The average cost of a gallon of unleaded gasoline in Greater Cincinnati was reported to be \(\$ 2.41(\text {The Cincinnati Enquirer, } \text { February } 3,2006\) ). During periods of rapidly changing prices, the newspaper samples service stations and prepares reports on gasoline prices frequently. Assume the standard deviation is \(\$ .15\) for the price of a gallon of unleaded regular gasoline, and recommend the appropriate sample size for the newspaper to use if it wishes to report a margin of error at \(95 \%\) confidence. a. Suppose the desired margin of error is \(\$ .07\) b. Suppose the desired margin of error is \(\$ .05\) c. Suppose the desired margin of error is \(\$ .03\)

Short Answer

Expert verified
The appropriate sample sizes for the newspaper to use for each desired margin of error at a 95% confidence level are: a) \(\$0.07\) margin of error: \(n = 30\) service stations. b) \(\$0.05\) margin of error: \(n = 84\) service stations. c) \(\$0.03\) margin of error: \(n = 279\) service stations.

Step by step solution

01

Identify the given information

We are given the following information: - Average cost of a gallon of unleaded gasoline: \(\$2.41\) - Standard deviation: \(\$0.15\) - Confidence level: \(95\%\) We are then asked to find the appropriate sample size for each of the following desired margins of error: a. \(\$0.07\) b. \(\$0.05\) c. \(\$0.03\)
02

Margin of error formula

The margin of error formula is given by: \(Margin\ of\ Error = Z_{\alpha/2} \times \frac{\sigma}{\sqrt{n}}\) Where: - \(Z_{\alpha/2}\) is the critical value of the standard normal distribution corresponding to a \(95\%\) confidence level, which can be found using a Z-table or calculator. For a \(95\%\) confidence level, \(Z_{\alpha/2} = 1.96\). - \(\sigma\) is the standard deviation, which is given as \(\$0.15\). - \(n\) is the sample size, which we need to find for each desired margin of error.
03

Solve for the sample size in each scenario

We will now use the margin of error formula to solve for the sample size in each scenario: a) Desired margin of error: \(\$0.07\) \(0.07 = 1.96 \times \frac{0.15}{\sqrt{n}}\) b) Desired margin of error: \(\$0.05\) \(0.05 = 1.96 \times \frac{0.15}{\sqrt{n}}\) c) Desired margin of error: \(\$0.03\) \(0.03 = 1.96 \times \frac{0.15}{\sqrt{n}}\)
04

Calculate the sample size for each scenario

Now, we will calculate the sample size for each scenario: a) \(0.07 = 1.96 \times \frac{0.15}{\sqrt{n}}\) \(n = \left(\frac{1.96\times 0.15}{0.07}\right)^2 = 29.89\) Since the sample size must be a whole number, we round up to the nearest whole number: \(n = 30\) b) \(0.05 = 1.96 \times \frac{0.15}{\sqrt{n}}\) \(n = \left(\frac{1.96\times 0.15}{0.05}\right)^2 = 83.72\) Again, we round up to the nearest whole number: \(n = 84\) c) \(0.03 = 1.96 \times \frac{0.15}{\sqrt{n}}\) \(n = \left(\frac{1.96\times 0.15}{0.03}\right)^2 = 278.07\) We round up to the nearest whole number: \(n = 279\)
05

Present the results

The appropriate sample size for the newspaper to use for each desired margin of error is: a) \(\$0.07\) margin of error: \(n = 30\) service stations. b) \(\$0.05\) margin of error: \(n = 84\) service stations. c) \(\$0.03\) margin of error: \(n = 279\) service stations.

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Key Concepts

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

Margin of Error
Understanding the margin of error is crucial when conducting surveys or studies. It refers to the extent of the range around a survey's result within which we can be confident that the true population parameter lies. For instance, if a poll shows that 60% of voters favor a candidate with a margin of error of 3%, the true proportion of the population favoring the candidate is likely between 57% and 63%.

