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Chapter 10: Problem 20

Energy Consumption In 2001 , the mean household expenditure for energy was \(\$ 1493,\) according to data from the U.S. Energy Information Administration. An economist wanted to know whether this amount has changed significantly from its 2001 level. In a random sample of 35 households, he found the mean expenditure (in 2001 dollars) for energy during the most recent year to be \(\$ 1618\), with a standard deviation \(\$ 321\). Construct a \(95 \%\) confidence interval about the mean energy expenditure. What does the interval imply?

Short Answer

Expert verified
The 95% confidence interval is \( (1511.67, 1724.33) \). It does not include \( \text{\textdollar} 1493 \), indicating a significant change in mean energy expenditure from 2001.

Step by step solution

01

Identify the Given Information

Given: - The mean household expenditure for energy in 2001: \(\text{\textdollar} 1493\)- Sample size \(n = 35\)- Sample mean expenditure for the recent year: \(\text{\textdollar} 1618\)- Sample standard deviation: \(\text{\textdollar} 321\)
02

Determine the Confidence Interval Formula

Use the formula for the confidence interval for the mean when the sample size is large (n > 30): \[\bar{x} \text{ 卤 } z \frac{s}{\text{sqrt}(n)}\]
03

Find the Critical Value

The confidence level is \(95\text{\textpercent}\), hence the critical value \(z\)-score for \(95\text{\textpercent}\) confidence level is \(1.96\).
04

Calculate the Margin of Error

Using the formula for margin of error (ME): \[ ME = z \frac{s}{\text{sqrt}(n)} \] Plug in the values: \[ ME = 1.96 \frac{321}{\text{sqrt}(35)} \] First, calculate \( \text{sqrt}(35) 鈮� 5.92\): Then: \[ ME 鈮� 1.96 \frac{321}{5.92} 鈮� 106.33 \]
05

Construct the Confidence Interval

The confidence interval is given by: \[\bar{x} \text{ 卤 } ME\] Plugging in the values: \[ 1618 \text{ 卤 } 106.33 \] So, the interval is: \[ (1618 - 106.33, 1618 + 106.33) 鈮� (1511.67, 1724.33) \]
06

Interpret the Results

The \(95\text{\textpercent}\) confidence interval for the mean energy expenditure is \( (1511.67, 1724.33) \). This interval does not include the 2001 mean expenditure of \( \text{\textdollar} 1493 \), suggesting that the mean energy expenditure has significantly changed from its 2001 level.

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Key Concepts

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

mean household expenditure
The mean household expenditure represents the average amount households spend on a specific item or category, such as energy. In this exercise, the mean energy expenditure for households in the most recent year was found to be \(\text{\textdollar} 1618\). This value is obtained by adding up all individual household expenditures and then dividing by the number of households sampled. Understanding the mean is crucial since it serves as a central value around which the confidence interval is calculated. The mean gives a good estimate of central tendency, helping us infer about the overall population from which the sample is drawn.
sample standard deviation
The sample standard deviation quantifies the amount of variation or dispersion within a set of data. It tells us how much household expenditures vary from the sample mean expenditure. In this case, the sample standard deviation is \(\text{\textdollar} 321\). This metric is vital because it helps in calculating the margin of error for the confidence interval. A higher standard deviation indicates that household expenditures vary widely, whereas a lower standard deviation indicates that expenditures are more consistent. To compute the sample standard deviation, you need to calculate the square root of the variance, which is the average of the squared differences from the mean.
critical value
The critical value is a key factor in determining the width of the confidence interval. For a given confidence level, it represents the number of standard deviations the mean is from the population mean. For a 95\(\%\) confidence level, the critical value (z-score) is 1.96. This value is derived from the standard normal distribution table. It indicates that 95\(\%\) of the data falls within 1.96 standard deviations from the mean. The higher the confidence level, the larger the critical value, which in turn widens the confidence interval.
margin of error
The margin of error quantifies the range within which the true population mean is expected to lie, given the sample mean and standard deviation. It is calculated as \[ ME = z \frac{s}{\text{sqrt}(n)} \] where \(z\) is the critical value, \(s\) is the sample standard deviation, and \(n\) is the sample size. In this case, \(ME 鈮� 106.33\). This means the actual mean household expenditure could be within \(106.33\) dollars above or below the sample mean. A smaller margin of error implies more confidence in the estimate, while a larger margin suggests more variability.
interpretation of confidence interval
The confidence interval provides a range of values within which we expect the true population mean to lie with a certain level of confidence. Here, the 95\(\%\) confidence interval for the mean energy expenditure is \( (1511.67, 1724.33)\). This interval suggests that there is a 95\(\%\) probability that the true mean household expenditure on energy falls within this range. Since the 2001 mean of \( \text{\textdollar} 1493 \) is not within this interval, we can infer that the mean energy expenditure has significantly changed from its 2001 level. The confidence interval thus offers not just an estimate of the mean but also an insight into the reliability of this estimate.

