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The following data give the one-way commuting times (in minutes) from home to work for a random sample of 30 workers. \(\begin{array}{llllllllll}23 & 17 & 34 & 26 & 18 & 33 & 46 & 42 & 12 & 37 \\ 44 & 15 & 22 & 19 & 28 & 32 & 18 & 39 & 40 & 48 \\ 25 & 36 & 23 & 39 & 42 & 46 & 29 & 17 & 24 & 31\end{array}\) a. Calculate the value of the point estimate of the mean one-way commuting time from home to work for all workers. b. Construct a \(99 \%\) confidence interval for the mean one-way commuting time from home to work for all workers.

Short Answer

Expert verified
a. The point estimate mean is the arithmetic mean of all the given times. b. The 99% confidence interval for the mean commuting time is calculated as (mean - critical value * standard error, mean + critical value * standard error).

Step by step solution

01

Compute the Point Estimate Mean

Sum up all the one-way commuting times, and then divide by the total number of samples. This will give the point estimate mean.
02

Calculate the Sample Standard Deviation

Subtract the mean from each data value and square each result. Then sum up these squared values and divide by the total number of samples minus 1. Finally, take the square root of the result. This will give the sample standard deviation.
03

Compute the Standard Error

Divide the standard deviation by the square root of the sample size. This will give the standard error.
04

Determine the Critical Value

To construct a 99% confidence interval, the critical value from the t-distribution table for 29 (which is the degree of freedom calculated as n-1) is needed. Take the critical value corresponding to 0.005 (since this is a two-tailed test, .01 should be divided by 2 to get .005).
05

Construct the Confidence Interval

Multiply the standard error by the critical value, then add and subtract the result from the sample mean. This will give the upper and lower limits for the 99% confidence interval.

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Key Concepts

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

Point Estimate
A point estimate provides a single value that serves as an estimate of a population parameter, such as the mean. In the context of statistical analysis, it often gives us the best guess of the true value based on sample data.

For instance, when estimating the mean one-way commuting time of a workforce based on a sample, the point estimate would be the sample mean itself. This point estimate helps summarize the data but comes with the caveat that it is susceptible to sample variability. This is why additional measures such as confidence intervals are often used alongside to provide a range within which the true mean likely falls.
Sample Mean
The sample mean is calculated by summing up all individual data points and dividing the total by the number of observations in the sample. In a way, it acts as our point estimate of the population mean.

This value is crucial because it provides a central measure around which other key statistics are calculated, such as variance, standard deviation, and confidence intervals. In our example, to find the sample mean, you would add all the commuting times together and divide by the total number of workers, which is 30. The simplicity of calculating the mean makes it a popular choice for analyzing data.
Standard Deviation
Standard deviation is a measure that indicates the amount of variation or dispersion within a set of data values. In essence, it tells us how much the data deviates from the mean, thereby providing insight into the reliability and consistency of the mean itself.

To calculate the sample standard deviation, follow these steps:
  • Subtract the sample mean from each data point and square each result.
  • Sum up all these squared differences.
  • Divide by the number of observations minus one (this is known as degrees of freedom).
  • Finally, take the square root of the result to obtain the standard deviation.
This computed value allows us to measure the spread of the data and is crucial for constructing confidence intervals.
T-Distribution
The t-distribution is a type of probability distribution that is symmetrical and bell-shaped, similar to the normal distribution but with heavier tails. This makes it useful for estimating population parameters when the sample size is small or when the population standard deviation is unknown.

When constructing a confidence interval, the t-distribution helps in determining the "critical value" that, when used alongside the standard error, dictates the width of the interval. The critical value is derived based on the desired confidence level (e.g., 99%) and the degrees of freedom, which is calculated as the sample size minus one. In our example, a critical value from the t-distribution is used to compute the margins of error in determining the 99% confidence interval for the mean commuting time.

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Most popular questions from this chapter

The U.S. Senate just passed a bill by a vote of \(55-45\) (with all 100 senators voting). A student who took an elementary statistics course last semester says, "We can use these data to make a confidence interval about \(p\). We have \(n=100\) and \(\hat{p}=55 / 100=.55\)." Hence, according to him, a \(95 \%\) confidence interval for \(p\) is \(\hat{p} \pm z \sigma_{\hat{p}}=.55 \pm 1.96 \sqrt{\frac{(.55)(.45)}{100}}=.55 \pm .098=.452\) to \(.648\) Does this make sense? If not, what is wrong with the student's reasoning?

a. Find the value of \(t\) for the \(t\) distribution with a sample size of 21 and area in the left tail equal to \(.10 .\) b. Find the value of \(t\) for the \(t\) distribution with a sample size of 14 and area in the right tail equal to \(.025 .\) c. Find the value of \(t\) for the \(t\) distribution with 45 degrees of freedom and \(.001\) area in the right tail. d. Find the value of \(t\) for the \(t\) distribution with 37 degrees of freedom and \(.005\) area in the left tail.

A department store manager wants to estimate the mean amount spent by all customers at this store at a \(98 \%\) confidence level. The manager knows that the standard deviation of amounts spent by all customers at this store is \(\$ 31\). What minimum sample size should he choose so that the estimate is within \(\$ 3\) of the population mean?

For a population, the value of the standard deviation is \(2.65\). A random sample of 35 observations taken from this population produced the following data. \(\begin{array}{lllllll}42 & 51 & 42 & 31 & 28 & 36 & 49 \\ 29 & 46 & 37 & 32 & 27 & 33 & 41 \\ 47 & 41 & 28 & 46 & 34 & 39 & 48 \\ 26 & 35 & 37 & 38 & 46 & 48 & 39 \\ 29 & 31 & 44 & 41 & 37 & 38 & 46\end{array}\) a. What is the point estimate of \(\mu\) ? b. Make a \(98 \%\) confidence interval for \(\mu\). c. What is the margin of error of estimate for part b?

Calculating a confidence interval for the proportion requires a minimum sample size. Calculate a confidence interval, using any confidence level of \(90 \%\) or higher, for the population proportion for each of the following. a. \(n=200\) and \(\hat{p}=.01\) b. \(n=160\) and \(\hat{p}=.9875\) Explain why these confidence intervals reveal a problem when the conditions for using the normal approximation do not hold.

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