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Given the information, the sampled population is normally distributed, \(n=55, \bar{x}=78.2,\) and \(\sigma=12:\) a. Find the 0.98 confidence interval for \(\mu\) b. Are the assumptions satisfied? Explain.

Short Answer

Expert verified
a. The 0.98 confidence interval for \(\mu\) is [74.04, 82.36] b. Yes, the assumptions are satisfied

Step by step solution

01

Determine the z-Score

From the standard normal distribution table, we can find the z-score corresponding to the given confidence level 0.98. In fact, since the table typically depicts the area to the left, (1+0.98)/2 = 0.99 or 99% should be used, given the symmetricity of the standard normal distribution. The z-score for 0.99 approximately equals 2.33.
02

Calculate Standard Error

Standard error(SE) is calculated by dividing the standard deviation, \(\sigma\), by the square root of the sample size, n. Therefore, \(SE = \sigma / \sqrt{n} = 12 / \sqrt{55} = 1.61 \).
03

Calculate the Confidence Interval

To find the confidence interval we use the formula: \(\bar{x} ± Z*(SE)\) . Substituting the values we get 78.2 ± 2.33*1.61. Therefore, the 0.98 confidence interval for \(\mu\) is [74.04, 82.36]
04

Validate the Assumptions

The assumptions for estimating a confidence interval for a population mean are: 1. Random Sampling: It's not specified in the problem, but usually assumed. 2. Either the population is normally distributed or the sample size is large enough(n>30) to apply Central Limit Theorem. The problem clearly specifies that the population is normally distributed. Hence, the assumptions are satisfied

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Key Concepts

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

Z-Score
A z-score is a statistical measurement that describes a value's relationship to the mean of a group of values. It is expressed in terms of standard deviations from the mean. If a value has a z-score of 0, it is the mean value. A positive z-score indicates the number of standard deviations an element is above the mean, whereas a negative z-score represents the element being below the mean.

For example, in our exercise, a z-score of approximately 2.33 tells us that the confidence interval's upper and lower boundaries are 2.33 standard deviations away from the mean of the sample. This is key for constructing a confidence interval, as it helps to determine how far from the sample mean the population mean \(\mu\) is likely to be with 98% confidence.
Standard Error
Standard error (SE) is a measure of how much discrepancy or 'error' to expect between a sample statistic, like the sample mean \(\bar{x}\), and the actual population parameter, such as the population mean \(\mu\).

The formula for SE is \(\sigma / \sqrt{n}\), where \(\sigma\) is the population standard deviation and \(n\) is the sample size. The smaller the standard error, the more representative the sample mean is of the population mean. In the problem, we've calculated a standard error of 1.61. This figure is critical in determining the precision of our estimate for the population mean when we are calculating the confidence interval.
Normal Distribution
Normal distribution, also known as the Gaussian distribution, is a probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean. It has a bell-shaped curve and is defined by its mean and standard deviation. The total area under the curve is 1.

In the context of our exercise, the assumption that the sampled population is normally distributed permits the use of z-scores for calculating the confidence interval. It also ensures that we're applying the z-score correctly because z-scores are based on the standard normal distribution, a special case of the normal distribution where the mean is 0 and the standard deviation is 1.
Central Limit Theorem
The Central Limit Theorem (CLT) states that the sampling distribution of the sample mean will be approximately normally distributed, regardless of the shape of the population distribution, provided the sample size is sufficiently large (usually n>30 is considered sufficient). As the sample size increases, the shape of the distribution of the sample means will approach a normal distribution.

Given that our sample size in this exercise is more than 30, even if the population were not normally distributed, we could still apply the CLT to justify the use of normal distribution methods for estimating the population mean. This theorem is foundational for the field of inferential statistics and allows for the estimation of confidence intervals even when little is known about the actual distribution of the population.
Population Mean Estimation
Estimating the population mean involves using sample data to make an informed guess about the mean value of the entire population. This is often done through inference methods such as confidence intervals, which provide a range of values within which the population mean is likely to fall.

The confidence interval calculated in our exercise, [74.04, 82.36], represents the range where we can be 98% confident that the true population mean lies. This interval is created using the sample mean, standard error, and the appropriate z-score for our confidence level, demonstrating how these concepts come together to inform our estimation of the population mean.

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Most popular questions from this chapter

Use a computer or calculator to select 36 random numbers from a normal distribution with mean 100 and standard deviation \(15 .\) Find the sample mean and \(z *\) for testing a two-tailed hypothesis test of \(\mu=100 .\) Using \(\alpha=0.05,\) state the decision. Repeat several times as in Table \(8.12 .\) Describe your findings.

Jack Williams is vice president of marketing for one of the largest natural gas companies in the nation. During the past 4 years, he has watched two major factors erode the profits and sales of the company. First, the average price of crude oil has been virtually flat, and many of his industrial customers are burning heavy oil rather than natural gas to fire their furnaces, regardless of added smokestack emissions. Second, both residential and commercial customers are still pursuing energy-conservation techniques (e.g., adding extra insulation, installing clockdrive thermostats, and sealing cracks around doors and windows to eliminate cold air infiltration). In previous years, residential customers bought an average of 129.2 mcf of natural gas from Jack's company \((\sigma=18 \mathrm{mcf})\) based on internal company billing records, but environmentalists have claimed that conservation is cutting fuel consumption up to \(3 \%\) per year. Jack has commissioned you to conduct a spot check to see if any change in annual usage has transpired before his next meeting with the officers of the corporation. A sample of 300 customers selected randomly from the billing records reveals an average of \(127.1 \mathrm{mcf}\) during the past 12 months. Is there a significant decline in consumption? a. Complete the appropriate hypothesis test at the 0.01 level of significance using the \(p\) -value approach so that you can properly advise Jack before his meeting. b. Because you are Jack's assistant, why is it best for you to use the \(p\) -value approach?

a. If \(\alpha\) is assigned the value \(0.001,\) what are we saying about the type I error? b. If \(\alpha\) is assigned the value \(0.05,\) what are we saying about the type I error? c. If \(\alpha\) is assigned the value \(0.10,\) what are we saying about the type I error?

Calculate the \(p\) -value, given \(H_{a}: \mu \neq 245\) and \(z \star=1.1\)

Ponemon Institute, along with Intel, published "The cost of a Lost Laptop" study in April 2009. With an increasingly mobile workforce carrying around more sensitive data on their laptops, the loss involves much more than the laptop itself. The average cost of a lost laptop based on cases from various industries is \(\$ 49,246 .\) This figure includes laptop replacement, data breach cost, lost productivity cost, and other legal and forensic costs. A separate study conducted with respect to 30 cases from health care industries produced a mean of \(\$ 67,873 .\) Assuming that \(\sigma=\$ 25,000,\) is there sufficient evidence to support the claim that health care laptop replacement costs are higher in general? Use a 0.001 level of significance.

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