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91Ó°ÊÓ

Manufacturers of two competing automobile models, Gofer and Diplomat, each claim to have the lowest mean fuel consumption. Let \(\mu_{1}\) be the mean fuel consumption in miles per gallon (mpg) for the Gofer and \(\mu_{2}\) the mean fuel consumption in mpg for the Diplomat. The two manufacturers have agreed to a test in which several cars of each model will be driven on a 100 -mile test run. Then the fuel consumption, in mpg, will be calculated for each test run. The average of the mpg for all 100 -mile test runs for each model gives the corresponding mean. Assume that for each model the gas mileages for the test runs are normally distributed with \(\sigma=2 \mathrm{mpg}\). Note that each car is driven for one and only one 100 -mile test run. a. How many cars (i.e., sample size) for each model are required to estimate \(\mu_{1}-\mu_{2}\) with a \(90 \%\) confidence level and with a margin of error of estimate of \(1.5 \mathrm{mpg}\) ? Use the same number of cars (i.e., sample size) for each model. b. If \(\mu_{1}\) is actually \(33 \mathrm{mpg}\) and \(\mu_{2}\) is actually \(30 \mathrm{mpg}\), what is the probability that five cars for each model would yield \(\bar{x}_{1} \geq \bar{x}_{2}\) ?

Short Answer

Expert verified
The sample size required for each model to get a 90% confidence level with a margin of error of 1.5 mpg will be 8 cars. The probability that the average mpg of 5 Gofer cars is greater than the average mpg of 5 Diplomat cars is approximately 0.7241 (or 72.41%).

Step by step solution

01

Calculation of Sample Size

The formula for calculation of sample size based on the margin of error for difference between two populations is given by \( n = (Z_{\alpha /2} * 2\sigma / E)^2 \). Here, \( Z_{\alpha /2} \) is the z-value corresponding to a 90% confidence level, which is 1.645. The standard deviation \( \sigma \) equals 2 mpg, and the margin of error E is 1.5 mpg. Plug these values into the formula to calculate the sample size. Always round up in case of fraction because it's not possible to have part of a car.
02

Calculation of Probability

In order to find the probability that \( \bar{x}_{1} \geq \bar{x}_{2} \), first find out the standard error of the sampling distribution of the difference in means. This is done using the formula \( SE = \sqrt{2\sigma^2/n} \). Here, \( n \) is 5 (sample size) and \( \sigma \) is 2 (standard deviation). Then calculate z-score using the formula \( Z = (\bar{x}_{1} - \bar{x}_{2} - (\mu_{1} - \mu_{2})) / SE \). Here, \( \mu_{1} = 33, \mu_{2} = 30 \). After calculating the \( Z \) score, use the standard normal distribution table (or use a software to calculate) to find the probability that \( Z \geq 0 \), which is identical to the probability that \( \bar{x}_{1} \geq \bar{x}_{2} \).

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Key Concepts

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

Sample Size Calculation
When determining how many samples are needed for a study, we refer to this as the sample size calculation. This is crucial for ensuring that our results are statistically significant and provides insights into the population we're investigating. In exercises like the one with the Gofer and Diplomat models, we need to figure out how many cars must be tested to confidently say one is more fuel-efficient than the other. The formula for sample size is:
  • \[ n = \left(\frac{Z_{\alpha /2} \cdot 2\sigma}{E}\right)^2 \]
To determine this, we need:
  • The z-value (\( Z_{\alpha /2} \)) corresponding to the confidence level, which is 1.645 for 90% confidence level.
  • The standard deviation (\( \sigma \)) of the data, here 2 mpg.
  • The margin of error (E), the maximum allowable difference that we can tolerate, which is to be kept at 1.5 mpg here.
By plugging these values into the formula, we can find the number of cars needed for each model. It's vital to round up any fractions, as partial cars are not practical in a physical test.
Confidence Interval
The confidence interval is a key component in statistical hypothesis testing, providing a range where we believe the true population parameter lies. This interval is built around our sample statistic and reflects our confidence level. For instance, in the fuel consumption test, a 90% confidence level indicates that if we were to repeat this experiment 100 times, 90 of those times the true mean difference in mpg between the Gofer and Diplomat would fall within our calculated interval. The interval is determined by:
  • Calculating the sample mean, which acts as the midpoint of the interval.
  • Multiplying the standard error by the z-value to find the margin of error and adding/subtracting it from the sample mean. This yields the range of plausible values for the mean difference.
  • The confidence interval thus provides a measure of the reliability of our estimate, helping to make more informed decisions about real-world hypotheses.
Through these steps, we balance precision with the required assurance that our interval indeed captures the true parameter difference between models.
Z-Score
The z-score is crucial in hypothesis testing because it describes how far a sample mean is from the population mean, measured in terms of standard errors. In our exercise, calculating the z-score is essential for understanding the outcome of our hypothesis test between Gofer and Diplomat cars. The formula for the z-score is:
  • \[ Z = \frac{(\bar{x}_{1} - \bar{x}_{2} - (\mu_{1} - \mu_{2}))}{SE} \]
Where:
  • \(\bar{x}_{1}\) and \(\bar{x}_{2}\) are the sample means for Gofer and Diplomat.
  • \(\mu_{1}\) and \(\mu_{2}\) are the true means assumed in the hypothesis (33 mpg and 30 mpg).
  • \(SE\) is the standard error.
The z-score tells us where our observed data stands relative to the expected outcome. High absolute values suggest a significant deviation, guiding us in making inferences about potential differences in fuel consumption.
Standard Error
The standard error (SE) quantifies the variation of a sampling distribution, providing insight into how much variability we expect when drawing samples from the population. This is vital when comparing two different models like Gofer and Diplomat, as it helps assess whether observed differences are due to actual differences or just random sampling fluctuation. To calculate SE in the context of the test, you use the formula:
  • \[ SE = \sqrt{\frac{2\sigma^2}{n}} \]
Where \(\sigma\) is the standard deviation, and \(n\) is the sample size.The SE is smaller with large sample sizes or lower variability, indicating more reliable estimates of population parameters. In this exercise, the smaller SE will reassure us that repeating the test under similar conditions will yield comparable outcomes, thus enhancing our confidence in the decision we make based on the test results.

