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Chapter 7: Problem 11

Basic Computation: Confidence Interval Suppose \(x\) has a mound-shaped distribution. A random sample of size 16 has sample mean 10 and sample standard deviation \(2 .\) (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 \(90 \%\) confidence interval for \(\mu .\) (c) Interpretation Explain the meaning of the confidence interval you computed.

Short Answer

Expert verified
Yes, use the t-distribution; 90% CI is (9.1235, 10.8765); the interval estimates the population mean with 90% confidence.

Step by step solution

01

Checking Requirements

To determine if it's appropriate to use the Student's \( t \) distribution, we check two key requirements: The underlying distribution should be approximately normal (mound-shaped), which is mentioned in the problem, and the sample size should be small, i.e., less than 30. Here, the sample size is 16, which is less than 30. Thus, using a \( t \) distribution is suitable.
02

Determine the t-Value

For a 90% confidence interval and a sample size of 16, we have 15 degrees of freedom (df = n - 1). Using a \( t \)-distribution table or calculator, find the critical \( t \)-value that corresponds to 90% confidence level and 15 df. Let's assume the critical \( t \) value is approximately 1.753.
03

Calculate the Standard Error

The standard error (SE) of the mean is calculated using the formula \( \text{SE} = \frac{s}{\sqrt{n}} \), where \( s \) is the sample standard deviation and \( n \) is the sample size. Substituting the given values, \( s = 2 \) and \( n = 16 \), we get \( \text{SE} = \frac{2}{\sqrt{16}} = \frac{2}{4} = 0.5 \).
04

Calculate the Confidence Interval

The confidence interval for the mean is calculated using the formula: \( \bar{x} \pm t \times \text{SE} \). Substituting the known values, \( \bar{x} = 10 \), \( t = 1.753 \), and \( \text{SE} = 0.5 \), the interval becomes: \[ 10 \pm 1.753 \times 0.5 = 10 \pm 0.8765 \] Thus, the confidence interval is \( (9.1235, 10.8765) \).
05

Interpretation of the Confidence Interval

The 90% confidence interval \((9.1235, 10.8765)\) means that we are 90% confident that the true population mean \( \mu \) lies within this interval. This does not mean there is a 90% probability that \( \mu \) is within this interval, but rather that if we were to take many samples and construct intervals in the same way, about 90% of those intervals would contain the true 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 used when estimating population parameters, especially when the sample size is small, usually less than 30. Unlike the normal distribution, the t-distribution accounts for the extra variability that often comes with smaller samples.

Key features of the t-distribution include:
  • It is symmetric and bell-shaped, similar to the normal distribution.
  • It has thicker tails, which accommodate the additional uncertainty with small samples.
  • The shape of the t-distribution changes depending on the degrees of freedom, becoming closer to a normal distribution as the degrees of freedom increase.
Understanding and employing the t-distribution is crucial when constructing confidence intervals for samples that are too small to rely on the normal distribution. This allows for accurate conclusions about the population mean.
Sample Size
In statistics, the sample size, denoted as \( n \), is the number of observations in a sample. It is a vital component because it directly impacts the reliability of statistical estimates.

Here鈥檚 why sample size matters:
  • Larger samples generally provide more precise estimates of population parameters.
  • When you have a smaller sample size, like 16 in this exercise, opting for a t-distribution is advisable because it accounts for additional estimation uncertainties.
  • The sample size influences the degrees of freedom, which affects the shape of the t-distribution.
In essence, aligning your statistical method with the sample size ensures more reliable and valid results.
Standard Error
The standard error (SE) is a measure of how much the sample mean is expected to vary from the actual population mean. When calculating the SE, you use the formula: \( \text{SE} = \frac{s}{\sqrt{n}} \) where \( s \) is the sample standard deviation and \( n \) is the sample size.

Understanding the standard error is important because:
  • A smaller standard error indicates that the sample mean is close to the population mean.
  • The SE is inversely related to the sample size, meaning larger samples yield smaller SE.
The standard error serves as a foundational element in calculating the confidence interval, providing insight into how accurately the sample mean estimates the population mean.
Population Mean
The population mean, often represented by the symbol \( \mu \), is the average of all measurements in a population. When you sample from a population, you aim to estimate this average because it provides a central tendency, or typical value, of the entire population.

Key considerations for the population mean involve:
  • While we cannot always measure the true population mean directly, we use the sample mean as an estimator.
  • The accuracy of the sample mean in representing the population mean depends heavily on the sample size and the standard error.
Confidence intervals are constructed around the sample mean to give a range where the true population mean is likely to lie, lending us a more complete picture of the entire population.
Degrees of Freedom
Degrees of freedom, often symbolized as df, represent the number of independent values in a statistical calculation that are free to vary. In this context, when estimating a population parameter, it typically equals the sample size minus one (\( n - 1 \)).

