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Lorraine computed a confidence interval for \(\mu\) based on a sample of size \(41 .\) Since she did not know \(\sigma\), she used \(s\) in her calculations. Lorraine used the normal distribution for the confidence interval instead of a Student's \(t\) distribution. Will her interval be longer or shorter than one obtained by using an appropriate Student's \(t\) distribution? Explain.

Short Answer

Expert verified
Lorraine's interval is shorter because the normal distribution underestimates variability compared to the \(t\)-distribution.

Step by step solution

01

Understand the Given Data

Lorraine calculated a confidence interval for the population mean \(\mu\) based on a sample size \(n = 41\). Since the population standard deviation \(\sigma\) is unknown, she used the sample standard deviation \(s\).
02

Review Distribution Choice

When calculating the confidence interval for a mean with unknown population standard deviation, the appropriate choice is the Student's \(t\)-distribution, especially with smaller sample sizes (typically \(n < 30\)). However, since \(n = 41\), it approaches normality, but the correct choice is still the \(t\)-distribution.
03

Compare Distributions

A normal distribution typically results in a narrower confidence interval for a given confidence level compared to a Student's \(t\)-distribution because the \(t\)-distribution accounts for the additional variability when using \(s\) instead of \(\sigma\). The \(t\)-distribution has heavier tails, reflecting this uncertainty.
04

Apply Distribution Characteristics

For \(n = 41\), the degrees of freedom for the \(t\)-distribution would be \(40\). Although approaching normality, the \(t\)-distribution for \(40\) degrees of freedom still has a slightly larger variance than the normal distribution, leading to a wider confidence interval.
05

Conclusion on Interval Length

Since Lorraine used the normal distribution instead of the appropriate Student's \(t\)-distribution for her sample size of \(41\), her confidence interval is shorter than it would be if she had used the correct \(t\)-distribution. This is due to the underestimated variability when not accounting for the added uncertainty with \(s\).

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Key Concepts

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

Student's t-distribution
The Student's t-distribution is an important concept when it comes to creating confidence intervals, especially when you don't know the population standard deviation and must rely on a sample standard deviation. This distribution accounts for the extra uncertainty that arises when estimating from a sample rather than knowing the whole population parameter.
The t-distribution is a bit different from a normal distribution, mainly because it has heavier tails. This means that it allows for a greater possibility of outlier values. As a result, when you're calculating a confidence interval using the t-distribution, you might notice that the interval is slightly wider compared to when you use a normal distribution. This is a protective measure to compensate for the additional variability and less reliable information gathered from a sample.
  • The t-distribution is used when the sample size is small (usually less than 30), but remains appropriate as the sample size grows, like in Lorraine's case of 41.
  • As the sample size increases, the t-distribution approaches the normal distribution.
Sample Standard Deviation
The sample standard deviation, denoted as \(s\), is a measure that quantifies the amount of variation or spread in a sample data set. It is calculated using the deviations of individual data points from the sample mean.
It's crucial in statistical analysis because it serves as an estimator for the population standard deviation when the latter is unknown. In Lorraine’s situation, since she didn’t know the population standard deviation, she had to use the sample standard deviation in her calculations.
  • The sample standard deviation can be seen as a counterpart to the population standard deviation, \(\sigma\), but based solely on sample data.
  • With the use of \(s\), there's more variability and thus more uncertainty compared to when \(\sigma\) is known.
Relying on \(s\) instead of \(\sigma\) introduces additional variability which is why it’s usually paired with the t-distribution in confidence interval calculations. This approach helps in providing a more cautious estimate, acknowledging the limited information at hand.
Normal Distribution
The normal distribution, sometimes called the Gaussian distribution, is one of the most commonly encountered distributions in statistics. It's a symmetric, bell-shaped distribution where most of the data points cluster around the mean.
In the context of confidence intervals, using the normal distribution is straightforward when the population standard deviation is known, and the sample size is large. When Lorraine used the normal distribution to calculate her confidence interval, she was assuming a less conservative approach.
  • The normal distribution is suitable for large sample sizes (often referred to as the Central Limit Theorem).
  • It results in narrower confidence intervals compared to the t-distribution because it does not account for the extra variability coming from using \(s\) instead of \(\sigma\).
In Lorraine's case, using the normal distribution instead of the Student’s t-distribution likely led to an underestimate of the confidence interval’s width, as it failed to accommodate the uncertainty due to the unknown \(\sigma\).
Degrees of Freedom
Degrees of freedom is a concept used in statistics to describe the number of independent values or quantities which can vary in an analysis without breaking any constraints. When dealing with the t-distribution, degrees of freedom play a significant role as they influence the shape of the distribution.
In Lorraine's analysis, the degrees of freedom would be determined by her sample size minus one, which is \(40\) (i.e., \(n - 1 = 41 - 1\)).
  • Degrees of freedom impact the critical values derived from the t-distribution, which in turn affects the confidence interval width.
  • More degrees of freedom make the t-distribution look more like a normal distribution; fewer degrees of freedom mean heavier tails and a wider confidence interval for the same confidence level.
As the degrees of freedom increase (as with Lorraine's sample size growing), the t-distribution tends to flatten and converge with the normal distribution, eventually leading to similar results. However, until that convergence is quite close, it's safer to use the t-distribution to account for sample variability.

