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Chapter 9: Problem 8

Construct the appropriate confidence interval. A simple random sample of size \(n=17\) is drawn from a population that is normally distributed. The sample mean is found to be \(\bar{x}=3.25,\) and the sample standard deviation is found to be \(s=1.17\). Construct a \(95 \%\) confidence interval for the population mean.

Short Answer

Expert verified
The 95% confidence interval is \[ 2.648 \text{ to } 3.852 \].

Step by step solution

01

- Identify the Given Information

Here, the sample size is given as \(n=17\), the sample mean is \(\bar{x} = 3.25\), and the sample standard deviation is \(s = 1.17\). These will be used to calculate the confidence interval.
02

- Determine the Critical Value

Since the sample size is less than 30 and the population is normally distributed, use the t-distribution. For a 95% confidence level and \(df = n - 1 = 16\), the critical value (t*) can be obtained from a t-table or statistical calculator. The critical value is approximately \(t_{0.025, 16} = 2.120\).
03

- Calculate the Standard Error

Calculate the standard error (SE) of the mean using the formula: \[ SE = \frac{s}{\sqrt{n}} \] Substituting the given values: \[ SE = \frac{1.17}{\sqrt{17}} \approx 0.284 \]
04

- Compute the Margin of Error

Calculate the margin of error (ME) using the formula: \[ ME = t^* \times SE \] Substituting the critical value and standard error: \[ ME = 2.120 \times 0.284 \approx 0.602 \]
05

- Construct the Confidence Interval

The confidence interval can be found by adding and subtracting the margin of error from the sample mean: \[ CI = \bar{x} \pm ME \] Substituting the mean and margin of error: \[ CI = 3.25 \pm 0.602 \] So, the interval is: \[ 3.25 - 0.602 \] to \[ 3.25 + 0.602 \] which simplifies to: \[ 2.648 \text{ to } 3.852 \]

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Key Concepts

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

t-distribution
When dealing with smaller sample sizes (typically less than 30) or when the population standard deviation is unknown, we use the t-distribution instead of the normal distribution. The t-distribution accounts for the extra variability that comes with smaller samples. Unlike the normal distribution, the t-distribution is slightly wider and has thicker tails. This means that it provides a bit more room to account for variability in small samples. The exact shape of the t-distribution depends on the degrees of freedom (df), which in this case is calculated as the sample size minus one.
margin of error
The margin of error (ME) quantifies the amount of random sampling error in a survey's results. It defines how much you expect the true population parameter to differ from the sample estimate. For instance, in this exercise, it's the range within which we expect the true population mean to lie, with a certain level of confidence. The formula for the margin of error is:
\[ ME = t^* \times SE \]
t* is the critical value from the t-distribution, and SE is the standard error. This ME gives us a range centered around our sample mean, accounting for the inherent variability and ensuring our confidence interval is reliable.
standard error
The standard error (SE) is a statistical term that measures the accuracy of a sample mean by quantifying the variation in sample estimates if the same population were sampled multiple times. It's calculated using the formula:
\[ SE = \frac{s}{\sqrt{n}} \]
where s is the sample standard deviation, and n is the sample size. In this exercise, our sample size (n) is 17, and the sample standard deviation (s) is 1.17. Substituting these values gives us the SE as approximately 0.284. A smaller SE indicates that the sample mean is more representative of the population mean.
sample mean
The sample mean (denoted as \( \bar{x} \)) represents the average value of a sample. It's calculated by summing up all the sample data points and then dividing by the number of data points. In this exercise, the sample mean is given as 3.25. The sample mean is a critical component in constructing confidence intervals because it serves as the center point. By adding and subtracting the margin of error to/from the sample mean, we establish a range within which we believe the true population mean lies.
degree of freedom
Degrees of freedom (df) are the number of independent values or quantities which can be assigned to a statistical distribution. For a sample, it's calculated as the sample size (n) minus one:
\[ df = n - 1 \]
In this exercise, with a sample size of 17, the degrees of freedom are 16. Degrees of freedom are crucial when using the t-distribution, as it affects the shape of the distribution. With more degrees of freedom, the t-distribution approaches the normal distribution. For fewer degrees of freedom, the t-distribution is wider, which means it provides more conservative estimates to account for the increased variability in smaller samples.

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Most popular questions from this chapter

In a Gallup poll, \(64 \%\) of the people polled answered yes to the following question: "Are you in favor of the death penalty for a person convicted of murder?" The margin of error in the poll was \(3 \%,\) and the estimate was made with \(95 \%\) confidence. At least how many people were surveyed?

You Explain It! Hours Worked In a survey conducted by the Gallup Organization, 1100 adult Americans were asked how many hours they worked in the previous week. Based on the results, a \(95 \%\) confidence interval for mean number of hours worked was lower bound: 42.7 and upper bound: \(44.5 .\) Which of the following represents a reasonable interpretation of the result? For those that are not reasonable, explain the flaw. (a) There is a \(95 \%\) probability the mean number of hours worked by adult Americans in the previous week was between 42.7 hours and 44.5 hours. (b) We are \(95 \%\) confident that the mean number of hours worked by adult Americans in the previous week was between 42.7 hours and 44.5 hours. (c) \(95 \%\) of adult Americans worked between 42.7 hours and 44.5 hours last week. (d) We are \(95 \%\) confident that the mean number of hours worked by adults in Idaho in the previous week was between 42.7 hours and 44.5 hours.

As the number of degrees of freedom in the \(t\) -distribution increases, the spread of the distribution ________ (increases/decreases).

The data sets represent simple random samples from a population whose mean is \(100 .\) $$ \begin{array}{rrrrr} & {\text { Data Set I }} \\ \hline 106 & 122 & 91 & 127 & 88 \\ \hline 74 & 77 & 108 & & \end{array} $$ $$ \begin{array}{rrrrr} \quad{\text { Data Set II }} \\ \hline 106 & 122 & 91 & 127 & 88 \\ \hline 74 & 77 & 108 & 87 & 88 \\ \hline 111 & 86 & 113 & 115 & 97 \\ \hline 122 & 99 & 86 & 83 & 102 \end{array} $$ $$ \begin{array}{rrrrr} {\text { Data Set III }} \\ \hline 106 & 122 & 91 & 127 & 88 \\ \hline 74 & 77 & 108 & 87 & 88 \\ \hline 111 & 86 & 113 & 115 & 97 \\ \hline 122 & 99 & 86 & 83 & 102 \\ \hline 88 & 111 & 118 & 91 & 102 \\ \hline 80 & 86 & 106 & 91 & 116 \end{array} $$ (a) Compute the sample mean of each data set. (b) For each data set, construct a \(95 \%\) confidence interval about the population mean. (c) What effect does the sample size \(n\) have on the width of the interval? For parts \((d)-(e),\) suppose that the data value 106 was accidentally recorded as \(016 .\) (d) For each data set, construct a \(95 \%\) confidence interval about the population mean using the incorrectly entered data. (e) Which intervals, if any, still capture the population mean, 100? What concept does this illustrate?

Suppose the following data represent the heights (in inches) of a random sample of males: 68,72,73,70,75,71 . Below are three bootstraps samples. For each bootstrap sample, determine the bootstrap sample mean. Bootstrap Sample 1: 68,72,68,73,72,75 Bootstrap Sample 2: 71,72,73,72,72,71 Bootstrap Sample 3: 72,73,68,68,73,71
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