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A machine produces parts with lengths that are normally distributed with \(\sigma=0.5 .\) A sample of 10 parts has a mean length of 75.92. a. Find the point estimate for \(\mu\) b. Find the \(98 \%\) confidence maximum error of estimate for \(\mu\) c. Find the \(98 \%\) confidence interval for \(\mu\)

Short Answer

Expert verified
a. The point estimate for \(\mu\) is 75.92. b. The 98% confidence maximum error of estimate for \(\mu\) is approximately 0.37. c. The 98% confidence interval for \(\mu\) is (75.55, 76.29).

Step by step solution

01

Find the Point Estimate for \(\mu\)

The best point estimate for the population mean \(\mu\) is simply the sample mean. Here, it is given as 75.92.
02

Find the Z-Score Corresponding to the Confidence Level

Next, we need to find the Z-score for a 98% confidence level. This can be found in a standard Z-table or using a Z-score calculator. For a 98% confidence level, the Z-score is approximately 2.33.
03

Calculate the Maximum Error of Estimate

The maximum error of estimate (E) is calculated by multiplying the Z-score for the given confidence level by the standard error of the mean. The standard error is calculated by dividing the standard deviation \(\sigma = 0.5\) by the square root of the sample size (n=10). Therefore, \(E = Z * \sigma / \sqrt{n} = 2.33 * 0.5/ \sqrt{10} \approx 0.37\)
04

Calculate the Confidence Interval for \(\mu\)

The confidence interval is found by adding and subtracting the maximum error of estimate from the point estimate. Therefore, the 98% confidence interval for \(\mu\) is \((75.92 - 0.37, 75.92 + 0.37) = (75.55, 76.29)\)

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Key Concepts

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

Point Estimate
In statistics, a point estimate is a single value given as the estimate of a population parameter that is not directly observable. For instance, the average length of a random sample of machine parts is used as the most reasonable guess for the true average length, or population mean, of all parts produced by the machine. Easy to understand, the exercise's point estimate for the population mean \( \mu \) is the sample mean of 75.92. This figure serves as the best guess or predicted value for the population mean from our sample data.

Point estimates are vital because they provide a summary of the vast amounts of data, allowing for a more manageable analysis. However, they are subject to sampling errors as they are based on sample data, not the entire population. Therefore, they are usually accompanied by a margin of error and a confidence interval, to indicate the reliability of the estimate.
Population Mean
The population mean, denoted as \( \mu \), is a measure representing the average of a given characteristic within an entire population. In the context of our exercise, the population mean would be the average length of all parts produced by the machine, not just the sample of 10 parts measured. The population mean is an essential concept in statistics because it provides a central location for the data set.

In practice, calculating the population mean is often impractical or impossible, as it would require measuring every single part produced. Hence, a sample mean, like the 75.92 we have, is used as an approximation of the population mean. To validate the approximation, statisticians use the sample mean to create a confidence interval, which gives a range where the true population mean is likely to lie.
Z-score
A Z-score is a statistical measurement that describes a value's relationship to the mean of a group of values, measured in terms of standard deviations from the mean. The Z-score for a 98% confidence level, found in our exercise, is approximately 2.33. This number tells us that the point estimate is 2.33 standard deviations away from the mean.

When constructing a confidence interval, the Z-score is used to determine the range of values around the point estimate that can capture the population mean with a certain level of confidence. A higher Z-score corresponds to a higher level of confidence, meaning the interval is more likely to contain the population mean. Calculating the right Z-score is critical when you aim to provide a precise and reliable confidence interval.
Standard Error
The standard error is a measure of the dispersion or spread of sample means around the population mean. It is essentially the standard deviation of the sampling distribution of the sample mean. In simpler terms, the standard error helps us understand how much we would expect sample means to fluctuate if we repeated our sampling process.

In our exercise, the standard error is calculated by dividing the standard deviation of the population \( \sigma \) by the square root of the sample size \( n \)—this turns out to be \( 0.5 / \sqrt{10} \). The smaller the standard error, the more precise the sample mean is as a point estimate of the population mean. This is why larger sample sizes, which result in a smaller standard error, are more desirable in statistical sampling.
Normal Distribution
The normal distribution is a continuous probability distribution that is symmetrical on both sides of the mean, so the right side of the distribution mirrors the left. It is also known as the Gaussian distribution, and it's one of the most important probability distributions in statistics because many variables are distributed normally in nature.

The machine parts' lengths in our exercise are said to be normally distributed, which implies a bell-shaped curve where most values cluster around the mean and fewer are found as you move away from it. This assumption of normal distribution is particularly useful because it allows us to use the Z-score in calculating our confidence interval, as Z-scores are based on the standard normal distribution—a special case of the normal distribution with a mean of zero and a standard deviation of one.

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Most popular questions from this chapter

From a population of unknown mean \(\mu\) and a standard deviation \(\sigma=5.0,\) a sample of \(n=100\) is selected and the sample mean 40.6 is found. Compare the concepts of estimation and hypothesis testing by completing the following: a. Determine the \(95 \%\) confidence interval for \(\mu\) b. Complete the hypothesis test involving \(H_{a}: \mu \neq 40\) using the \(p\) -value approach and \(\alpha=0.05\) c. Complete the hypothesis test involving \(H_{a}: \mu \neq 40\) using the classical approach and \(\alpha=0.05\) d. On one sketch of the standard normal curve, locate the interval representing the confidence interval from part a; the \(z \star, p\) -value, and \(\alpha\) from part b; and the \(z \\#\) and critical regions from part c. Describe the relationship between these three separate procedures.

Use a computer and generate 50 random samples, each of size \(n=28,\) from a normal probability distribution with \(\mu=19\) and \(\sigma=4\) a. Calculate the \(z\) * corresponding to each sample mean that would result when testing the null hypothesis \(\mu=18\) b. In regard to the \(p\) -value approach, find the proportion of \(50 z \star\) values that are "more extreme" than the \(z=-1.04\) that occurred in Exercise \(8.201\left(H_{a}: \mu \neq 18\right)\) Explain what this proportion represents. c. In regard to the classical approach, find the critical values for a two- tailed test using \(\alpha=0.01 ;\) find the proportion of \(50 z \star\) values that fall in the noncritical region. Explain what this proportion represents.

Waiting times (in hours) at a popular restaurant are believed to be approximately normally distributed with a variance of 2.25 during busy periods. a. A sample of 20 customers revealed a mean waiting time of 1.52 hours. Construct the \(95 \%\) confidence interval for the population mean. b. Suppose that the mean of 1.52 hours had resulted from a sample of 32 customers. Find the \(95 \%\) confidence interval. c. What effect does a larger sample size have on the confidence interval?

A manufacturing process produces ball bearings with diameters having a normal distribution and a standard deviation of \(\sigma=0.04 \mathrm{cm} .\) Ball bearings that have diameters that are too small or too large are undesirable. To test the null hypothesis that \(\mu=0.50 \mathrm{cm},\) a sample of 25 is randomly selected and the sample mean is found to be 0.51. a. Design null and alternative hypotheses such that rejection of the null hypothesis will imply that the ball bearings are undesirable. b. Using the decision rule established in part a, what is the \(p\) -value for the sample results? c. If the decision rule in part a is used with \(\alpha=0.02\) what is the critical value for the test statistic?

For each of the following pairs of values, state the decision that will occur and why. a. \(\quad p\) -value \(=0.014, \alpha=0.02\) b. \(\quad p\) -value \(=0.118, \alpha=0.05\) c. \(\quad p\) -value \(=0.048, \alpha=0.05\) d. \(\quad p\) -value \(=0.064, \alpha=0.10\)

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