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A computer company that recently developed a new software product wanted to estimate the mean time taken to learn how to use this software by people who are somewhat familiar with computers. A random sample of 12 such persons was selected. The following data give the times taken (in hours) by these persons to learn how to use this software. $$ \begin{array}{llllll} 1.75 & 2.25 & 2.40 & 1.90 & 1.50 & 2.75 \\ 2.15 & 2.25 & 1.80 & 2.20 & 3.25 & 2.60 \end{array} $$ Construct a \(95 \%\) confidence interval for the population mean. Assume that the times taken by all persons who are somewhat familiar with computers to learn how to use this software are approximately normally distributed.

Short Answer

Expert verified
The confidence interval gives a range of values for the mean time it might take for people who are somewhat familiar with computers to learn how to use the software. The specific interval will be determined after performing the outlined steps, which revolve around basic statistics and probability theory.

Step by step solution

01

Calculate the Sample Mean and Sample Standard Deviation

First, the sample mean (also called the sample average) and the sample standard deviation have to be calculated. The sample mean is the sum of all the data values divided by the number of data values. The sample standard deviation is the square root of the sum of the squared differences between each data value and the sample mean divided by the (number of data values - 1).
02

Find the Z-Score

The z-score for a 95% confidence interval is the value that cuts off the upper 2.5% of the standard normal distribution, since 5% is to be cut off from the total area under the curve, leaving 95% for the central area. The z-score for two-tailed test at 95% confidence level is approximately 1.96.
03

Calculate the Confidence Interval

The formula to calculate a confidence interval is (X̄ - Z * (s/√n), X̄ + Z * (s/√n)), where X̄ is the sample mean, Z is the z-score, s is the sample standard deviation, and n is the number samples. Substitute the previously calculated values into this formula to find the confidence interval.

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Key Concepts

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

Sample Mean Calculation
The sample mean, often represented as \( \bar{X} \), is a central value that gives us an idea of the average time it takes for individuals to learn the software in our given sample.
To calculate the sample mean, we simply sum up all the individual data points and divide by the number of data points.
Mathematically, this is expressed as:
  • \( \bar{X} = \frac{1.75 + 2.25 + 2.40 + 1.90 + 1.50 + 2.75 + 2.15 + 2.25 + 1.80 + 2.20 + 3.25 + 2.60}{12} \)
This results in an average time which represents the central tendency of our data.
By calculating the sample mean, we aim to approximate the population mean, which is often unknown in practice.
Sample Standard Deviation
The sample standard deviation provides us an understanding of how spread out the data points are in relation to the sample mean.
It represents the average distance between each data point and the mean. This helps to understand the variability in learning times.
The formula for the sample standard deviation \( s \) is:
  • \( s = \sqrt{\frac{\sum (x_i - \bar{X})^2}{n-1}} \)
Where:
  • \( x_i \) represents each data point
  • \( \bar{X} \) is the sample mean
  • \( n \) is the number of data points, which in this case is 12
The calculation involves finding the square of the difference between each data point and the mean, summing those squares, dividing by \( n-1 \), and finally taking the square root.
This concept of measuring spread is crucial, as it affects our confidence interval width.
Z-Score
The z-score is an essential statistical tool that helps us determine how data points relate to the standard normal distribution.
In constructing confidence intervals, we use it to identify the appropriate cut-off points on the normal curve.
For a 95% confidence interval, we need a z-score that includes the middle 95% of the distribution.
This leaves 2.5% in each tail, and the corresponding z-score is approximately 1.96 for a two-tailed test.
  • This systematic selection of the z-score makes sure the confidence interval covers the true population mean with 95% certainty.
  • The z-score is critical as it adjusts the interval width based on desired confidence levels.
Understanding the statistical significance of the z-score ensures that we can correctly interpret and scale our intervals for real-world applications.
Normal Distribution
Normal distribution, often called the bell curve, is a fundamental concept in statistics describing a symmetric, bell-shaped curve where most of the observations cluster around the central peak.
In this context, we assume our data is approximately normally distributed, which helps in constructing valid confidence intervals.
This assumption allows us to use the z-score method effectively.
  • Normal distribution has mean \( \mu \) and standard deviation \( \sigma \).
  • The majority of data falls within 68% of one standard deviation from the mean, 95% within two, and 99.7% within three standard deviations, which is the Empirical Rule.
Recognizing the properties of a normal distribution is crucial since it affects statistical procedures, such as hypothesis testing and confidence intervals.
This assumption is especially key when the sample size is small, like in this exercise.

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Most popular questions from this chapter

You want to estimate the percentage of students at your college or university who are satisfied with the campus food services. Briefly explain how you will make such an estimate. Select a sample of 30 students and ask them whether or not they are satisfied with the campus food services. Then calculate the percentage of students in the sample who are satisfied. Using this information, find the confidence interval for the corresponding population percentage. Select your own confidence level.

Determine the most conservative sample size for the estimation of the population proportion for the following. a. \(E=.025\), confidence level \(=95 \%\) b. \(E=.05, \quad\) confidence level \(=90 \%\) c. \(E=.015\), confidence level \(=99 \%\)

At Farmer's Dairy, a machine is set to fill 32 -ounce milk cartons. However, this machine does not put exactly 32 ounces of milk into each carton; the amount varies slightly from carton to carton. It is known that when the machine is working properly, the mean net weight of these cartons is 32 ounces. The standard deviation of the amounts of milk in all such cartons is always equal to \(.15\) ounce. The quality control department takes a sample of 25 such cartons every week, calculates the mean net weight of these cartons, and makes a \(99 \%\) confidence interval for the population mean. If either the upper limit of this confidence interval is greater than \(32.15\) ounces or the lower limit of this confidence interval is less than \(31.85\) ounces, the machine is stopped and adjusted. A recent sample of 25 such cartons produced a mean net weight of \(31.94\) ounces. Based on this sample, will you conclude that the machine needs an adjustment? Assume that the amounts of milk put in all such cartons have a normal distribution.

A bank manager wants to know the mean amount owed on credit card accounts that become delinquent. A random sample of 100 delinquent credit card accounts taken by the manager produced a mean amount owed on these accounts equal to \(\$ 2640\). The population standard deviation was \(\$ 578\). a. What is the point estimate of the mean amount owed on all delinquent credit card accounts at this bank? b. Construct a \(97 \%\) confidence interval for the mean amount owed on all delinquent credit card accounts for this bank.

For each of the following, find the area in the appropriate tail of the \(t\) distribution. a. \(t=-1.302\) and \(d f=42\) b. \(t=2.797\) and \(n=25\) c. \(t=1.397\) and \(n=9\) d. \(t=-2.383\) and \(d f=67\)

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