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Chapter 9: Problem 14

The Buffalo. New York, Chamber of Commerce wants to estimate the mean time workers who are employed in the downtown area spend getting to work. A sample of 15 workers reveals the following number of minutes spent traveling. $$\begin{array}{llllllll}14 & 24 & 24 & 19 & 24 & 7 & 31 & 20 \\ 26 & 23 & 23 & 28 & 16 & 15 & 21 &\end{array}$$ Develop a \(98 \%\) confidence interval for the population mean. Interpret the result.

Short Answer

Expert verified
The 98% confidence interval is (14.45, 24.89) minutes.

Step by step solution

01

Calculate the Sample Mean

First, we need to calculate the mean of the given sample data. The sample data consists of the following travel times in minutes: 14, 24, 24, 19, 24, 7, 31, 20, 26, 23, 23, 28, 16, 15, 21. Sum these numbers and then divide by the number of data points, which is 15.The sum of these values is: 295.The sample mean \( \bar{x} \) is calculated as:\[ \bar{x} = \frac{295}{15} = 19.67 \text{ minutes} \]
02

Calculate the Sample Standard Deviation

Next, calculate the sample standard deviation. Use the formula:\[ s = \sqrt{\frac{1}{n-1} \sum_{i=1}^{n}(x_i - \bar{x})^2} \]First, calculate the variance by finding each squared difference from the mean, summing them, and dividing by \( n-1 \), where \( n = 15 \).The squared differences are: \[ (14-19.67)^2, (24-19.67)^2, \,\ldots,\, (21-19.67)^2 \]Sum of squared differences: \( 645.33 \).The sample variance is then:\[ \text{Variance} = \frac{645.33}{14} = 46.09 \] The sample standard deviation \( s \) is:\[ s = \sqrt{46.09} = 6.79 \text{ minutes} \]
03

Determine the T-Value

Since the population standard deviation is unknown and the sample size is small, use the t-distribution to find the critical t-value. For a 98% confidence interval and degrees of freedom \( df = n-1 = 14 \), use a t-table or calculator to find the t-value.For a two-tailed test with 98% confidence level, the t-value is approximately 2.977.
04

Calculate the Margin of Error

Compute the margin of error (ME) using the formula:\[ ME = t \times \frac{s}{\sqrt{n}} \]where \( t = 2.977 \), \( s = 6.79 \), and \( n = 15 \).Calculate:\[ ME = 2.977 \times \frac{6.79}{\sqrt{15}} \approx 5.22 \]
05

Determine the Confidence Interval

Calculate the confidence interval using the formula:\[ \bar{x} \pm ME \]Substitute the values:\[ 19.67 \pm 5.22 \]The confidence interval is:\[ (14.45, 24.89) \]
06

Interpret the Confidence Interval

The 98% confidence interval for the mean time workers spend getting to work in downtown Buffalo is between 14.45 and 24.89 minutes. This means we are 98% confident that the true mean commute time for all downtown workers falls within this range.

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Key Concepts

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

Sample Mean
In statistics, the sample mean is a crucial measure that represents the average of a dataset. It helps summarize the central tendency of a group of numbers.
To calculate the sample mean, sum all the data values and divide by the number of observations. In this exercise, the sample represents the travel times for workers in Buffalo. By summing the times (14, 24, 24, etc.) and dividing by the number of observations (15), we obtain a sample mean of 19.67 minutes.

The sample mean (\( \bar{x} \)) is an estimate of the population mean and is often used when the selection is random and unbiased. It can be influenced by extreme values but still provides valuable insights into the overall dataset.
Sample Standard Deviation
The sample standard deviation is a measure of the amount of variation or dispersion in a set of values.
It quantifies how much the individual data points deviate from the sample mean. A higher standard deviation indicates more variation, while a lower value suggests the data points are closer to the mean.

To compute the sample standard deviation, calculate the variance first. This involves finding the squared differences of each data point from the mean, summing them up, and then dividing by one less than the number of observations (\( n-1 \)). For our exercise, this step generates a variance of 46.09. The standard deviation, obtained by taking the square root of variance, is approximately 6.79 minutes.
This gives us an understanding of how travel times are spread out.
T-Value
When we work with a small sample size and don't know the population standard deviation, the t-distribution becomes a key tool.
The t-value is a critical factor used to determine the confidence interval, accounting for the increased variability in small samples. For this exercise, the sample size is 15, which entails using degrees of freedom (df) of 14 (\( n-1 \)).

Using a statistical table or calculator, the t-value for a 98% confidence level and df = 14 is approximately 2.977. This value helps determine the margin of error, as it affects the range for the confidence interval.
Margin of Error
The margin of error is an expression of the amount of random sampling error in a survey's results.
It's crucial in statistics to measure the uncertainty around a sample mean. It reflects the extent to which the sample mean might differ from the true population mean.

To calculate the margin of error (\( ME \)), multiply the t-value by the sample standard deviation divided by the square root of the sample size (\( \frac{s}{\sqrt{n}} \)). In our exercise, this calculation (\( 2.977 \times \frac{6.79}{\sqrt{15}} \)) results in approximately 5.22 minutes. The margin of error helps in constructing the confidence interval, providing a range that is likely to encompass the population mean.

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Most popular questions from this chapter

You plan to conduct a survey to find what proportion of the workforce has two or more jobs. You decide on the \(95 \%\) confidence level and a margin of error of \(2 \% .\) A pilot survey reveals that 5 of the 50 sampled hold two or more jobs. How many in the workforce should be interviewed to meet your requirements?

The U.S. Dairy Industry wants to estimate the mean yearly milk consumption. A sample of 16 people reveals the mean yearly consumption to be 45 gallons with a standard deviation of 20 gallons. Assume the population distribution is normal. a. What is the value of the population mean? What is the best estimate of this value? b. Explain why we need to use the \(t\) distribution. What assumption do you need to make? c. For a 90\% confidence interval, what is the value of \(t ?\) d. Develop the \(90 \%\) confidence interval for the population mean. e. Would it be reasonable to conclude that the population mean is 48 gallons?

Past surveys reveal that 30% of tourists going to Las Vegas to gamble spend more than \(\$ 1,000 .\) The Visitor's Bureau of Las Vegas wants to update this percentage. a. How many tourists should be randomly selected to estimate the population proportion with a 90% confidence level and a 1% margin of error? b. The Bureau feels the sample size determined above is too large. What can be done to reduce the sample? Based on your suggestion, recalculate the sample size.

An important factor in selling a residential property is the number of times real estate agents show a home. A sample of 15 homes recently sold in the Buffalo, New York, area revealed the mean number of times a home was shown was 24 and the standard deviation of the sample was 5 people. Develop a \(98 \%\) confidence interval for the population mean.

As part of their business promotional package, the Milwaukee Chamber of Commerce would like an estimate of the mean cost per month to lease a one- bedroom apartment. The mean cost per month for a random sample of 40 apartments currently available for lease was 884. The standard deviation of the sample was 50. a. Develop a 98% confidence interval for the population mean. b. Would it be reasonable to conclude that the population mean is 950 per month?
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