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Most popular questions from this chapter

The home run percentage is the number of home runs per 100 times at bat. A random sample of 43 professional baseball players gave the following data for home run percentages (Reference: The Baseball Encyclopedia, Macmillan). $$ \begin{array}{llllllllll} 1.6 & 2.4 & 1.2 & 6.6 & 2.3 & 0.0 & 1.8 & 2.5 & 6.5 & 1.8 \\ 2.7 & 2.0 & 1.9 & 1.3 & 2.7 & 1.7 & 1.3 & 2.1 & 2.8 & 1.4 \\ 3.8 & 2.1 & 3.4 & 1.3 & 1.5 & 2.9 & 2.6 & 0.0 & 4.1 & 2.9 \\ 1.9 & 2.4 & 0.0 & 1.8 & 3.1 & 3.8 & 3.2 & 1.6 & 4.2 & 0.0 \\ 1.2 & 1.8 & 2.4 & & & & & & & \end{array} $$ (a) Use a calculator with mean and standard deviation keys to verify that \(\bar{x} \approx 2.29\) and \(s \approx 1.40 .\) (b) Compute a \(90 \%\) confidence interval for the population mean \(\mu\) of home run percentages for all professional baseball players. Hint: If you use Table 6 of Appendix II, be sure to use the closest \(d\). \(f\). that is smaller. (c) Compute a \(99 \%\) confidence interval for the population mean \(\mu\) of home run percentages for all professional baseball players. (d) The home run percentages for three professional players are Tim Huelett, \(2.5 \quad\) Herb Hunter, \(2.0 \quad\) Jackie Jensen, \(3.8\) Examine your confidence intervals and describe how the home run percentages for these players compare to the population average. (e) In previous problems, we assumed the \(x\) distribution was normal or approximately normal. Do we need to make such an assumption in this problem? Why or why not? Hint: See the central limit theorem in Section \(7.2 .\)

Results of a poll of a random sample of 3003 American adults showed that \(20 \%\) do not know that caffeine contributes to dehydration. The poll was conducted for the Nutrition Information Center and had a margin of error of \(\pm 1.4 \%\). (a) Does the margin of error take into account any problems with the wording of the survey question, interviewer errors, bias from sequence of questions, and so forth? (b) What does the margin of error reflect?

Case studies showed that out of 10,351 convicts who escaped from U.S. prisons, only 7867 were recaptured (The Book of Odds, by Shook and Shook, Signet). (a) Let \(p\) represent the proportion of all escaped convicts who will eventually be recaptured. Find a point estimate for \(p\). (b) Find a \(99 \%\) confidence interval for \(p .\) Give a brief statement of the meaning of the confidence interval. (c) Is use of the normal approximation to the binomial justified in this problem? Explain.

Most married couples have two or three personality preferences in common (see reference in Problem 13). Myers used a random sample of 375 married couples and found that 132 had three preferences in common. Another random sample of 571 couples showed that 217 had two personality preferences in common. Let \(p_{1}\) be the population proportion of all married couples who have three personality preferences in common. Let \(p_{2}\) be the population proportion of all married couples who have two personality preferences in common. (a) Find a \(90 \%\) confidence interval for \(p_{1}-p_{2}\). (b) Examine the confidence interval in part (a) and explain what it means in the context of this problem. Does the confidence interval contain all positive, all negative, or both positive and negative numbers? What does this tell you about the proportion of married couples with three personality preferences in common compared with the proportion of couples with two preferences in common (at the \(90 \%\) confidence level)?

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