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Gamma Corporation is considering the installation of govemors on cars driven by its sales staff. These devices would limit the car speeds to a preset level, which is expected to improve fuel economy. The company is planning to test several cars for fuel consumption without governors for 1 week. Then governors would be installed in the same cars, and fuel consumption will be monitored for another week. Gamma Corporation wants to estimate the mean difference in fuel consumption with a margin of error of estimate of 2 mpg with a 90 \% confidence level. Assume that the differences in fuel consumption are normally distributed and that previous studies suggest that an estimate of \(s_{d}=3 \mathrm{mpg}\) is reasonable. How many cars should be tested? (Note that the critical value of \(t\) will depend on \(n\), so it will be necessary to use trial and error.)

Short Answer

Expert verified
The sample size \(n\) that Gamma Corporation needs can be calculated using the formula for sample size estimation and by an iterative approach, which involves guessing numbers for \(n\), refining the t-value and then recalculating the sample size until an acceptable level of precision is achieved.

Step by step solution

01

Define the variables

First, define the variables from the given problem. Here, \(d = 2 \, mpg\) is the margin of error, \(s_{d}=3 \, mpg\) is the estimated standard deviation, and the confidence level is \(90\% = 1 - \alpha = 0.9\), so \(\alpha = 0.1\). Since this is a two-tailed test, \(\alpha/2 = 0.05, \, \text{and then}\, t_{\alpha/2} = t_{0.05}\). The value of \(t_{0.05}\) will depend on the degrees of freedom which is \(df = n - 1\). Since \(n\) is unknown, the exact value of \(t_{0.05}\) can't be determined at this stage.
02

Setup the equation

We'll use the formula for the sample size, which for this problem is \(n = (\frac{t_{\alpha/2} \times s_{d}}{d})^2\). Substituting the known values we have \(n = (\frac{t_{0.05} \times 3}{2})^2\). We can't solve for \(n\) yet because \(t_{0.05}\) depends on \(n\).
03

Iterations to find \(n\)

As the exact \(n\) number is unknown, we have to make guesses for \(n\) and, as a result, refine the \(t_{0.05}\) values. Start by guessing \(n=10\), with \(df=9\), check the t-table for \(t_{0.05,9}=1.833\). Substitute this value in the formula and find the calculation of \(n\). Keep adjusting the guess for \(n\) until the computed \(n\) and the guessed \(n\) are close enough.
04

Final calculation

By going through the loop of steps and refining the guesses, we can find the minimum sample size required to ensure a 90 \% confidence level and a margin of error of 2 mpg.

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Key Concepts

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

Margin of Error
The margin of error is a crucial concept when estimating statistics. It represents the range within which the true value of a parameter (like the mean) is expected to lie. In our exercise, the margin of error is 2 miles per gallon (mpg). This means that the estimated mean difference in fuel consumption might vary by up to 2 mpg from the true mean difference.

The margin of error is determined by the standard deviation of the data, the sample size, and the confidence level. In simpler terms, it's how much wiggle room we allow around our estimated mean. A smaller margin of error means our estimate is more precise, requiring a larger sample size or greater certainty about other factors.
  • Wide margins mean less precision.
  • Narrow margins require more data or are more precise.
Confidence Level
Confidence level tells us how sure we are that the true parameter falls within our margin of error. In this exercise, we are using a 90% confidence level. This means there's a 90% chance that the true mean difference in fuel consumption is within our calculated range, given our sample data.

Higher confidence levels lead to broader intervals, as we need more certainty, which can require a larger sample size.
  • A 90% confidence level is commonly used for moderate assurance.
  • Higher levels, like 95%, offer more certainty but also widen the estimation range, possibly needing larger samples.
Selecting the right confidence level depends on how much certainty is needed for the conclusions being drawn.
T-Distribution
The t-distribution is a statistical distribution used when the sample size is small and the population variance is unknown. It helps estimate the population parameters, especially when dealing with smaller sample sizes. In this exercise, the t-distribution is key to estimating the sample size needed to achieve our margin of error and confidence level.

The shape of the t-distribution changes based on degrees of freedom, which is related to sample size ( - 1 in this case). Unlike the standard normal distribution, the t-distribution is broader, accommodating the greater variability in estimates from smaller samples.
  • T-distribution adjusts for small sample sizes.
  • Critical value of t changes with degrees of freedom.
It's essential for accurate sample size calculations in our exercise.
Fuel Efficiency Testing
Fuel efficiency testing, like in this exercise, involves assessing how modifications to vehicles impact fuel consumption. Gamma Corporation is evaluating whether installing governors can improve this.

