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Home sales again In the previous exercise, you found a \(95 \%\) confidence interval to estimate the average loss in home value. a. Suppose the standard deviation of the losses had been \(\$ 3000\) instead of \(\$ 1500 .\) What would the larger standard deviation do to the width of the confidence interval (assuming the same level of confidence)? b. Your classmate suggests that the margin of error in the interval could be reduced if the confidence level were changed to \(90 \%\) instead of \(95 \%\). Do you agree with this statement? Why or why not? c. Instead of changing the level of confidence, would it be more statistically appropriate to draw a bigger sample?

Short Answer

Expert verified
a. The larger standard deviation would widen the confidence interval. b. Yes, the classmate is correct--lowering the confidence level to \(90\%\) would decrease the margin of error and narrow the confidence interval. c. Yes, increasing the sample size would likely be more statistically appropriate than changing the level of confidence, as it enhances the precision of the estimate without compromising accuracy.

Step by step solution

01

- Impact of Larger Standard Deviation on Confidence Interval

The formula for a confidence interval for the population mean (\(\mu\)) is \(\bar{x} \pm z \frac{s}{\sqrt{n}}\), where \(\bar{x}\) is the sample mean, \(s\) is the sample standard deviation, and \(n\) is the sample size.\n\nGiven that the standard deviation \(s\) has increased from \(\$1500\) to \(\$3000\), the margin of error will therefore also increase, leading to a wider confidence interval. The precise increase could be calculated if more information were provided, such as the sample mean, the value of \(z\) (which depends on the desired confidence level), and the sample size.
02

- Effect of Changing Confidence Level on Margin of Error

The z-value in the margin of error formula corresponds to the desired level of confidence. A \(95\%\) confidence level corresponds to a z-value of approximately 1.96, and a \(90\%\) confidence level corresponds to a z-value of approximately 1.645. Because \(1.96 > 1.645\), the margin of error would be smaller with a \(90\%\) confidence interval than with a \(95\%\) confidence interval. In other words, decreasing the confidence level decreases the margin of error, leading to a narrower confidence interval, assuming that all other factors remain the same.
03

- Improving Confidence Interval via Sample Size

The formula for the margin of error of a confidence interval includes the standard deviation divided by the square root of the sample size \(\frac{s}{\sqrt{n}}\). Therefore, given a fixed level of confidence and standard deviation, increasing the sample size \(n\) will decrease the margin of error, leading to a narrower confidence interval. This is more statistically appropriate than modifying the confidence level because it improves the precision of the estimate without sacrificing the accuracy.

