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For a data set obtained from a sample, \(n=81\) and \(\bar{x}=48.25\). It is known that \(\sigma=4.8\). a. What is the point estimate of \(\mu ?\) b. Make a \(95 \%\) confidence interval for \(\mu .\) c. What is the margin of error of estimate for part b?

Short Answer

Expert verified
a. The point estimate of \(\mu\) is 48.25. b. The 95% confidence interval for \(\mu\) is \([47.57, 48.93]\). c. The margin of error is 0.68.

Step by step solution

01

Compute the point estimate of \(\mu\)

The point estimate of the population mean (\(\mu\)) is the same as the mean of the sample. Therefore, \(\mu\) = \(\bar{x}\) = 48.25.
02

Calculate the Confidence Interval

To find the 95% confidence interval for \(\mu\), use the formula \(\bar{x} \pm z*\frac{\sigma}{\sqrt{n}}\), where \(\sigma\) = 4.8, \(n\) = 81, \(\bar{x}\) = 48.25 and \(z\) = 1.96. Substitute these values into the formula to find that the interval is \([48.25 - 1.96 * 4.8/\sqrt{81}, 48.25 + 1.96 * 4.8/\sqrt{81}]\), which simplifies to \([47.57, 48.93]\).
03

Calculate the Margin of Error

The margin of error is the half-width of the confidence interval and is calculated as \( z*\frac{\sigma}{\sqrt{n}}= 1.96 * 4.8/\sqrt{81}= 0.68\).

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Key Concepts

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

Point Estimate
A point estimate is a single value that serves as an estimate for a population parameter, such as the mean or proportion. In the context of our exercise, we are dealing with the estimation of the population mean \( \mu \).
The point estimate for \( \mu \) using a sample is simply the sample mean, represented by \( \bar{x} \). This is because the sample mean is the best estimate of the population mean based on the data available.
In the given exercise, the point estimate equals the sample mean \( \bar{x} = 48.25 \). Remember, it's like taking a snapshot of the sample to make a guess about the whole population.
Margin of Error
The margin of error provides insight into how much the sample mean \( \bar{x} \) is expected to differ from the true population mean \( \mu \). It gives us a range around our point estimate, reflecting the uncertainty associated with the sampling process.
To compute the margin of error, you need the critical value (\( z \)), standard deviation (\( \sigma \)), and sample size (\( n \)). In our example, the critical value for a 95% confidence level is \( z = 1.96 \). The formula for the margin of error is:
\[ z \frac{\sigma}{\sqrt{n}} \]
This comes out to \( 1.96 \times \frac{4.8}{\sqrt{81}} = 0.68 \).
It's important because it defines the distance from the point estimate to the endpoints of the confidence interval.
Statistical Inference
Statistical inference involves making predictions or generalizations about a population based on a sample of data. One of its vital tools is the confidence interval, which helps in understanding how sample statistics approximate the real population parameter.
In our exercise, a 95% confidence interval is calculated to estimate the true population mean \( \mu \). This interval is constructed using the sample mean \( \bar{x} \), the margin of error, and a critical value, providing a range that is likely to contain the population mean.
The confidence interval was calculated as:
\[ \bar{x} \pm z \frac{\sigma}{\sqrt{n}} = [47.57, 48.93] \]
By providing a specific confidence level, it shows that if we were to take many samples and build intervals in the same way, 95% of those intervals would capture the true population mean. This plays a crucial role in decision-making and hypothesis testing.

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Most popular questions from this chapter

A department store manager wants to estimate at a \(90 \%\) confidence level the mean amount spent by all customers at this store. The manager knows that the standard deviation of amounts spent by all customers at this store is \(\$ 31\). What sample size should he choose so that the estimate is within \(\$ 3\) of the population mean?

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