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Lorraine was in a hurry when she computed a confidence interval for \(\mu .\) Because \(\sigma\) was not known, she used a Student's \(t\) distribution. However, she accidentally used degrees of freedom \(n\) instead of \(n-1 .\) Will her confidence interval be longer or shorter than one found using the correct degrees of freedom \(n-1 ?\) Explain.

Short Answer

Expert verified
Lorraine's confidence interval will be shorter because using \( df = n \) lowers the critical t-value needed for the interval.

Step by step solution

01

Understanding Degrees of Freedom

In statistical terms, degrees of freedom (df) typically refer to the number of independent values that can vary in an analysis without breaking any given constraints. When making a confidence interval for a population mean with an unknown c3, we use the Student's t-distribution. The degrees of freedom in this context are calculated as sample size minus one, \( df = n - 1 \).
02

Degrees of Freedom Affect on t-distribution

The Student's t-distribution becomes closer to the normal distribution as the degrees of freedom increase (as \( n \) gets larger). With smaller degrees of freedom, the t-distribution is wider and has heavier tails. This implies that using \( df = n \), she may have used a t-distribution that was slightly more constrained than the correct version (\( df = n - 1 \)).
03

Comparing the t-values

For a given confidence level (say 95%), the critical t-value over a given distribution dictates the margin of error. Since \( df = n \) is greater than \( df = n - 1 \), the critical t-value she used would be smaller than the one that should have been used. This would normally result in a narrower interval since a lower critical t-value reduces the margin of error.
04

Conclusion about the Interval Length

Therefore, a higher degree of freedom results in a lower critical t-value when calculating intervals, making the interval length shorter. Since Lorraine used \( df = n \) instead of \( df = n - 1 \), her confidence interval is shorter.

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Key Concepts

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

Degrees of Freedom
When calculating a statistical measure such as a confidence interval, it's crucial to understand the concept of degrees of freedom (df). Degrees of freedom refer to the number of independent pieces of information available to estimate other parameters. Think of it this way—if you were given a choice of 10 numbers to sum to 100, you can freely choose 9 numbers, but the 10th number is dictated by your previous choices. This idea of restriction is what degrees of freedom illustrate. In the context of estimating a population mean where the population standard deviation is unknown, degrees of freedom are determined by the formula \( df = n - 1 \)... "n" representing the sample size. This serves as a correction factor, accounting for the uncertainty introduced by estimating unknown parameters. Understanding degrees of freedom is vital for choosing the correct statistical distribution, like the Student's t-distribution, which is used instead of the normal distribution when the sample size is small or the standard deviation is unknown.
Student's t-Distribution
The Student's t-distribution is a statistical distribution used extensively throughout inferential statistics, particularly when the sample size is small, and the population standard deviation is unknown. This distribution was introduced by William Sealy Gosset under the pseudonym "Student". The shape of the t-distribution is symmetric and bell-shaped, quite similar to the normal distribution; however, it possesses heavier tails. The degree of heaviness in the tails is determined by the degrees of freedom. A smaller sample size (lower degrees of freedom) leads to heavier tails. This means there is more room for extreme values, making the t-distribution versatile in handling variability within small data sets. Conversely, as the sample size increases, the degrees of freedom rise, and the t-distribution starts resembling the normal distribution more closely. Recognizing the correct distribution is crucial, especially when constructing confidence intervals or hypothesis testing.
Critical t-value
The critical t-value is an essential component when calculating confidence intervals using the t-distribution. It represents the cutoff point beyond which the probability of finding the sample statistic is rare under the assumed distribution. In simpler terms, it's used to determine how "far away" your sample statistic can be from the population parameter and still be considered reasonably probable. To find this critical value, you need two pieces of information: the desired confidence level (like 95% or 99%) and the degrees of freedom. Together, these determine the critical t-value from a t-distribution table. If the degrees of freedom are miscalculated—as was Lorraine's case—then the critical t-value isn't accurate, leading to incorrect margins of error. In her scenario, using \( df = n \) rather than \( df = n - 1 \) caused her to use a smaller critical t-value, which subsequently shortened her confidence interval. This is a good reminder to always ensure calculations are based on the correct degrees of freedom for accurate results.
Margin of Error
Margin of error is a measure of the potential error in a statistical analysis. It indicates how much uncertainty exists around the sample estimate of a population parameter. In the context of a confidence interval, it determines how wide the interval is around the sample mean, hence influencing the precision and reliability of the interval estimate. The formula for the margin of error in a confidence interval is: \[ \text{Margin of Error} = \text{Critical t-value} \times \text{Standard Error} \] A larger margin of error suggests less precision, whereas a smaller one indicates more certainty. This margin is defined by the critical t-value and the standard error of the mean. If the critical t-value is underestimated due to incorrect degrees of freedom, as when using \( df = n \) instead of \( n - 1 \), it results in a smaller margin of error, producing a narrower confidence interval. Consequently, Lorraine's incorrect degrees of freedom calculation led to a less accurate portrayal of the true variability possible in the estimate.

