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Chapter 9: Problem 10

Construct the appropriate confidence interval. A simple random sample of size \(n=210\) is drawn from a population. The sample mean is found to be \(\bar{x}=20.1,\) and the sample standard deviation is found to be \(s=3.2\). Construct a \(90 \%\) confidence interval for the population mean.

Short Answer

Expert verified
The \(90\%\) confidence interval for the population mean is \((19.736, 20.464)\).

Step by step solution

01

Identify the given values

From the problem, the sample size is given as \(n = 210\), the sample mean is \(\bar{x} = 20.1\), and the sample standard deviation is \(s = 3.2\). The confidence level is \(90\text{\%}\).
02

Find the critical value

For a \(90\text{\%}\) confidence interval, the critical value \(z_{\frac{\alpha}{2}}\) (which corresponds to the standard normal distribution) can be found using a z-table or calculator. \(z_{\frac{\alpha}{2}} = 1.645\).
03

Calculate the standard error

The standard error (SE) of the mean is calculated by dividing the sample standard deviation \(s\) by the square root of the sample size \(n\): \[ SE = \frac{s}{\sqrt{n}} = \frac{3.2}{\sqrt{210}} \approx 0.221\].
04

Determine the margin of error

The margin of error (ME) is found by multiplying the critical value by the standard error: \[ ME = z_{\frac{\alpha}{2}} \times SE = 1.645 \times 0.221 \approx 0.364\].
05

Construct the confidence interval

The confidence interval is calculated by adding and subtracting the margin of error from the sample mean: \[ (\bar{x} - ME, \bar{x} + ME) = (20.1 - 0.364, 20.1 + 0.364) = (19.736, 20.464)\].

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Key Concepts

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

Sample Mean
The sample mean, often represented as \(\bar{x}\), is a fundamental concept in statistics. It is calculated as the sum of all the individual observations divided by the number of observations in the sample (n). In simple terms, it is the average value of the sample data.
Here is the formula for the sample mean: \[ \bar{x} = \frac{\text{Sum of all sample values}}{n} \]
This value gives us a point estimate of the population mean, which is the average value of the entire population. Point estimates, however, are subject to sampling variability.
In our exercise, the sample mean is given as \(\bar{x} = 20.1\). This indicates that the average value of the observations in our sample of 210 is 20.1.
Standard Error
The standard error (SE) is a measure of the variability or spread of the sample mean. It gives an indication of how much the sample mean is expected to fluctuate from one sample to another.
The standard error is calculated using the sample standard deviation (s) and the sample size (n) with the following formula: \[ SE = \frac{s}{\frac{\text{Sum of all sample values}}{n}} \]
In our example, the sample standard deviation (s) is 3.2, and our sample size (n) is 210. Therefore, the standard error is approximately 0.221.
A lower standard error indicates that the sample mean is a more precise estimate of the population mean.
Margin of Error
The margin of error (ME) tells us how much we expect our sample mean to vary from the true population mean. It is a crucial part of constructing a confidence interval.
The margin of error is calculated by multiplying the critical value (z_{\frac{\alpha}{2}}) by the standard error (SE): \[ ME = z_{\frac{\alpha}{2}} \times SE \]
For a 90% confidence interval, the critical value (z_{\frac{\alpha}{2}}) is 1.645. Given our standard error of 0.221, the margin of error is approximately 0.364.
This means that we can expect the true population mean to fall within a range of our sample mean 卤 0.364 with 90% confidence.
In this example, our confidence interval is therefore (19.736, 20.464), providing a range where we believe the true population mean lies.

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Most popular questions from this chapter

Determine the point estimate of the population mean and margin of error for each confidence interval. Lower bound: \(20,\) upper bound: 30

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?

The Sullivan Statistics Survey I asks, "Is a language other than English the primary language spoken in your home?" Treat the survey respondents as a random sample of adult Americans. Go to www.pearsonhighered.com/sullivanstats to obtain the data file SullivanSurveyI using the file format of your choice for the version of the text you are using. The column "Language" has survey responses. Construct and interpret a \(95 \%\) confidence interval for the proportion of adult Americans whose primary language spoken at home is something other than English.

In March 2014, Harris Interactive conducted a poll of a random sample of 2234 adult Americans 18 years of age or older and asked, "Which is more annoying to you, tailgaters or slow drivers who stay in the passing lane?" Among those surveyed, 1184 were more annoyed by tailgaters. (a) Explain why the variable of interest is qualitative with two possible outcomes. What are the two outcomes? (b) Verify the requirements for constructing a \(90 \%\) confidence interval for the population proportion of all adult Americans who are more annoyed by tailgaters than slow drivers in the passing lane. (c) Construct a \(90 \%\) confidence interval for the population proportion of all adult Americans who are more annoyed by tailgaters than slow drivers in the passing lane.

The exponential probability distribution can be used to model waiting time in line or the lifetime of electronic components. Its density function is skewed right. Suppose the wait-time in a line can be modeled by the exponential distribution with \(\mu=\sigma=5\) minutes. (a) Simulate obtaining a random sample of 15 wait-times. (b) Explain why constructing a \(95 \%\) confidence interval using Student's \(t\) -distribution is a bad idea. Nonetheless, construct a \(95 \%\) confidence interval using Student's \(t\) -distribution. (c) Use the data to construct a \(95 \%\) confidence interval for the mean using 1000 resamples.
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