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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 about the population mean.

Short Answer

Expert verified
The 95% confidence interval is (2.65, 3.85).

Step by step solution

01

Identify the given information

First, identify the given information from the problem: the sample size (), sample mean (\bar{x}), sample standard deviation (si.e, and the confidence level.\...n}\[n = 17\] \[\bar{x} = 3.25\] \[s = 1.17\] The confidence level is 95%.
02

Determine the t-critical value

Since the sample size is small (<30) and drawn from a normally distributed population, use the t-distribution. Determine the t-critical value (_{t^{*}}) for a 95% confidence level with degrees of freedom (df) equal to -1.Use a t-table or statistical software to find the value.\[df = n - 1 = 17 - 1 = 16\] For a 95% confidence interval, and df = 16, the t-critical value is approximately _{0.025,16} = 2.120.
03

Calculate the margin of error

Use the t-critical value and the standard deviation to compute the margin of error (E).The formula is S s/sqrt(n).\[E = t_{0.025,16} * \frac{s}{\sqrt{n}}\] Substituting the values:\[E = 2.120 * \frac{1.17}{\sqrt{17}} 鈮� 2.120 * 0.2837 鈮� 0.600\]
04

Construct the confidence interval

Now, use the sample mean and the margin of error to find the confidence interval.The formula is \bar{x} m s s -E, w E.\[CI = (\bar{x} - E, \bar{x} + E)\] Substitute the values to obtain the interval:\[CI = (3.25 - 0.600, 3.25 + 0.600)\] \[CI = (2.65, 3.85)\]

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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 \(\bar{x}\) is the average value of a sample. It's an estimate of the average value of the population, but calculated from the sample data. In the given problem, the sample mean is found to be \(\bar{x} = 3.25\).
t-Distribution
The t-distribution is a statistical distribution used when the sample size is small (typically less than 30), or when the population standard deviation is unknown. This distribution is similar to the normal distribution but has heavier tails. It's often applied to confidence intervals for the mean in small samples. For our problem, we have a sample size of 17, and we use the t-distribution to find the t-critical value necessary for constructing the confidence interval.
Margin of Error
The margin of error (E) quantifies the amount of random sampling error in a survey's results. It's calculated using the t-critical value, the sample standard deviation, and the sample size. For the given problem, the margin of error is calculated as: \[E = t_{0.025,16} * \frac{s}{\sqrt{n}} \] Substituting the values, we get: \[ E = 2.120 * \frac{1.17}{\sqrt{17}} \approx 2.120 * 0.2837 \approx 0.600\] This value is then used to construct the confidence interval.
Sample Size
Sample size (n) refers to the number of observations in a sample. The larger the sample size, the more reliable your estimate of the population parameters will be. In the given exercise, the sample size is 17. This small sample size is why the t-distribution is used instead of the normal distribution.
Standard Deviation
Standard deviation (s) indicates how much the values in the sample deviate from the sample mean on average. It's a measure of the dispersion or spread of the sample data. In our example, the sample standard deviation is 1.17. This value helps us quantify the variability in our sample, which is critical when calculating the margin of error and the confidence interval.

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Most popular questions from this chapter

A Gallup poll conducted May \(20-22,2005,\) asked 1006 Americans, "During the past year, about how many books, either hardcover or paperback, did you read either all or part of the way through?" Results of the survey indicated that \(\bar{x}=13.4\) books and \(s=16.6\) books. Construct a \(99 \%\) confidence interval for the mean number of books Americans read either all or part of during the preceding year. Interpret the interval.

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The mean age of the 42 presidents of the United States on the day of inauguration is 54.8 years with a standard deviation of 6.2 years. A researcher constructed a \(95 \%\) confidence interval for the mean age of presidents on inauguration day. He wrote that he was \(95 \%\) confident the mean age of the president on inauguration day is between 53.0 and 56.7 years of age. What is wrong with the researcher's analysis?

Home Ownership An urban economist wishes to estimate the proportion of Americans who own their house. What size sample should be obtained if he wishes the estimate to be within 0.02 with 90\% confidence if (a) he uses an estimate of 0.675 from the fourth quarter of 2000 obtained from the U.S. Census Bureau? (b) he does not use any prior estimates?

Many companies pay a portion of their profits to shareholders in the form of dividends. A stock's dividend yield is defined as the dividend paid by the company divided by its stock price. A stock analyst wants to estimate the mean dividend yield of financial stocks. He obtains a simple random sample of 11 financial stocks and obtains the following dividend yields. The data are in percent, so 3.17 represents \(3.17 \% .\) Construct and interpret a \(90 \%\) confidence interval for the dividend yield of financial stocks. An analysis of the data indicates that it is reasonable to conclude dividend yields follow a normal distribution. The data set has no outliers. $$\begin{array}{llllll} 3.19 & 1.86 & 0 & 0.54 & 1.76 & 2.08 \\ \hline 0.67 & 0.67 & 1.57 & 0 & 2.12 & \\ \hline \text {Source} \text { Morningstar } & & & & \\ \hline \end{array}$$
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