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Merrill Lynch Securities and Health Care Retirement, Inc., are two large employers in downtown Toledo, Ohio. They are considering jointly offering child care for their employees. As a part of the feasibility study, they wish to estimate the mean weekly child-care cost of their employees. A sample of 10 employees who use child care reveals the following amounts spent last week. $$\begin{array}{llllllllll}\hline 107 & \$ 92 & \$ 97 & \$ 95 & \$ 105 & \$ 101 & \$ 91 & \$ 99 & \$ 95 & \$ 104 \\\\\hline\end{array}$$ Develop a 90 percent confidence interval for the population mean. Interpret the result.

Short Answer

Expert verified
The 90% confidence interval for the mean weekly child-care cost is $95.39 to $101.81.

Step by step solution

01

Calculate the Sample Mean

Add up all the given child-care costs and divide by the number of observations. The sample values are: 107, 92, 97, 95, 105, 101, 91, 99, 95, 104. First, find the sum: 107 + 92 + 97 + 95 + 105 + 101 + 91 + 99 + 95 + 104 = 986. Next, divide by the number of samples, which is 10. Sample Mean \( \bar{x} = \frac{986}{10} = 98.6 \).
02

Calculate the Sample Standard Deviation

Use the formula for the sample standard deviation:\[ s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}} \]Compute each deviation from the mean, square it, sum all squared deviations, then divide by \(n-1\) where \(n\) is 10.Calculate deviations and their squares:- \((107 - 98.6)^2 = 70.56\)- \((92 - 98.6)^2 = 43.56\)- \((97 - 98.6)^2 = 2.56\)- \((95 - 98.6)^2 = 12.96\)- \((105 - 98.6)^2 = 40.96\)- \((101 - 98.6)^2 = 5.76\)- \((91 - 98.6)^2 = 57.76\)- \((99 - 98.6)^2 = 0.16\)- \((95 - 98.6)^2 = 12.96\)- \((104 - 98.6)^2 = 29.16\)Sum of squares: 276.4Now divide by \(n-1 = 10-1 = 9\):\[ s = \sqrt{\frac{276.4}{9}} = \sqrt{30.7111} = 5.54 \approx 5.54 \]
03

Calculate the Standard Error of the Mean

The standard error of the mean (SEM) is given by: \[ SEM = \frac{s}{\sqrt{n}} \] Where \(s\) is the sample standard deviation and \(n\) is the number of samples.\[ SEM = \frac{5.54}{\sqrt{10}} = \frac{5.54}{3.16} \approx 1.75 \]
04

Determine the t-Critical Value for 90% Confidence Interval

With a sample size of 10, degrees of freedom \(df = n-1 = 9\). Using a t-distribution table, find the t-value for a 90% confidence interval and 9 degrees of freedom: \( t \approx 1.833 \) (this value can vary slightly depending on the table/source).
05

Calculate the Confidence Interval

The formula for the confidence interval is: \[ \bar{x} \pm t \times SEM \]Substitute the known values:\( 98.6 \pm 1.833 \times 1.75 \)Calculate the margin of error:\( 1.833 \times 1.75 = 3.208 \approx 3.21 \)Thus, the confidence interval is:\[ (98.6 - 3.21, 98.6 + 3.21) = (95.39, 101.81) \]
06

Interpret the Confidence Interval

The 90% confidence interval means that we are 90% confident that the true population mean weekly child-care cost for the employees lies between $95.39 and $101.81.

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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 an essential measure in statistics that helps us understand the central tendency of a data set. It provides a quick summary of where the data points tend to cluster. To compute the sample mean, sum up all the observations in your sample and divide by the total number of observations. It's expressed as \( \bar{x} = \frac{\sum x_i}{n} \), where \( x_i \) are your data points and \( n \) is the number of observations. In our exercise, the sum of child-care costs was 986, resulting in a sample mean of 98.6 when divided by the 10 observations. This value represents the average amount spent by the employees on child care, offering a base for further statistical analysis.
Standard Deviation
The standard deviation is crucial for understanding data variability. It quantifies how much the individual data points deviate from the sample mean. A larger standard deviation indicates that the data points are spread out over a wider range, whereas a smaller one suggests that they are closer to the mean. To calculate the standard deviation for a sample, use the formula: \[ s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}} \] Here, \( x_i \) represents each data point, and \( \bar{x} \) is the sample mean. The subtraction from each data point to the mean, then squaring, helps in identifying the spread of the dataset. The results are summed up and divided by \( n-1 \) (i.e., degrees of freedom), and finally, the square root is taken. Our calculated standard deviation was approximately 5.54, indicating a moderate level of variability in child-care costs among the employees.
Standard Error
The standard error of the mean (SEM) quantifies the precision of the sample mean as an estimate of the true population mean. It tells us how much the sample mean is expected to vary from the actual population mean if we were to take multiple samples. The SEM is calculated by dividing the standard deviation by the square root of the sample size:\[ SEM = \frac{s}{\sqrt{n}} \] In our example, with a sample size \( n = 10 \) and standard deviation \( s = 5.54 \), the SEM is calculated to be approximately 1.75. A smaller SEM indicates that the sample mean is a more accurate reflection of the population mean, suggesting relatively reliable results from our sample.
t-Distribution
When we work with smaller samples, such as 10 in our example, we use the t-distribution instead of the normal distribution to calculate confidence intervals. The t-distribution adjusts for the sample size, ensuring more accuracy when estimating the population mean. It requires knowing the degrees of freedom, calculated as \( n-1 \). With 9 degrees of freedom and a 90% confidence level, we can find the t-critical value using a t-distribution table.For our example, the t-critical value was approximately 1.833. This factor helps adjust the margin of error when constructing the confidence interval. The t-distribution is particularly useful for smaller datasets, offering a safety net by accounting for more variability than the typical normal distribution.

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Most popular questions from this chapter

The attendance at the Savannah Colts minor league baseball game last night was \(400 .\) A random sample of 50 of those in attendance revealed that the mean number of soft drinks consumed per person was 1.86 with a standard deviation of \(0.50 .\) Develop a 99 percent confidence interval for the mean number of soft drinks consumed per person.

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The estimate of the population proportion is to be within plus or minus .05, with a 95 percent level of confidence. The best estimate of the population proportion is . \(15 .\) How large a sample is required?

The Huntington National Bank, like most other large banks, found that using automatic teller machines (ATMs) reduces the cost of routine bank transactions. Huntington installed an ATM in the corporate offices of the Fun Toy Company. The ATM is for the exclusive use of Fun's 605 employees. After several months of operation, a sample of 100 employees revealed the following use of the ATM machine by Fun employees in a month. $$\begin{array}{|cc|}\hline \text { Number of Times } & \\\\\text { ATM Used } & \text { Frequency } \\\\\hline 0 & 25 \\\1 & 30 \\\2 & 20 \\\3 & 10 \\\4 & 10 \\\5 & 5 \\\\\hline\end{array}$$ a. What is the estimate of the proportion of employees who do not use the ATM in a month? b. Develop a 95 percent confidence interval for this estimate. Can Huntington be sure that at least 40 percent of the employees of Fun Toy Company will use the ATM? c. How many transactions does the average Fun employee make per month? d. Develop a 95 percent confidence interval for the mean number of transactions per month. e. Is it possible that the population mean is 0? Explain.

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