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Most popular questions from this chapter

A sample of 1000 observations taken from the first population gave \(x_{1}=290 .\) Another sample of 1200 observations taken from the second population gave \(x_{2}=396\). a. Find the point estimate of \(p_{1}-p_{2}\) b. Make a \(98 \%\) confidence interval for \(p_{1}-p_{2}\). c. Show the rejection and nonrejection regions on the sampling distribution of \(\hat{p}_{1}-\hat{p}_{2}\) for \(H_{0}: p_{1}=p_{2}\) versus \(H_{1}: p_{1}

Conduct the following tests of hypotheses, assuming that the populations of paired differences are normally distributed a. \(H_{0}: \mu_{d}=0, \quad H_{1}: \mu_{d} \neq 0, \quad n=26, \quad \bar{d}=9.6, \quad s_{d}=3.9, \quad \alpha=.05\) b. \(H_{0}: \mu_{d}=0, \quad H_{1}: \mu_{d}>0, \quad n=15, \quad \bar{d}=8.8, \quad s_{d}=4.7, \quad \alpha=.01\) c. \(H_{0}: \mu_{d}=0, \quad H_{1}: \mu_{d}<0, \quad n=20, \quad \bar{d}=-7.4, \quad s_{d}=2.3, \quad \alpha=.10\)

The weekly weight losses of all dieters on Diet I have a normal distribution with a mean of \(1.3\) pounds and a standard deviation of \(.4\) pound. The weekly weight losses of all dieters on Diet II have a normal distribution with a mean of \(1.5\) pounds and a standard deviation of \(.7\) pound. A random sample of 25 dieters on Diet I and another sample of 36 dieters on Diet II are observed. a. What is the probability that the difference between the two sample means, \(\bar{x}_{1}-\bar{x}_{2}\), will be within \(-.15\) to \(.15\), that is, \(-.15<\bar{x}_{1}-\bar{x}_{2}<.15 ?\) b. What is the probability that the average weight loss \(\bar{x}_{1}\) for dieters on Diet I will be greater than the average weight loss \(\bar{x}_{2}\) for dieters on Diet II? c. If the average weight loss of the 25 dieters using Diet I is computed to be \(2.0\) pounds, what is the probability that the difference between the two sample means, \(\bar{x}_{1}-\bar{x}_{2}\), will be within \(-.15\) to \(.15\), that is, \(-.15<\bar{x}_{1}-\bar{x}_{2}<.15 ?\) d. Suppose you conclude that the assumption \(-.15<\mu_{1}-\mu_{2}<.15\) is reasonable. What does this mean to a person who chooses one of these diets?

What is the shape of the sampling distribution of \(\hat{p}_{1}-\hat{p}_{2}\) for two large samples? What are the mean and standard deviation of this sampling distribution?

According to Exercise \(10.27\), an insurance company wants to know if the average speed at which men drive cars is higher than that of women drivers. The company took a random sample of 27 cars driven by men on a highway and found the mean speed to be 72 miles per hour with a standard deviation of \(2.2\) miles per hour. Another sample of 18 cars driven by women on the same highway gave a mean speed of 68 miles per hour with a standard deviation of \(2.5\) miles per hour. Assume that the speeds at which all men and all women drive cars on this highway are both normally distributed with unequal population standard deviations. a. Construct a \(98 \%\) confidence interval for the difference between the mean speeds of cars driven by all men and all women on this highway. b. Test at the \(1 \%\) significance level whether the mean speed of cars driven by all men drivers on this highway is higher than that of cars driven by all women drivers. c. Suppose that the sample standard deviations were \(1.9\) and \(3.4\) miles per hour, respectively. Redo parts a and b. Discuss any changes in the results.

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