/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 9 A random sample of 10 houses hea... [FREE SOLUTION] | 91Ó°ÊÓ

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A random sample of 10 houses heated with natural gas in a particular area, is selected, and the amount of gas (in therms) used during the month of January is determined for each house. The resulting observations are as follows: \(\begin{array}{llllllllll}103 & 156 & 118 & 89 & 125 & 147 & 122 & 109 & 138 & 99\end{array}\) a. Let \(\mu_{J}\) denote the average gas usage during January by all houses in this area. Compute a point estimate of \(\mu_{J}\) b. Suppose that 10,000 houses in this area use natural gas for heating. Let \(\tau\) denote the total amount of gas used by all of these houses during January. Estimate \(\tau\) using the given data. What statistic did you use in computing your estimate? c. Use the data in Part (a) to estimate \(p\), the proportion of all houses that used at least 100 therms. d. Give a point estimate of the population median us- age based on the sample of Part (a). Which statistic did you use?

Short Answer

Expert verified
The point estimate of \(\mu_{J}\) is the arithmetic mean of the data set. The estimate of \(\tau\) is obtained by multiplying the mean by the total number of houses. The estimate of \(p\) is the proportion of observations that are at least 100. The point estimate of the population median is obtained by averaging the two central numbers of the sorted data set.

Step by step solution

01

Calculate the point estimate of \(\mu_{J}\)

The point estimate of \(\mu_{J}\), the average usage of gas during January, is simply the arithmetic mean of the provided data set. That is, sum up all the 10 observations and divide by 10 to get the average.
02

Estimate \(\tau\)

To estimate \(\tau\), the total amount of gas used by all houses, multiply the point estimate of \(\mu_{J}\) obtained in Step 1 by the total number of houses, which is 10,000 in this case. This is because \(\mu_{J}\) is considered representative of the average consumption per house in the locality.
03

Estimate the proportion \(p\)

To estimate \(p\), the proportion of houses that used at least 100 therms, count the number of observations greater than or equal to 100 and divide by the total number of observations, i.e. 10.
04

Estimate the population median usage

To estimate the population median usage, sort the data set in ascending order, and since it contains 10 observations which is an even number, the median is a mean of the two central numbers. That is, take the average of the 5th and the 6th observation.

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Key Concepts

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

Sample Mean
Understanding the sample mean is fundamental when analyzing data. The sample mean, often represented by \( \bar{x} \), is a very common point estimate used to infer the average of a population. In the context of a real-world scenario - for example, the average gas usage by houses during January - we use the observations from our sample to calculate it. To find the sample mean, you sum up all the observed values and divide by the number of observations.

For the exercise provided, you would add the gas usage of all 10 houses together and divide this total by 10. It's a straightforward approach, but it plays a crucial role in statistics as it sets the stage for more complex analyses like regressions and hypothesis testing. It's important to remember, though, the sample mean is only an estimate of the true population mean, and it’s accuracy depends on how representative the sample is of the entire population.
Population Total Estimation
When we talk about population total estimation, we're venturing beyond our sample to make predictions about the total quantity of interest within an entire population. This typically involves multiplying the sample mean by the population size.

In our exercise, after calculating the average gas usage from our sample, we estimate the total gas usage, \( \tau \), for all 10,000 houses by taking our sample mean as a representation of the average usage per house. This extrapolates the data from the 10 observed houses to all the houses using gas in the area. It's an efficient method for estimation, but we must be cautious and consider the variability and potential bias within our sample. If our sample is not representative, our estimate might be less accurate.
Proportion Estimate
Estimating proportions is another key statistical task that helps us understand the characteristics of our population. A proportion estimate is simply the fraction of the sample that meets a certain criterion.

In this exercise, we're asked to estimate \( p \), the proportion of houses that used at least 100 therms of gas. To do this, we count how many houses in our sample used 100 or more therms, and divide by the total number of houses in the sample. Proportion estimates like this can provide a quick snapshot of a particular subset within your data, and are especially useful in opinion polls and quality control.
Population Median
The population median is a measure of central tendency that indicates the middle value of a dataset when ordered from smallest to largest. Estimating the population median from a sample involves sorting your sample data and identifying the middle value or, if the sample size is even, taking the average of the two middle values.