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Most popular questions from this chapter

Statistical Significance versus Practical Significance The manufacturer of a daily dietary supplement claims that its product will help people lose weight. The company obtains a random sample of 950 adult males aged 20 to 74 who take the supplement and finds their mean weight loss after 8 weeks to be 0.9 pound with standard deviation weight loss of 7.2 pounds. (a) State the null and alternative hypotheses. (b) Test the hypothesis at the \(\alpha=0.1\) level of significance. Is a mean weight loss of 0.9 pound significant? (c) Do you think that a mean weight loss of 0.9 pound is worth the expense and commitment of a daily dietary supplement? In other words, does the weight loss have any practical significance? (d) Test the hypothesis at the \(\alpha=0.1\) level of significance with \(n=40\) subjects. Assume the same sample statistics. Is a sample mean weight loss of 0.9 pound significantly more than 0 pound? What do you conclude about the impact of large samples on the hypothesis test?

A simple random sample of size \(n=16\) is drawn from a population that is normally distributed. The sample variance is found to be 13.7 . Test whether the population variance is greater than 10 at the \(\alpha=0.05\) level of significance.

Professors Honey Kirk and Diane Lerma of Palo Alto College developed a "learning community curriculum that blended the developmental mathematics and the reading curriculum with a structured emphasis on study skills." In a typical developmental mathematics course at Palo Alto College, \(50 \%\) of the students complete the course with a letter grade of \(\mathrm{A}, \mathrm{B},\) or \(\mathrm{C} .\) In the experimental course, of the 16 students enrolled, 11 completed the course with a letter grade of \(\mathrm{A}, \mathrm{B},\) or \(\mathrm{C} .\) Do you believe the experimental course was effective at the \(\alpha=0.05\) level of significance? (a) State the appropriate null and alternative hypotheses. (b) Verify that the normal model may not be used to estimate the P-value (c) Explain why this is a binomial experiment. (d) Determine the \(P\) -value using the binomial probability distribution. State your conclusion to the hypothesis test. (e) Suppose the course is taught with 48 students and 33 complete the course with a letter grade of \(\mathrm{A}, \mathrm{B},\) or \(\mathrm{C}\). Verify the normal model may now be used to estimate the \(P\) -value. (f) Use the normal model to obtain and interpret the \(P\) -value. State your conclusion to the hypothesis test. (g) Explain the role that sample size plays in the ability to reject statements in the null hypothesis.

NCAA rules require the circumference of a softball to be \(12 \pm 0.125\) inches. Suppose that the NCAA also requires that the standard deviation of the softball circumferences not exceed 0.05 inch. A representative from the NCAA believes the manufacturer does not meet this requirement. She collects a random sample of 20 softballs from the production line and finds that \(s=0.09\) inch. Is there enough evidence to support the representative's belief at the \(\alpha=0.05\) level of significance?

Study Time Go to www.pearsonhighered.com/sullivanstats to obtain the data file \(10_{-} 3_{-} 26\) using the file format of your choice for the version of the text you are using. The data represent the amount of time students in Sullivan's online statistics course spent studying for Section 4.1 -Scatter Diagrams and Correlation. (a) Draw a histogram of the data. Describe the shape of the distribution. (b) Draw a boxplot of the data. Are there any outliers? (c) Based on the shape of the histogram and boxplot, explain why a large sample size is necessary to perform inference on the mean using the normal model. (d) According to MyStatLab, the mean time students would spend on this assignment nationwide is 95 minutes. Treat the data as a random sample of all Sullivan online statistics students. Do the sample data suggest that Sullivan's students are any different from the country as far as time spent on Section 4.1 goes? Use an \(\alpha=0.05\) level of significance.
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