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Key Concepts

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

Standard Deviation
Standard deviation is a crucial measure in statistics that tells us about the spread or dispersion of a dataset. It's an indicator of how much individual data points deviate from the mean or average of the dataset. This measure is very important in understanding how varied or dispersed your data is. In the context of confidence intervals, a larger standard deviation signifies that the data points are spread out further from the mean. This increased spread means the margin of error will be larger, leading to a wider confidence interval. If the standard deviation were to increase from $1500 to $3000, as mentioned in the exercise, the confidence interval's width would increase.
  • The main effect: A greater standard deviation increases uncertainty, which translates to a wider confidence interval.
  • Importance in analysis: It reflects more variability in data, making predictions less precise.
Understanding the standard deviation helps you gauge the reliability and certainty of your estimates.
Margin of Error
The margin of error is a measure describing the range within which we can expect the true population parameter, such as the mean, to fall. It provides an understanding of the confidence interval's precision and reliability. Calculating the margin of error involves the formula: \[\text{Margin of Error} = z \frac{s}{\sqrt{n}}\]where \(z\) is the z-score linked to the desired confidence level, \(s\) is the sample standard deviation, and \(n\) the sample size.A key point highlighted in the exercise is how changing the confidence level affects the margin of error. A shift from a 95% confidence level to a 90% confidence level reduces the z-score from 1.96 to 1.645, respectively:
  • Lowers the z-value: A decrease in confidence level reduces the margin of error.
  • Outcome: The confidence interval becomes narrower, suggesting higher precision but less confidence.
This illustrates the trade-off between confidence level and precision; a smaller margin of error offers more precise estimates but at a lower confidence level.
Sample Size
Sample size, denoted as \(n\), plays a direct role in determining the margin of error in confidence intervals. A larger sample size results in a smaller margin of error and thereby a more precise confidence interval. The relationship is found through the term \(\frac{s}{\sqrt{n}}\) in the margin of error formula.Increasing the sample size improves the statistical accuracy of your estimates without altering the confidence level:
  • Effect: A larger \(n\) reduces the margin of error, narrowing the confidence interval.
  • Benefit: Greater precision without sacrificing confidence.
A larger sample better represents the population, reducing the random error in the estimate. Therefore, when seeking more accurate predictions, considering a larger sample is often the best approach.
Confidence Level
The confidence level is the probability that the confidence interval contains the true population parameter. Common levels include 90%, 95%, and 99%, reflecting how sure you are about your estimate. A higher confidence level implies more certainty that the interval has the true parameter but results in a wider interval due to a larger z-value. In the exercise example, comparing a 95% to a 90% confidence level shows that:
  • Higher confidence (95%) involves a longer interval but more certainty (z-value 1.96).
  • Lower confidence (90%) provides a shorter, more precise interval but less certainty (z-value 1.645).
The choice of confidence level depends on the context: a higher level suits scenarios needing more reliability, but where precision is more valued, a lower level might be used to provide sharper estimates.

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Most popular questions from this chapter

32\. Parking II Suppose that, for budget planning purposes, the city in Exercise 30 ?needs a better estimate of the mean daily income from parking fees. a. Someone suggests that the city use its data to create a \(95 \%\) confidence interval instead of the \(90 \%\) interval first created. How would this interval be better for the city? (You need not actually create the new interval.) b. How would the \(95 \%\) interval be worse for the planners? c. How could they achieve an interval estimate that would better serve their planning needs?

Tips A waiter believes the distribution of his tips has a model that is slightly skewed to the right, with a mean of \(\$ 9.60\) and a standard deviation of \(\$ 5.40 .\) a. Explain why you cannot determine the probability that a given party will tip him at least \(\$ 20\). b. Can you estimate the probability that the next 4 parties will tip an average of at least \(\$ 15 ?\) Explain. c. Is it likely that his 10 parties today will tip an average of at least \(\$ 15 ?\) Explain.

At work Some business analysts estimate that the length of time people work at a job has a mean of 6.2 years and a standard deviation of 4.5 years. a. Explain why you suspect this distribution may be skewed to the right. b. Explain why you could estimate the probability that 100 people selected at random had worked for their employers an average of 10 years or more, but you could not estimate the probability that an individual had done so.

Popcorn Yvon Hopps ran an experiment to determine optimum power and time settings for microwave popcorn. His goal was to find a combination of power and time that would deliver high-quality popcorn with less than \(10 \%\) of the kernels left unpopped, on average. After experimenting with several bags, he determined that power 9 at 4 minutes was the best combination. To be sure that the method was successful, he popped 8 more bags of popcorn (selected at random) at this setting. All were of high quality, with the following percentages of uncooked popcorn: \(7,13.2,10,6,7.8,2.8,2.2,5.2 .\) Does the \(95 \%\) confidence interval suggest that he met his goal of an average of no more than \(10 \%\) uncooked kernels? Explain.

Ruffles Students investigating the packaging of potato chips purchased 6 bags of Lay's Ruffles marked with a net weight of 28.3 grams. They carefully weighed the contents of each bag, recording the following weights (in grams): 29.3 , \(28.2,29.1,28.7,28.9,28.5 .\) a) Do these data satisfy the assumptions for inference? Explain. b) Find the mean and standard deviation of the weights. c) Create a \(95 \%\) confidence interval for the mean weight of such bags of chips. d) Explain in context what your interval means. e) Comment on the company's stated net weight of 28.3 grams. *f) Why might finding a bootstrap confidence interval not be a good idea for these data?

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