/*! 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 61 A company randomly selected nine... [FREE SOLUTION] | 91Ó°ÊÓ

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A company randomly selected nine office employees and secretly monitored their computers for one month. The times (in hours) spent by these employees using their computers for non-job-related activities (playing games, personal communications, etc.) during this month are given below. \(\begin{array}{lllllllll}7 & 1 & 29 & 8 & 1 & 14 & 1 & 41 & 6\end{array}\) Assuming that such times for all employees are normally distributed, make a \(95 \%\) confidence interval for the corresponding population mean for all employees of this company.

Short Answer

Expert verified
95% Confidence Interval for mean time spent by all employees on non-job-related activities is \([ \bar{x} - E , \bar{x} + E]\)

Step by step solution

01

Calculating Sample Mean

Firstly, calculate the sample mean. This is done by summing all sample data points and dividing by the number of data points. The result for sample mean \(\bar{x}\) will be \(\frac{7 + 1 + 29 + 8 + 1 + 14 + 1 + 41 + 6}{9}\)
02

Calculating Standard Deviation

Once the sample mean is calculated, the next step is to calculate the sample standard deviation (s). This involves subtracting each observation from the mean, squaring the result, summing these squared values up, dividing this sum by the sample size minus one, and finally taking the square root of the result.
03

Determining the Z-value

For a 95% confidence interval, and sample size above 30 or assuming population is normally distributed, the z-value (z*) from the standard normal distribution is 1.96.
04

Calculating Margin of Error

After you determine the z-value, calculate the margin of error (E) by multiplying the z-value by the standard deviation divided by the square root of the sample size \(E = z* \cdot \frac{s}{\sqrt{n}}\)
05

Constructing Confidence Interval

Finally, the confidence interval is constructed by creating a range from the sample mean minus the margin of error to the sample mean plus the margin of error. \([- \bar{x} - E , \bar{x} + E]\)

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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 is a way to measure the central tendency of your data. It's like finding the average of your dataset. You simply add up all the data points and divide by the number of data points you have.
In the given exercise, the times spent on non-job-related activities are added: 7, 1, 29, 8, 1, 14, 1, 41, and 6. That's a total of 108.
Since there are 9 employees, you divide 108 by 9. The sample mean \(\bar{x}\) is 12.
The sample mean helps us understand the average behavior of the group being studied. In this exercise, it tells us the average time spent on non-job-related activities per employee.
Standard Deviation
Standard deviation is a way to measure how spread out your data is. When data points are close to the mean, the standard deviation is small. When they are spread out, it's large.
To find the standard deviation, follow these steps:
  • Subtract the sample mean from each data point.
  • Square each result to eliminate negative values.
  • Add all the squared results together.
  • Divide by the number of data points minus one (this gives you the variance).
  • Take the square root of the variance to find the standard deviation.
This measure tells us how much variation there is from the sample mean. In the context of the exercise, it indicates how consistent employees' non-job-related computer use is.
Z-value
The Z-value is a statistical measure that tells us how many standard deviations away an element is from the mean. It's crucial for creating confidence intervals.
For a 95% confidence interval, which is common in statistical problems, the Z-value from the standard normal distribution is typically 1.96. This value is based on the assumption that the population from which you are sampling is normally distributed.
Note that before using a Z-value, it's essential to ensure your sample size is large enough or that the population is already assumed to be normally distributed, as stated in the original exercise.
This Z-value is used later, alongside the standard deviation and sample size, to calculate the margin of error.
Margin of Error
The margin of error represents the range within which we expect the true population mean to lie. It quantifies the uncertainty or possible error in our sample estimate.
To calculate it, multiply the Z-value by the standard deviation divided by the square root of the sample size: \(E = z* \cdot \frac{s}{\sqrt{n}}\).
For the exercise, this calculation gives us a buffer around the sample mean to account for sampling variability.
The confidence interval itself is then the sample mean plus or minus this margin of error. It provides a range where the true average time spent on non-job-related activities is likely located, considering the sample data.

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Most popular questions from this chapter

Refer to Exercise 8.92. Assume that a preliminary study has shown that \(93 \%\) of all Tony's pizzas are delivered within 30 minutes. How large should the sample size be so that the \(99 \%\) confidence interval for the population proportion has a margin of error of \(.02 ?\)

A random sample of 300 female members of health clubs in Los Angeles showed that they spend, on average, \(4.5\) hours per week doing physical exercise with a standard deviation of \(.75\) hour. Find a \(98 \%\) confidence interval for the population mean.

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 $$ \hat{p} \pm z \sigma_{\hat{p}}=.55 \pm 1.96 \sqrt{\frac{(.55)(.45)}{100}}=.55 \pm .098=.452 \text { to } .648 $$ Does this make sense? If not, what is wrong with the student's reasoning?

Salaried workers at a large corporation receive 2 weeks' paid vacation per year. Sixteen randomly selected workers from this corporation were asked whether or not they would be willing to take a \(3 \%\) reduction in their annual salaries in return for 2 additional weeks of paid vacation. The following are the responses of these workers. \(\begin{array}{llllllll}\text { No } & \text { Yes } & \text { No } & \text { No } & \text { Yes } & \text { No } & \text { No } & \text { Yes } \\ \text { Yes } & \text { No } & \text { No } & \text { No } & \text { Yes } & \text { No } & \text { No } & \text { No }\end{array}\) Construct a \(97 \%\) confidence interval for the percentage of all salaried workers at this corporation who would accept a \(3 \%\) pay cut in return for 2 additional weeks of paid vacation.

Briefly explain the similarities and the differences between the standard normal distribution and the \(t\) distribution.

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