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The interval from \(-2.33\) to \(1.75\) captures an area of \(.95\) under the \(z\) curve. This implies that another largesample \(95 \%\) confidence interval for \(\mu\) has lower limit \(\bar{x}-2.33 \frac{\sigma}{\sqrt{n}}\) and upper limit \(\bar{x}+1.75 \frac{\sigma}{\sqrt{n}} .\) Would you recommend using this \(95 \%\) interval over the \(95 \%\) interval \(\bar{x} \pm 1.96 \frac{\sigma}{\sqrt{n}}\) discussed in the text? Explain. (Hint: Look at the width of each interval.)

Short Answer

Expert verified
We recommend using the 95% confidence interval \(\bar{x} \pm 1.96 \frac{\sigma}{\sqrt{n}}\) because it has a narrower width (3.92 vs 4.08), indicating a more precise estimate of the population mean \(\mu\).

Step by step solution

01

Understand the Formulae

Firstly, understanding the formulas used for calculating the confidence intervals is crucial. The standard formula for a confidence interval in this context is given by \(\bar{x} \pm Z \frac{\sigma}{\sqrt{n}}\), where \(\bar{x}\) is the sample mean, \(Z\) is the Z-score which depends on the desired confidence level, \(\sigma\) is the population standard deviation, and \(n\) is the size of the sample.
02

Calculate Width of the First Interval

For the first interval, the formula is \(\bar{x}-2.33 \frac{\sigma}{\sqrt{n}}\) (lower limit) and \(\bar{x}+1.75 \frac{\sigma}{\sqrt{n}}\) (upper limit). Subtract the lower limit from the upper limit to find the width of the interval: width = \((\bar{x}+1.75 \frac{\sigma}{\sqrt{n}}) - (\bar{x}-2.33 \frac{\sigma}{\sqrt{n}}) = 4.08 \frac{\sigma}{\sqrt{n}}\)
03

Calculate Width of the Second Interval

For the second interval, the formula is \(\bar{x}-1.96 \frac{\sigma}{\sqrt{n}}\) (lower limit) and \(\bar{x}+1.96 \frac{\sigma}{\sqrt{n}}\) (upper limit). Subtract the lower limit from the upper limit to calculate the width: width =\((\bar{x}+1.96 \frac{\sigma}{\sqrt{n}}) - (\bar{x}-1.96 \frac{\sigma}{\sqrt{n}}) = 3.92 \frac{\sigma}{\sqrt{n}}\)
04

Compare the Widths and Draw Conclusion

Comparing the widths of the two intervals, it can be seen that the interval with width 3.92 is narrower than the interval with width 4.08. A narrower confidence interval is better as it indicates a more precise estimate of the population parameter. Therefore, the 95% interval \(\bar{x} \pm 1.96 \frac{\sigma}{\sqrt{n}}\) from the text is recommended because it gives a more precise estimate.

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Key Concepts

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

Z-score
The Z-score is a statistical measurement that describes a value's relationship to the mean of a group of values. It is measured in terms of standard deviations from the mean. If a Z-score is 0, it indicates that the data point's score is identical to the mean score. A Z-score of 1.0 would indicate a value that is one standard deviation from the mean. Z-scores may be positive or negative, with a negative value indicating the score is below the mean, and a positive score indicating it is above the mean.

For confidence intervals, the Z-score gives us a way to determine how far from the sample mean we should look to ensure we've captured a certain proportion of the data. This becomes the 'Z multiplier' in our interval formula. For example, in a 95% confidence interval, we frequently use a Z-score of 1.96, which corresponds to the Z-score that encloses 95% of the data under a standard normal distribution curve.
Standard Deviation
Standard deviation is a measure of how spread out numbers are around the mean. It is a typical way to measure the variability or dispersion of a data set. A low standard deviation indicates that the data points tend to be close to the mean, whereas a high standard deviation indicates that the data points are spread out over a wide range of values.

In the context of confidence intervals, standard deviation plays a critical role since it reflects the degree of uncertainty in our sample's estimate of the population parameter. In the formula for a confidence interval, we divide the standard deviation by the square root of the sample size to come up with the standard error, which adjusts the standard deviation to account for the size of the sample. Larger samples have smaller standard errors, making for tighter confidence intervals.
Sample Mean
The sample mean, denoted as \( \bar{x} \), is the average of all the numbers in a sample. It serves as an estimate of the population mean, which is the average you would expect if you could look at every single value in the population. The sample mean is important in the calculation of confidence intervals as it is the starting point -- we build the interval around the sample mean, using it to estimate the population mean.

Understanding the sample mean is important for interpreting confidence intervals. For instance, when we say we have a 95% confidence interval, we mean that if we were to take many samples and build a confidence interval from each one, we'd expect about 95% of those intervals to contain the true population mean. The sample mean is at the heart of those intervals.
Population Parameter Estimation
Population parameter estimation involves using sample data to make inferences about a population. A population parameter could be a mean, proportion, standard deviation, or any other metric that describes the entire population. Since it's often impractical or impossible to measure the entire population, we take a sample and then use this sample to estimate the parameter.

The precision of our estimate is expressed in terms of a confidence interval, which provides a range of values within which the true population parameter is likely to lie with a certain degree of confidence. This range is calculated using the Z-score, standard deviation, and sample mean to create upper and lower bounds. The width of this interval gives us an idea of how precise our estimate is – narrower intervals suggest more precision, while wider intervals suggest less precision. When making decisions based on these estimates, one must consider the confidence level and the width of the confidence interval.

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Most popular questions from this chapter

The eating habits of 12 bats were examined in the article "Foraging Behavior of the Indian False Vampire Bat" (Biotropica \([1991]: 63-67) .\) These bats consume insects and frogs. For these 12 bats, the mean time to consume a frog was \(\bar{x}=21.9 \mathrm{~min}\). Suppose that the standard deviation was \(s=7.7 \mathrm{~min} .\) Construct and interpret a \(90 \%\) confidence interval for the mean suppertime of a vampire bat whose meal consists of a frog. What assumptions must be reasonable for the one-sample \(t\) interval to be appropriate?

Five hundred randomly selected working adults living in Calgary, Canada were asked how long, in minutes, their typical daily commute was (Calgary Herald Traffic Study, Ipsos, September 17,2005 ). The resulting sample mean and standard deviation of commute time were \(28.5\) minutes and \(24.2\) minutes, respectively. Construct and interpret a \(90 \%\) confidence interval for the mean commute time of working adult Calgary residents.

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Despite protests from civil libertarians and gay rights activists, many people favor mandatory AIDS testing of certain at-risk groups, and some people even believe that all citizens should be tested. What proportion of the adults in the United States favor mandatory testing for all citizens? To assess public opinion on this issue, researchers conducted a survey of 1014 randomly selected adult U.S. citizens ("Large Majorities Continue to Back AIDS Testing," Gallup Poll Monthly [1991]: 25-28). The article reported that 466 of the 1014 people surveyed believed that all citizens should be tested. Use this information to estimate \(\pi\), the true proportion of all U.S. adults who favor AIDS testing of all citizens.

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