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Basic Computation: Confidence Interval A random sample of size 81 has sample mean 20 and sample standard deviation \(3 .\) (a) Check Requirements Is it appropriate to use a Student's \(t\) distribution to compute a confidence interval for the population mean \(\mu\) ? Explain. (b) Find a \(95 \%\) confidence interval for \(\mu\). (c) Interpretation Explain the meaning of the confidence interval you computed.

Short Answer

Expert verified
(19.337, 20.663); 95% confident the mean is within this interval.

Step by step solution

01

Check Sample Size

To determine if it's appropriate to use a Student's t-distribution, we need to check the sample size. For a sample size greater than 30, a t-distribution is appropriate, even if the population distribution is not normal. Here, the sample size is 81, which is greater than 30.
02

Check Population Standard Deviation

We also use the t-distribution when the population standard deviation is unknown, and the sample standard deviation is provided instead. Here, we were given the sample standard deviation of 3, not the population standard deviation.
03

Conclude Usage of t-distribution

Since the sample size is larger than 30 and the population standard deviation is unknown, it is appropriate to use the Student's t-distribution for constructing the confidence interval.
04

Determine Critical Value

For a 95% confidence interval and 80 degrees of freedom (since the sample size is 81), lookup or use a t-distribution table/calculator to find the critical value. This value is approximately 1.990 at 95% confidence for our sample size.
05

Calculate Margin of Error

The formula to calculate the margin of error (ME) is \[ ME = t \times \left( \frac{s}{\sqrt{n}} \right) \] where \( t = 1.990 \), \( s = 3 \), and \( n = 81 \). Calculating gives \[ ME = 1.990 \times \left( \frac{3}{\sqrt{81}} \right) = 1.990 \times \frac{3}{9} = 1.990 \times 0.333 = 0.663 \].
06

Compute Confidence Interval

The confidence interval is given by \( \bar{x} \pm ME \). Using the sample mean \( \bar{x} = 20 \) and the margin of error \( ME = 0.663 \), we find the interval: \[ 20 - 0.663 = 19.337 \] and \[ 20 + 0.663 = 20.663 \]. Thus, the confidence interval is \( (19.337, 20.663) \).
07

Interpret the Confidence Interval

The confidence interval \((19.337, 20.663)\) means we are 95% confident that the true population mean \( \mu \) lies within this interval. Another way to say it is that if we were to take 100 different samples and compute intervals in the same way, 95 of them would contain the true population mean.

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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 a probability distribution that is very useful when we work with small sample sizes or when we do not know the population standard deviation. It is used to estimate population parameters when the sample size is small, and the population distribution is normal or nearly normal. When the sample size is larger than 30, the distribution of the sample mean can be approximated well by the normal distribution. But, here's the twist – even if the population data is skewed or it doesn't perfectly follow a normal distribution, the t-distribution works well for larger samples.
In our exercise, we had a sample size of 81. This is quite large, and therefore, we can use the t-distribution to calculate the confidence interval. This approach is quite forgiving, especially when we don't have the population standard deviation, which is actually quite common.
Ultimately, the t-distribution gives us a reliable way to make estimates about the population mean even when our data or statistical knowledge isn't perfect.
Population Mean
The population mean is a measure of central tendency that represents the average value of a set of data from the entire population. However, it's important to note that often the true population mean is unknown. Instead, we use the sample mean to estimate it. The real crux of the matter is our belief about the sample mean being a good estimate of the population mean.
In the context of the original exercise, our sample mean was 20. Using the sample data, statistical methods enable us to determine how close our sample mean might be to the actual population mean. Through constructing a confidence interval using techniques like the Student's t-distribution, statisticians gain insight into the potential values that the true population mean could take. Essentially, this concept transforms raw data into meaningful insights, giving students a clearer picture of the actual scenario.
Margin of Error
The margin of error (ME) is a critical concept when calculating confidence intervals. It quantifies the uncertainty or the range within which the true population parameter is expected to lie. Effectively, it tells us how much we expect our sample statistic to deviate from the actual population value.
The formula to calculate the margin of error is \[ ME = t \times \left( \frac{s}{\sqrt{n}} \right) \]where:
  • \( t \) is the critical value from the t-distribution, based on the desired confidence level
  • \( s \) is the sample standard deviation
  • \( n \) is the sample size
In our example, the critical value \( t \) was 1.990, \( s \) was 3, and the sample size \( n \) was 81. Calculating the margin of error provides a numerical insight into the possible range of error in estimating the population mean. In essence, a smaller margin of error indicates a more precise estimation. Thus, understanding the margin of error is key to interpreting the confidence interval accurately.
Sample Standard Deviation
The sample standard deviation is a vital measure in statistics, representing the dispersion or spread of data points in a sample around the sample mean. It captures how much the individual data points differ from the mean of the sample. In other words, it helps us understand the variability inherent in the specific data set we are examining.
For our problem, the sample standard deviation was given as 3. This value was used alongside the sample size to calculate the margin of error, which in turn helped us establish a confidence interval. The sample standard deviation plays an important role because it influences the width of the confidence interval. A larger sample standard deviation can lead to a broader confidence interval, indicating more variability and thus less certainty about the estimate of the population mean. From this, we see that comprehending the sample standard deviation enriches our understanding and interpretation of statistical data.

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Most popular questions from this chapter

Expand Your Knowledge: Sample Size, Difference of Proportions What about the sample size \(n\) for confidence intervals for the difference of proportions \(p_{1}-p_{2}\) ? Let us make the following assumptions: equal sample sizes \(n=n_{1}=n_{2}\) and all four quantities \(n_{1} \hat{p}_{1}, n_{1} \hat{q}_{1}, n_{2} \hat{p}_{2}\), and \(n_{2} \hat{q}_{2}\) are greater than \(5 .\) Those readers familiar with algebra can use the procedure outlined in Problem 28 to show that if we have preliminary estimates \(\hat{p}_{1}\) and \(\hat{p}_{2}\) and a given maximal margin of error \(E\) for a specified confidence level \(c\), then the sample size \(n\) should be at least $$ n=\left(\frac{z_{c}}{E}\right)^{2}\left(\hat{p}_{1} \hat{q}_{1}+\hat{p}_{2} \hat{q}_{2}\right) $$ However, if we have no preliminary estimates for \(\hat{p}_{1}\) and \(\hat{p}_{2}\), then the theory similar to that used in this section tells us that the sample size \(n\) should be at least $$ n=\frac{1}{2}\left(\frac{z_{c}}{E}\right)^{2} $$ (a) In Problem 17 (Myers-Briggs personality type indicators in common for married couples), suppose we want to be \(99 \%\) confident that our estimate \(\hat{p}_{1}-\hat{p}_{2}\) for the difference \(p_{1}-p_{2}\) has a maximal margin of error \(E=0.04\). Use the preliminary estimates \(\hat{p}_{1}=289 / 375\) for the proportion of couples sharing two personality traits and \(\hat{p}_{2}=23 / 571\) for the proportion having no traits in common. How large should the sample size be (assuming equal sample size-i.e., \(n=n_{1}=n_{2}\) )? (b) Suppose that in Problem 17 we have no preliminary estimates for \(\hat{p}_{1}\) and \(\hat{p}_{2}\) and we want to be \(95 \%\) confident that our estimate \(\hat{p}_{1}-\hat{p}_{2}\) for the difference \(p_{1}-p_{2}\) has a maximal margin of error \(E=0.05 .\) How large should the sample size be (assuming equal sample size-i.e., \(n=n_{1}=n_{2}\) )?

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