This was exactly the process used in the exercise, where after arranging the gas usage amounts in ascending order, we calculated the median by averaging the 5th and 6th observations. The median is less affected by outliers and skewed data compared to the mean, and thus it provides another perspective on the center of the data. It’s a valuable statistic when you're dealing with data that isn’t symmetrically distributed or when you're interested in the typical experience in a population, rather than the overall level.

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Most popular questions from this chapter

"Tongue Piercing May Speed Tooth Loss. Researchers Say" is the headline of an article that appeared in the San Luis Obispo Tribune (June 5,2002 ). The article describes a study of 52 young adults with pierced tongues. The researchers found receding gums, which can lead to tooth loss, in 18 of the participants. Construct a \(95 \%\) confidence interval for the proportion of young adults with pierced tongues who have receding gums. What assumptions must be made for use of the \(z\) confidence interval to be appropriate?

The article “Kids Digital Day: Almost 8 Hours" (USA Today, January 20,2010 ) summarized results from a national survey of 2002 Americans age 8 to 18 . The sample was selected in a way that was expected to result in a sample representative of Americans in this age group. a. Of those surveyed, 1321 reported owning a cell phone. Use this information to construct and interpret a \(90 \%\) confidence interval estimate of the proportion of all Americans age 8 to 18 who own a cell phone. b. Of those surveyed, 1522 reported owning an MP3 music player. Use this information to construct and interpret a \(90 \%\) confidence interval estimate of the proportion of all Americans age 8 to 18 who own an MP3 music player. c. Explain why the confidence interval from Part (b) is narrower than the confidence interval from Part (a) even though the confidence level and the sample size used to compute the two intervals was the same.

The article "The Association Between Television Viewing and Irregular Sleep Schedules Among Children Less Than 3 Years of Age" (Pediatrics [2005]: \(851-856\) ) reported the accompanying \(95 \%\) confidence intervals for average TV viewing time (in hours per day) for three different age groups. $$\begin{array}{lc} \text { Age Group } & 95 \% \text { Confidence Interval } \\ \hline \text { Less than } 12 \text { months } & (0.8,1.0) \\ 12 \text { to } 23 \text { months } & (1.4,1.8) \\ 24 \text { to } 35 \text { months } & (2.1,2.5) \\ \hline \end{array}$$ a. Suppose that the sample sizes for each of the three age group samples were equal. Based on the given confidence intervals, which of the age group samples had the greatest variability in TV viewing time? Explain your choice. b. Now suppose that the sample standard deviations for the three age group samples were equal, but that the three sample sizes might have been different. Which of the three age-group samples had the largest sample size? Explain your choice. c. The interval \((.768,1.032)\) is either a \(90 \%\) confidence interval or a \(99 \%\) confidence interval for the mean TV viewing time computed using the sample data for children less than 12 months old. Is the confidence level for this interval \(90 \%\) or \(99 \%\) ? Explain your choice.

In a survey on supernatural experiences, 722 of 4013 adult Americans surveyed reported that they had seen or been with a ghost ("What Supernatural Experiences We've Had," USA Today, February 8,2010 ). a. What assumption must be made in order for it to be appropriate to use the formula of this section to construct a confidence interval to estimate the proportion of all adult Americans who have seen or been with a ghost? b. Construct and interpret a \(90 \%\) confidence interval for the proportion of all adult Americans who have seen or been with a ghost. c. Would a \(99 \%\) confidence interval be narrower or wider than the interval computed in Part (b)? Justify your answer.

The interval from \(-2.33\) to \(1.75\) captures an area of \(.95\) under the \(z\) curve. This implies that another large-sample \(95 \%\) confidence interval for \(\mu\) has lower limit \(\bar{x}=2.33 \frac{\sigma}{\sqrt{n}}\) and upper limit \(\bar{x}\) + \(1.75 \frac{\sigma}{\sqrt{n}}\). Would you recommend using this \(95 \%\) interval over the \(95 \%\) interval \(\bar{x} \pm 1.96 \frac{\sigma}{\sqrt{n}}\) discussed in the text? Explain. (Hint: Look at the width of each interval.)

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