The process includes testing fuel consumption before and after the modifications, measuring any differences. By estimating the mean difference, the company can infer the effectiveness of the governors. This testing relies on carefully controlled conditions to ensure accuracy and relevance.
  • Comparative testing between normal and modified states.
  • Statistical analysis determines the modification's impact.
Understanding the exact impact on fuel efficiency helps in making data-driven decisions for vehicle usage strategies.

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Most popular questions from this chapter

In parts of the eastern United States, whitctail deer are a major nuisance to farmers and homeowners, frequently damaging crops, gardens, and landscaping. A consumer organization arranges a test of two of the leading deer repellents \(\mathrm{A}\) and \(\mathrm{B}\) on the market. Fifty-six unfenced gardens in areas having high concentrations of deer are used for the test. Twenty-nine gardens are chosen at random to receive repellent \(\mathrm{A}\), and the other 27 receive repellent \(\mathrm{B}\). For each of the 56 gardens, the time elapsed between application of the repellent and the appearance in the garden of the first deer is recorded. For repellent \(\mathrm{A}\), the mean time is 101 hours. For repellent \(B\), the mean time is 92 hours. Assume that the two populations of elapsed times have normal distributions with population standard deviations of 15 and 10 hours, respectively. a. Let \(\mu_{1}\) and \(\mu_{2}\) be the population means of elapsed times for the two repellents, respectively. Find the point estimate of \(\mu_{1}-\mu_{2}\). b. Find a \(97 \%\) confidence interval for \(\mu_{1}-\mu_{2}\) c. Test at a \(2 \%\) significance level whether the mean elapsed times for repellents \(A\) and \(B\) are different. Use both approaches, the critical-value and \(p\) -value, to perform this test.

One type of experiment that might be performed by an exercise physiologist is as follows: Each person in a random sample is tested in a weight room to determine the heaviest weight with which he or she can perform an incline press five times with his or her dominant arm (defined as the hand that a person uses for writing). After a significant rest period, the same weight is determined for each individual's nondominant arm. The physiologist is interested in the differences in the weights pressed by each arm. The following data represent the maximum weights (in pounds) pressed by cach arm for a random sample of 18 fifteen-year old girls. Assume that the differences in weights pressed by each arm for all fifteenyear old girls are approximately normally distributed. \begin{tabular}{cccccc} \hline Subject & Dominant Arm & Nondominant Arm & Subject & Dominant Arm & Nondominant Arm \\ \hline 1 & 59 & 53 & 10 & 47 & 38 \\ 2 & 32 & 30 & 11 & 40 & 35 \\ 3 & 27 & 24 & 12 & 36 & 36 \\ 4 & 18 & 20 & 13 & 21 & 25 \\ 5 & 42 & 40 & 14 & 51 & 48 \\ 6 & 12 & 12 & 15 & 30 & 30 \\ 7 & 29 & 24 & 16 & 32 & 31 \\ 8 & 33 & 34 & 17 & 14 & 14 \\ 9 & 22 & 22 & 18 & 26 & 27 \\ \hline \end{tabular} a. Make a \(99 \%\) confidence interval for the mean of the paired differences for the two populations, where a paired difference is equal to the maximum weight for the dominant arm minus the maximum weight for the nondominant arm. b. Using a \(1 \%\) significance level, can you conclude that the average paired difference as defined in part a is positive?

Maria and Ellen both specialize in throwing the javelin. Maria throws the javelin a mean distance of 200 feet with a standard deviation of 10 feet, whereas Ellen throws the javelin a mean distance of 210 feet with a standard deviation of 12 feet. Assume that the distances cach of these athletes throws the javelin are normally distributed with these population means and standard deviations. If Maria and Ellen each throw the javelin once, what is the probability that Maria's throw is longer than Ellen's?

Assuming that the two populations are normally distributed with unequal and unknown population standard deviations, construct a \(95 \%\) confidence interval for \(\mu_{1}-\mu_{2}\) for the following. $$ \begin{array}{lll} n_{1}=14 & \bar{x}_{1}=109.43 & s_{1}=2.26 \\ n_{2}=15 & \bar{x}_{2}=113.88 & s_{2}=5.84 \end{array} $$

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