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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 $5689.18.

Step by step solution

01

Calculate the t-score for the given confidence level

First, we need to find the t-score corresponding to the desired confidence level (90%). The degrees of freedom (\(df\)) in this case is \(n-1 = 36 - 1 = 35\). Using a t-distribution table or calculator, look up the t-score with cumulative probability of 0.95 (since we want a 90% upper confidence bound), for 35 degrees of freedom, which is approximately 1.69.
02

Calculate the standard error of the sample mean

Next, we need to calculate the standard error of the sample mean. The standard error (\(SE\)) is the standard deviation (\(s\)) divided by the square root of the sample size (\(n\)). Using the given sample standard deviation (\(s = \$ 764\)) and sample size (\(n=36\)), the standard error is: \(SE = \frac{s}{\sqrt{n}} = \frac{764}{\sqrt{36}} = 764 / 6 \approx \$ 127.33\).
03

Calculate the margin of error

Now, we can calculate the margin of error (\(ME\)) using the t-score obtained in step 1 and the standard error calculated in step 2. The margin of error is: \(ME = t \cdot SE = 1.69 \cdot 127.33 \approx \$ 215.18\).
04

Find the upper confidence bound

Finally, we can find the 90% upper confidence bound by adding the margin of error to the sample mean. The upper confidence bound is: \(\bar{x} + ME = 5474 + 215.18 \approx \$ 5689.18\). So, the 90% upper confidence bound for the company's mean monthly expenses is approximately $5689.18.

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Key Concepts

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

Understanding the T-Distribution
The t-distribution is pivotal in the field of statistics, especially when managing small sample sizes and when the population standard deviation is unknown. It resembles the normal distribution but has heavier tails, which means it is more prone to producing values that fall far away from the mean. This property allows it to accommodate the added uncertainty that comes with smaller sample sizes.

The shape of the t-distribution varies according to the number of degrees of freedom (df), which correlates with the sample size. In our example, with 35 degrees of freedom, the t-distribution curve would be slightly more spread out than a normal distribution curve, reflecting greater variability. As the sample size increases, the t-distribution approaches the normal distribution. This convergence is essential when we calculate confidence intervals, using the t-distribution to estimate the range within which the true population mean is likely to lie with a certain level of confidence.
Decomposing Standard Error
The term 'standard error' (SE) reflects the variability or error present in an estimation process. Specifically, in our example, it measures the extent of variation one can expect in sample means across different samples from the same population. The formula for SE is essentially the sample standard deviation (s) divided by the square root of the sample size (n). This calculation adjusts for the size of the sample—the larger the sample, the smaller the error, due to the principle of the law of large numbers.

For instance, with a standard deviation of \(764 and a sample size of 36, we arrive at a standard error of approximately \)127.33. This figure tells us that if we took multiple samples of 36 months and calculated each of their means, the standard error is the average amount those means would vary from the true population mean. Understanding SE is crucial because it's directly proportional to the margin of error; a larger SE implies a wider confidence interval.
Degrees of Freedom Explained
Degrees of freedom (df) is a concept that surfaces often in statistical analyses, particularly those involving sample data and estimations. It refers to the number of independent values or quantities that can vary in an analysis without breaking any constraints. In the context of the t-distribution, the degrees of freedom equals the sample size minus one (df = n - 1).

In our exercise, the sample size is 36, which leaves us with 35 degrees of freedom. This value is critical when determining the exact shape of the t-distribution we need to use to calculate the confidence interval. Degrees of freedom account for the sample size and are a corrective measure to the standard error calculation, ensuring that the t-distribution appropriately reflects the sample's information and the accuracy of our estimations improves.
Margin of Error's Role
The margin of error (ME) represents a range above and below the sample mean within which we can expect the true population mean to lie, considering a specific confidence level. In simpler terms, ME gives us a cushion that reflects the expected amount of sampling error.

Note that ME increases with a higher standard error or a higher t-value (which can reflect a higher confidence level or less certain data). For our company's monthly operating expenses, given the standard error and the t-score, we calculated a margin of error of approximately \(215.18. This value is then used to establish an upper confidence bound, hinting that the company's actual mean monthly expenses would not exceed \)5689.18, with 90% confidence. Such information is invaluable for financial planning and uncertainty quantification in business processes.

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Most popular questions from this chapter

You want to estimate the difference in grade point averages between two groups of college students accurate to within .2 grade point, with probability approximately equal to \(.95 .\) If the standard deviation of the grade point measurements is approximately equal to .6, how many students must be included in each group? (Assume that the groups will be of equal size.)

Auto Accidents A recent year's records of auto accidents occurring on a given section of highway were classified according to whether the resulting damage was \(\$ 3000\) or more and to whether a physical injury resulted from the accident. The data follows: $$\begin{array}{lcc}\hline & \text { Under } \$ 3000 & \$ 3000 \text { or More } \\\\\hline \text { Number of Accidents } & 32 & 41 \\\\\text { Number Involving Injuries } & 10 & 23\end{array}$$ a. Estimate the true proportion of accidents involving injuries when the damage was \(\$ 3000\) or more for similar sections of highway and find the margin of error. b. Estimate the true difference in the proportion of accidents involving injuries for accidents with damage under \(\$ 3000\) and those with damage of \(\$ 3000\) or more. Use a \(95 \%\) confidence interval.

Independent random samples were selected from two binomial populations, with sample sizes and the number of successes given in Exercises \(5-6 .\) Find the best point estimate for the difference in the population proportion of successes and calculate the margin of error: $$n_{1}=60, n_{2}=60, x_{1}=43, x_{2}=36$$

Refer to Exercise 18 (Section 8.5). The percentage of people catching a cold when exposed to a cold virus is shown in the table, for a group of people with only a few social contacts and a group with six or more activities. \(^{21}\) $$\begin{array}{lcc}\hline & \text { Few Social Outlets } & \text { Many Social Outlets } \\\\\hline \text { Sample Size } & 96 & 105 \\\\\text { Percent with Colds } & 62 \% & 35 \% \\\\\hline\end{array}$$ Construct a \(95 \%\) lower confidence bound for the difference in population proportions. Does it appear that when exposed to a cold virus, a greater proportion of those with fewer social contacts contracted a cold? Is this counter-intuitive?

Use the information given in Exercises 9-15 to find the necessary confidence interval for the population mean \(\mu .\) Interpret the interval that you have constructed. A \(90 \%\) confidence interval, \(n=50, \bar{x}=21.9\), \(s^{2}=3.44\)

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