Their importance includes:
  • Degrees of freedom influence the critical \( t \)-value obtained from t-distribution tables.
  • A higher number of degrees of freedom results in a t-distribution that more closely resembles a normal distribution.
  • Understanding degrees of freedom is essential to ensure you're using the correct distribution for statistical inference.
They are a critical part of hypothesis testing and confidence intervals since they adjust the precision of your estimates based on the sample size.

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Most popular questions from this chapter

Expand Your Knowledge: Alternate Method for Confidence Intervals 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?

Small Business: Bankruptcy The National Council of Small Businesses is interested in the proportion of small businesses that declared Chapter 11 bankruptcy last year. Since there are so many small businesses, the National Council intends to estimate the proportion from a random sample. Let \(p\) be the proportion of small businesses that declared Chapter 11 bankruptcy last year. (a) If no preliminary sample is taken to estimate \(p\), how large a sample is necessary to be \(95 \%\) sure that a point estimate \(\hat{p}\) will be within a distance of \(0.10\) from \(p\) ? (b) In a preliminary random sample of 38 small businesses, it was found that six had declared Chapter 11 bankruptcy. How many more small businesses should be included in the sample to be \(95 \%\) sure that a point estimate \(\hat{p}\) will be within a distance of \(0.10\) from \(p\) ?

Pro Football and Basketball: Heights of Players Independent random samples of professional football and basketball players gave the following information (References: Sports Encyclopedia of Pro Football and Official NBA Basketball Encyclopedia). Note: These data are also available for download at the Online Study Center. $$ \begin{aligned} &\text { Heights (in } \mathrm{ft} \text { ) of pro football players: } x_{1} ; n_{1}=45\\\ &\begin{array}{llllllllll} 6.33 & 6.50 & 6.50 & 6.25 & 6.50 & 6.33 & 6.25 & 6.17 & 6.42 & 6.33 \\ 6.42 & 6.58 & 6.08 & 6.58 & 6.50 & 6.42 & 6.25 & 6.67 & 5.91 & 6.00 \\ 5.83 & 6.00 & 5.83 & 5.08 & 6.75 & 5.83 & 6.17 & 5.75 & 6.00 & 5.75 \\ 6.50 & 5.83 & 5.91 & 5.67 & 6.00 & 6.08 & 6.17 & 6.58 & 6.50 & 6.25 \\ 6.33 & 5.25 & 6.67 & 6.50 & 5.83 & & & & & \end{array} \end{aligned} $$ $$ \begin{aligned} &\text { Heights (in } \mathrm{ft} \text { ) of pro basketball players: } x_{2} ; n_{2}=40\\\ &\begin{array}{llllllllll} 6.08 & 6.58 & 6.25 & 6.58 & 6.25 & 5.92 & 7.00 & 6.41 & 6.75 & 6.25 \\ 6.00 & 6.92 & 6.83 & 6.58 & 6.41 & 6.67 & 6.67 & 5.75 & 6.25 & 6.25 \\ 6.50 & 6.00 & 6.92 & 6.25 & 6.42 & 6.58 & 6.58 & 6.08 & 6.75 & 6.50 \\ 6.83 & 6.08 & 6.92 & 6.00 & 6.33 & 6.50 & 6.58 & 6.83 & 6.50 & 6.58 \end{array} \end{aligned} $$ (a) Use a calculator with mean and standard deviation keys to verify that \(\bar{x}_{1} \approx 6.179, s_{1} \approx 0.366, \bar{x}_{2} \approx 6.453\), and \(s_{2} \approx 0.314 .\) (b) Let \(\mu_{1}\) be the population mean for \(x_{1}\) and let \(\mu_{2}\) be the population mean for \(x_{2}\). Find a \(90 \%\) confidence interval for \(\mu_{1}-\mu_{2}\). (c) Interpretation Examine the confidence interval and explain what it means in the context of this problem. Does the interval consist of numbers that are all positive? all negative? of different signs? At the \(90 \%\) level of confidence, do professional football players tend to have a higher population mean height than professional basketball players? (d) Check Requirements Which distribution (standard normal or Student's \(t)\) did you use? Why? Do you need information about the height distributions? Explain.

Confidence Intervals: Sample Size A random sample is drawn from a population with \(\sigma=12\). The sample mean is 30 . (a) Compute a \(95 \%\) confidence interval for \(\mu\) based on a sample of size 49 . What is the value of the margin of error? (b) Compute a \(95 \%\) confidence interval for \(\mu\) based on a sample of size 100 . What is the value of the margin of error? (c) Compute a \(95 \%\) confidence interval for \(\mu\) based on a sample of size 225 . What is the value of the margin of error? (d) Compare the margins of error for parts (a) through (c). As the sample size increases, does the margin of error decrease? (e) Critical Thinking Compare the lengths of the confidence intervals for parts (a) through (c). As the sample size increases, does the length of a \(90 \%\) confidence interval decrease?

Law Enforcement: Escaped Convicts Case studies showed that out of 10,351 convicts who escaped from U.S. prisons, only 7867 were recaptured (The Book of \(\mathrm{O} d d s\) 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) Check Requirements Is use of the normal approximation to the binomial justified in this problem? Explain.
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