To narrow the margin of error, we can increase the sample size, which reduces the uncertainty of our estimate. In the context of the gasoline price survey, by desiring a smaller margin of error (such as \(0.03 instead of \)0.07), there is a need for a larger sample size. Larger sample sizes provide more data points and thus a clearer picture of the true average cost, but they also require more resources to collect.
Standard Deviation
The standard deviation is a measure of the amount of variation or dispersion in a set of values. A low standard deviation indicates that the values tend to be close to the mean of the set, while a high standard deviation indicates that the values are spread out over a wider range.

In the exercise, the standard deviation of the price of gasoline is $0.15, which indicates the typical deviation from the average cost. When calculating the sample size necessary to achieve a specific margin of error, the standard deviation plays a pivotal role. It is factored in to account for how much variation can be expected in the data—in this case, the variation in gasoline prices across different service stations.
Confidence Level
The confidence level is the probability that the margin of error contains the true parameter value. Commonly expressed as a percentage, typical confidence levels in statistics are 90%, 95%, and 99%. In the gasoline prices example, a confidence level of 95% was selected. This implies that if the same sampling method is repeated numerous times, we can expect about 95% of the calculated confidence intervals to contain the true average cost of gasoline.

Choosing a higher confidence level generally results in a wider margin of error, therefore needing a larger sample size to maintain the same margin of error. However, it increases confidence in the results.
Normal Distribution
The normal distribution, also known as the Gaussian distribution, is a symmetrical, bell-shaped distribution that describes how the values of a variable are distributed. It is central to the concept of sampling and statistical inference. In many situations, including the exercise on gasoline prices, we assume that the sample mean follows a normal distribution, especially as the sample size increases, due to the Central Limit Theorem.

This distribution is used to find the critical z-value (e.g., 1.96 for 95% confidence) that determines how many standard deviations away from the mean the sample mean is expected to be within a certain confidence level. The normal distribution is essential for calculating margins of error and sample sizes.
Statistical Significance
Statistical significance indicates whether the results of a study or survey are likely not due to chance. It shows whether an effect or difference observed is likely to be genuine or if it could simply be the result of random variations in the data. In the gasoline price scenario, if the newspaper conducted multiple surveys and consistently found a difference greater than the margin of error, it might suggest a significant change in gasoline prices.

To establish statistical significance, researchers often set a significance level (denoted as alpha, \( \alpha \)) before the study, which is the probability of rejecting the null hypothesis when it is actually true. Commonly used significance levels are 0.05 (5%), 0.01 (1%), and 0.10 (10%). Ensuring statistical significance is essential for the credibility and reliability of any study's conclusions.

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Most popular questions from this chapter

Which would be hardest for you to give up: Your computer or your television? In a survey of 1677 U.S. Internet users, \(74 \%\) of the young tech elite (average age of 22 ) say their computer would be very hard to give up (PC Magazine, February 3, 2004). Only 48\% say their television would be very hard to give up. a. Develop a \(95 \%\) confidence interval for the proportion of the young tech elite that would find it very hard to give up their computer. b. Develop a \(99 \%\) confidence interval for the proportion of the young tech elite that would find it very hard to give up their television. c. In which case, part (a) or part (b), is the margin of error larger? Explain why.

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The National Center for Education Statistics reported that \(47 \%\) of college students work to pay for tuition and living expenses. Assume that a sample of 450 college students was used in the study. a. Provide a \(95 \%\) confidence interval for the population proportion of college students who work to pay for tuition and living expenses. b. Provide a \(99 \%\) confidence interval for the population proportion of college students who work to pay for tuition and living expenses. c. What happens to the margin of error as the confidence is increased from \(95 \%\) to \(99 \% ?\)

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Mileage tests are conducted for a particular model of automobile. If a \(98 \%\) confidence interval with a margin of error of 1 mile per gallon is desired, how many automobiles should be used in the test? Assume that preliminary mileage tests indicate the standard deviation is 2.6 miles per gallon.

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