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Chapter 8: Problem 109

Multiple choice: CI property Increasing the confidence level causes the margin of error of a confidence interval to \((\) a) increase, \((b)\) decrease, \((c)\) stay the same.

Short Answer

Expert verified
(a) increase.

Step by step solution

01

Understanding Confidence Intervals

A confidence interval gives a range of values that is likely to contain the population parameter with a certain level of confidence. The margin of error determines the width of this interval.
02

Defining Margin of Error

The margin of error is impacted by the confidence level and sample size. It is the range of values above and below the sample statistic in a confidence interval.
03

Impact of Increasing Confidence Level

As the confidence level increases, you become more confident that the interval contains the population parameter. To achieve this, the interval must be wider, which happens when the margin of error increases.
04

Analyzing the Options

With an increased confidence level and a wider interval, the margin of error does not decrease or stay the same; it increases. Therefore, option (a) is the correct choice.

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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 component in the calculation of confidence intervals. It represents the range within which the true population parameter is estimated to lie. Essentially, the margin of error defines how much we expect our sample results can differ from the actual population value. A smaller margin of error means our estimate is more precise.

Factors Affecting Margin of Error:
  • Sample Size: Larger sample sizes tend to reduce the margin of error because they provide more data, making our estimates closer to the actual population value.
  • Confidence Level: Higher confidence levels increase the margin of error. This is because to be more confident that the interval contains the population parameter, the range must be wider.
Think of the margin of error as a safety net for our estimates. The amount of spread or "buffer" we add to our sample statistics impacts the certainty of our conclusions.
Confidence Level
The Confidence Level is an expression of how certain we are that the confidence interval computed from a sample contains the true population parameter. Common confidence levels include 90%, 95%, and 99%. These percentages represent how often the true parameter would fall within the calculated interval if we were to take numerous samples and compute an interval from each.

Why Confidence Level Matters:
  • Higher Confidence: With a higher confidence level, such as 99%, we are more sure that our interval includes the true parameter. However, this increases the margin of error, making the interval wider.
  • Balancing Act: There is a trade-off between confidence level and precision. Selecting a very high confidence level will give us a wider interval, which reduces precision. Conversely, a lower confidence level narrows the interval but increases the risk of not capturing the true parameter.
Thus, choosing the right confidence level depends on how much uncertainty you're willing to accept in your estimates.
Population Parameter
A Population Parameter is a value that represents a certain characteristic of the entire population. Unlike sample statistics (which estimate a parameter based on a sample), a parameter is a constant value describing some aspect of the population, such as mean, median, or proportion.

Importance of Population Parameters:
  • True Representation: Parameters provide the true baseline from which we can compare sample data. Unfortunately, they're often unknown because it can be impractical to measure the entire population.
  • Guide for Samples: When calculating a confidence interval, the objective is to use sample data to make inferences about the population parameter, assuming the sample is representative.
By understanding parameters, we can appreciate the purpose and construction of confidence intervals, which aim to capture these elusive values. Despite them being unknowable in many cases, confidence intervals provide a meaningful way to understand the possible range for these parameters.

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Most popular questions from this chapter

Using \(t\) -table Using Table \(B\) or software or a calculator, report the \(t\) -score which you multiply by the standard error to form the margin of error for a a. \(95 \%\) confidence interval for a mean with 5 observations. b. \(95 \%\) confidence interval for a mean with 15 observations. c. \(99 \%\) confidence interval for a mean with 15 observations.

Heights of seedlings Exercise 8.6 reported heights (in \(\mathrm{mm}\) ) of \(55.5,60.3,60.6,62.1,65.5,\) and 69.2 for six seedlings fourteen days after germination. a. Using software or a calculator, verify that the \(95 \%\) confidence interval for the population mean is (57.3,67.1) . b. Name two things you could do to get a narrower interval than the one in part a. c. Construct a \(99 \%\) confidence interval. Why is it wider than the \(95 \%\) interval? d. On what assumptions is the interval in part a based? Explain how important each assumption is.

Help the poor? One question (called NATFAREY) on the General Social Survey for the year 2008 asks, "Are we spending too much, too little, or about the right amount on assistance to the poor?" Of the 998 people who responded in 2008 , 695 said too little, 217 said about right, and 86 said too much. a. Find the point estimate of the population proportion who would answer "about right." b. The margin of error of this estimate is \(0.05 .\) Explain what this represents.

Anorexia in teenage girls A study \(^{6}\) compared various therapies for teenage girls suffering from anorexia, an eating disorder. For each girl, weight was measured before and after a fixed period of treatment. The variable measured was the change in weight, \(X=\) weight at the end of the study minus weight at the beginning of the study. The therapies were designed to aid weight gain, corresponding to positive values of \(X .\) For the sample of 17 girls receiving the family therapy, the changes in weight during the study were 11,11,6,9,14,-3,0,7,22,-5,-4,13,13,9,4,6,11 a. Plot these with a dot plot or box plot, and summarize. b. Using a calculator or software, show that the weight changes have \(\bar{x}=7.29\) and \(s=7.18\) pounds. c. Using a calculator or software, show that the standard error of the sample mean was se \(=1.74\). d. To use the \(t\) distribution, explain why the \(95 \%\) confidence interval uses the \(t\) -score equal to 2.120 . e. Let \(\mu\) denote the population mean change in weight for this therapy. Using results from parts \(\mathrm{b}, \mathrm{c},\) and \(\mathrm{d}\), show that the \(95 \%\) confidence interval for \(\mu\) is \((3.6,11.0) .\) Explain why this suggests that the true mean change in weight is positive, but possibly quite small.

Why called "degrees of freedom"? You know the sample mean \(\bar{x}\) of \(n\) observations. Once you know \((n-1)\) of the observations, show that you can find the remaining one. In other words, for a given value of \(\bar{x},\) the values of \((n-1)\) observations determine the remaining one. In summarizing scores on a quantitative variable, having \((n-1)\) degrees of freedom means that only that many observations are independent. (If you have trouble with this, try to show it for \(n=2,\) for instance showing that if you know that \(\bar{x}=80\) and you know that one observation is \(90,\) then you can figure out the other observation. The \(d f\) value also refers to the divisor in \(\left.s^{2}=\Sigma(x-\bar{x})^{2} /(n-1) .\right)\)
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