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A random sampling of a company's monthly operating expenses for \(n=36\) months produced a sample mean of \(\$ 5474\) and a standard deviation of \(\$ 764 .\) Find a \(90 \%\) upper confidence bound for the company's mean monthly expenses.

Short Answer

Expert verified
Answer: The 90% upper confidence bound for the company's mean monthly expenses is approximately $5683.54.

Step by step solution

01

Identify given data

We are given the following data: - \(n = 36\) (number of months in the sampling) - \(\overline{X} = \$ 5474\) (sample mean) - \(\sigma = \$ 764\) (standard deviation)
02

Find the Z-score for the 90% confidence level

We are trying to find a \(90\%\) confidence bound, which means that we want to find the Z-score corresponding to the \(90\%\) probability area in the middle of the Z-distribution. Since we are looking only for an upper bound, we need to look for a Z-score corresponding to the \(95\%\) of the area. Using a Z-table or calculator, we find that the Z-score is approximately \(1.645\).
03

Plug the values into the upper confidence bound formula

Now we can plug the values into the formula: $$\mu_{upper} = \overline{X} + Z\frac{\sigma}{\sqrt{n}}$$ Substituting the values: $$\mu_{upper} = 5474 + 1.645\frac{764}{\sqrt{36}}$$
04

Calculate the upper confidence bound

Now we can use a calculator to perform the arithmetic operations: $$\mu_{upper} = 5474 + 1.645\frac{764}{6}$$ $$\mu_{upper} = 5474 + 1.645(127.333)$$ $$\mu_{upper} = 5474 + 209.535$$ Round the final answer to two decimal places: $$\mu_{upper} \approx \$ 5683.54$$ Hence, the \(90\%\) upper confidence bound for the company's mean monthly expenses is \(\$ 5683.54\).

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Key Concepts

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

Upper Confidence Bound
When working with statistical data, an upper confidence bound offers insight into the highest probable value for a population parameter, like the mean. It's especially useful when we want to assert that the true mean is not above this value with a certain level of confidence, here being 90%.

To calculate an upper confidence bound, we need:
  • The sample mean \( \overline{X} \)
  • The standard deviation \( \sigma \)
  • The sample size \( n \)
  • The Z-score corresponding to the desired confidence level
In this example, we calculated the upper bound using the formula:\[ \mu_{upper} = \overline{X} + Z\frac{\sigma}{\sqrt{n}} \]Substituting our known values allows us to determine the maximum expected mean of the company's monthly expenses, which was found to be approximately \( \$5683.54 \). This value says that we are 90% confident the true mean will not exceed this amount.
Sample Mean
The sample mean is a fundamental statistic used to summarize a set of data with a single value, representing the average of that data set. In our context, the sample mean \( \overline{X} \) is \( \$5474 \).

This value serves as an estimate of the population mean, providing us a midpoint to build our confidence interval around. It is calculated by summing all observations and dividing by the number of observations:\[\overline{X} = \frac{\sum X_i}{n} \]Where \( \sum X_i \) is the sum of all sample values, and \( n \) is the total number of values in the sample.

The concept of a sample mean helps us to explore trends and make decisions based on data, such as evaluating expenses or predicting future costs.
Standard Deviation
The standard deviation \( \sigma \) is a measure of how spread out numbers in a data set are. It reflects how much variability or dispersion exists from the average (mean).

In the example, the given standard deviation is \( \\(764 \). This means that the monthly expenses typically vary \( \\)764 \) from the average expense of \( \$5474 \).

A smaller standard deviation would suggest expenses are closely huddled around the mean, while a larger one indicates a wider spread.
  • Calculating the standard deviation helps in understanding data variability
  • It plays a crucial role in forming confidence intervals, as it adjusts the width of those intervals
Using the standard deviation along with the mean and Z-score allows us to determine the confidence bounds, which become critical for making reliable predictions based on sample data.

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Most popular questions from this chapter

The meat department of a local supermarket chain packages ground beef using meat trays of two sizes: one designed to hold approximately 1 pound of meat, and one that holds approximately 3 pounds. A random sample of 35 packages in the smaller meat trays produced weight measurements with an average of 1.01 pounds and a standard deviation of .18 pound. a. Construct a \(99 \%\) confidence interval for the average weight of all packages sold in the smaller meat trays by this supermarket chain. b. What does the phrase "99\% confident" mean? c. Suppose that the quality control department of this supermarket chain intends that the amount of ground beef in the smaller trays should be 1 pound on average. Should the confidence interval in part a concern the quality control department? Explain.

Find a \(99 \%\) lower confidence bound for the binomial proportion \(p\) when a random sample of \(n=400\) trials produced \(x=196\) successes.

Suppose you wish to estimate a binomial parameter \(p\) correct to within \(.04,\) with probability equal to .95. If you suspect that \(p\) is equal to some value between 1 and .3 and you want to be certain that your sample is large enough. how large should \(n\) be?

A quality-control engineer wants to estimate the fraction of defectives in a large lot of printer ink cartridges. From previous experience, he feels that the actual fraction of defectives should be somewhere around .05. How large a sample should he take if he wants to estimate the true fraction to within .01, using a \(95 \%\) confidence interval?

Independent random samples of size \(n_{1}=n_{2}=\) 500 were selected from each of two populations. The mean and standard deviations for the two samples were \(\bar{x}_{1}=125.2, \bar{x}_{2}=123.7, s_{1}=5.6,\) and \(s_{2}=6.8\) a. Find a point estimate for the difference in the two population means. Calculate the margin of error. b. Based on the results in part a, can you conclude that there is a difference in the two population means? Explain.

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