mu \pm (z*\frac{\sigma}{\sqrt{n}}) \). Using the mean and standard deviation calculated in step 2, plug in the values to the formula to get the confidence interval.
04

Interpret the Confidence Interval

The result from step 3 is a range of values. If the range includes the number on the SUV's sticker (18 mpg), then the sticker is consistent with the collected data.

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Key Concepts

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

Normality Assumption
When we talk about the normality assumption in statistics, we're essentially discussing whether a dataset resembles a normal distribution, which is a bell-shaped curve when plotted. This is important for many statistical methods because they require the data to be approximately normally distributed to produce reliable results.

In the context of the fuel economy data for the SUV, we would typically look at a histogram or a Q-Q plot to assess normality. However, even without these visual tools, we can rely on an important statistical principle—the Central Limit Theorem—to help justify the assumption of normality. If a sample is sufficiently large, usually considered to be more than 30 observations, the distribution of the sample means tends to be normal, regardless of the shape of the population distribution. In our case, with a sample size of 50, it's reasonable to believe the data is approximately normally distributed.

This normality assumption is crucial because it allows us to use certain statistical methods that rely on this characteristic. Without it, calculations such as confidence intervals for the mean mileage per gallon would be less reliable or might require non-parametric methods instead.
Central Limit Theorem
The Central Limit Theorem (CLT) is a statistical theory that plays a pivotal role in the field of inferential statistics. It tells us that when independent random variables are added, their sum tends toward a normal distribution, even if the original variables themselves are not normally distributed. This theorem applies as long as two conditions are met:

  • The sample size must be large enough (commonly n > 30).
  • The observations must be independent.

This theorem is the reason we could comfortably make a normality assumption for the SUV's fuel economy data set. With a sample size of 50, which is larger than the typical threshold of 30, the CLT suggests that the distribution of the sample mean will approximate a normal distribution. Thus, even without graphical analysis, we can proceed with statistical procedures that assume normality, such as constructing confidence intervals.
Population Mean
When we calculate the mean, or average, of a set of data, we're attempting to find the center or the 'typical' value of that data set. In statistics, when we talk about the population mean (represented by the Greek letter \( \mu \)), we're referring to the average of all possible values in the population.

In our exercise, the population mean represents the true average mileage per gallon for all SUVs of the model in consideration. Since checking every single SUV is impractical, we use sample means to estimate the population mean. We calculated the sample mean based on the SUV mileage per tank data and then constructed a 95% confidence interval around this mean. This interval is our best estimate of where the true population mean lies.

By comparing this confidence interval to what's claimed on the SUV's window sticker, we can assess how well the vehicle's stated fuel economy aligns with our sample data. If the interval includes the value on the sticker, it suggests that the vehicle's stated fuel economy is consistent with real-world data collected from the sample.

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Most popular questions from this chapter

How important is the assumption "The sampled population is normally distributed" to the use of Student's \(t\) -distribution? Using a computer, simulate drawing 100 samples of size 10 from each of three different types of population distributions, namely, a normal, a uniform, and an exponential. First generate 1000 data values from the population and construct a histogram to see what the population looks like. Then generate 100 samples of size 10 from the same population; each row represents a sample. Calculate the mean and standard deviation for each of the 100 samples. Calculate \(t \neq\) for each of the 100 samples. Construct histograms of the 100 sample means and the 100 t \(\star\) values. (Additional details can be found in the Student Solutions Manual.) For the samples from the normal population: a. Does the \(\bar{x}\) distribution appear to be normal? Find percentages for intervals and compare them with the normal distribution. b. Does the distribution of \(t \star\) appear to have a \(t\) -distribution with df \(=9 ?\) Find percentages for intervals and compare them with the \(t\) -distribution. For the samples from the rectangular or uniform population: c. Does the \(\bar{x}\) distribution appear to be normal? Find percentages for intervals and compare them with the normal distribution. d. Does the distribution of \(t \star\) appear to have a \(t\) -distribution with df \(=9 ?\) Find percentages for intervals and compare them with the \(t\) -distribution. For the samples from the skewed (exponential) population: e. Does the \(\bar{x}\) distribution appear to be normal? Find percentages for intervals and compare them with the normal distribution. f. Does the distribution of \(t \star\) appear to have a \(t\) -distribution with df \(=9 ?\) Find percentages for intervals and compare them with the \(t\) -distribution. In summary: g. In each of the preceding three situations, the sampling distribution for \(\bar{x}\) appears to be slightly different from the distribution of \(t \star .\) Explain why. h. Does the normality condition appear to be necessary in order for the calculated test statistic \(t \star\) to have a Student's \(t\) -distribution? Explain.

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