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Most popular questions from this chapter

A random sample of 5792 physicians in Colorado showed that 3139 provided at least some charity care (i.e., treated poor people at no cost). These data are based on information from State Health Care Data: Utilization, Spending, and Characteristics (American Medical Association). (a) Let \(p\) represent the proportion of all Colorado physicians who provide some charity care. Find a point estimate for \(p\). (b) Find a \(99 \%\) confidence interval for \(p .\) Give a brief explanation of the meaning of your answer in the context of this problem. (c) Is the normal approximation to the binomial justified in this problem? Explain.

Sam computed a \(90 \%\) confidence interval for \(\mu\) from a specific random sample of size \(n\). He claims that at the \(90 \%\) confidence level, his confidence interval contains \(\mu\). Is this claim correct? Explain.

How hard is it to reach a businessperson by phone? Let \(p\) be the proportion of calls to businesspeople for which the caller reaches the person being called on the first try. (a) If you have no preliminary estimate for \(p\), how many business phone calls should you include in a random sample to be \(80 \%\) sure that the point estimate \(\hat{p}\) will be within a distance of \(0.03\) from \(p ?\) (b) The Book of Odds, by Shook and Shook (Signet), reports that businesspeople can be reached by a single phone call approximately \(17 \%\) of the time. Using this (national) estimate for \(p\), answer part (a).

When \(\sigma\) is unknown and the sample is of size \(n \geq 30\), there are two methods for computing confidence intervals for \(\mu\). Method 1: Use the Student's \(t\) distribution with \(d . f .=n-1 .\) This is the method used in the text. It is widely employed in statistical studies. Also, most statistical software packages use this method. Method 2: When \(n \geq 30\), use the sample standard deviation \(s\) as an estimate for \(\sigma\), and then use the standard normal distribution. This method is based on the fact that for large samples, \(s\) is a fairly good approximation for \(\sigma .\) Also, for large \(n\), the critical values for the Student's \(t\) distribution approach those of the standard normal distribution. Consider a random sample of size \(n=31\), with sample mean \(\bar{x}=45.2\) and sample standard deviation \(s=5.3\). (a) Compute \(90 \%, 95 \%\), and \(99 \%\) confidence intervals for \(\mu\) using Method 1 with a Student's \(t\) distribution. Round endpoints to two digits after the decimal. (b) Compute \(90 \%, 95 \%\), and \(99 \%\) confidence intervals for \(\mu\) using Method 2 with the standard normal distribution. Use \(s\) as an estimate for \(\sigma\). Round endpoints to two digits after the decimal. (c) Compare intervals for the two methods. Would you say that confidence intervals using a Student's \(t\) distribution are more conservative in the sense that they tend to be longer than intervals based on the standard normal distribution? (d) Repeat parts (a) through (c) for a sample of size \(n=81\). With increased sample size, do the two methods give respective confidence intervals that are more similar?

Sam computed a \(95 \%\) confidence interval for \(\mu\) from a specific random sample. His confidence interval was \(10.1<\mu<12.2 .\) He claims that the probability that \(\mu\) is in this interval is \(0.95\). What is wrong with his claim?

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