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A research firm conducted a survey of 49 randomly selected Americans to determine the mean amount spent on coffee during a week. The sample mean was 20 per week. The population distribution is normal with a standard deviation of 5 a. What is the point estimate of the population mean? Explain what it indicates. b. Using the 95% level of confidence, determine the confidence interval for mu . Explain what it indicates.

Short Answer

Expert verified
a) The point estimate is $20. b) The 95% confidence interval is $(18.6, 21.4).$

Step by step solution

01

Identify the Point Estimate

The point estimate of the population mean is the sample mean. For this exercise, the sample mean is given as 20. The point estimate indicates the best single estimate of the population mean for the amount spent on coffee.
02

Determine the Standard Error

The standard error (SE) of the mean is calculated using the formula \( SE = \frac{\sigma}{\sqrt{n}} \), where \( \sigma \) is the population standard deviation, and \( n \) is the sample size. Here, \( \sigma = 5 \) and \( n = 49 \), so \( SE = \frac{5}{\sqrt{49}} = \frac{5}{7} \approx 0.7143 \).
03

Find the Z-score for 95% Confidence Level

A 95% confidence level means the Z-score is approximately 1.96, which signifies the range within which we expect the true population mean to fall with 95% certainty.
04

Calculate the Confidence Interval

The confidence interval is calculated using the formula \( \bar{x} \pm Z \times SE \), where \( \bar{x} \) is the sample mean, \( Z \) is the Z-score, and \( SE \) is the standard error. Substituting, we have \( 20 \pm 1.96 \times 0.7143 \). This calculates to \( 20 \pm 1.4 \), resulting in an interval of \( (18.6, 21.4) \).
05

Interpret the Confidence Interval

The interval \( (18.6, 21.4) \) means that we are 95% confident that the true population mean amount spent on coffee per week falls between \(18.60 and \)21.40.

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Key Concepts

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

Point Estimate
In statistics, the point estimate is a single value that serves as an estimate of a population parameter. It provides the best guess or approximation of the unknown parameter based on the sample data we have collected.

For the given problem, the point estimate of the population mean for the amount spent on coffee is simply the sample mean. In this exercise, we found the sample mean to be \(20\). This means that \(20\) is our best single estimate of the average amount spent on coffee by the entire population of Americans in a week.

The point estimate is crucial because it gives researchers a starting point to understand the larger population. However, keep in mind that while a point estimate is helpful, it may not always capture the true population mean precisely, prompting the need for additional statistical tools like confidence intervals.
Standard Error
The standard error (SE) is a measure of how much the sample mean is expected to vary from the true population mean. It essentially shows the amount of variation or "error" you might expect when using a sample's results to estimate a population parameter.

To calculate the standard error in our scenario, we use the formula: \[SE = \frac{\sigma}{\sqrt{n}}\]Here, \(\sigma\) is the population standard deviation, which is \(5\), and \(n\) is the sample size, \(49\). Substituting these values, we get:\[SE = \frac{5}{\sqrt{49}} = \frac{5}{7} \approx 0.7143\]

This result means that the sample mean is expected to vary by approximately \(0.7143\) units from the true population mean. A smaller standard error indicates more precision in our estimate of the population mean.
95% Confidence Level
The 95% confidence level is a common benchmark in statistics. It means that if we were to repeat the study multiple times, about 95% of the confidence intervals calculated from those samples would contain the true population parameter. This confidence level indicates how certain we are about our interval estimates.

In our exercise, we use a Z-score of approximately \(1.96\), which corresponds to a 95% confidence level. This Z-score represents the number of standard errors away from the mean that captures the central 95% of the normal distribution.

When you see a 95% confidence interval, you can interpret it as there being a 95% chance that the interval calculated from the sample encompasses the true population mean. It's a statistical way of expressing our certainty about how close the sample mean is to the true mean.
Z-score
The Z-score is a statistical measurement that describes a value's position in relation to the mean of a group of values. It tells us how many standard deviations an element is from the mean.

In the context of confidence intervals, the Z-score helps us determine the range within which the true population mean is likely to fall. For a 95% confidence level, the Z-score is generally about \(1.96\). This Z-score indicates we can be 95% confident that the population mean falls within this range of our sample mean.

With our data, the confidence interval calculation involves multiplying the standard error by the Z-score (\(1.96\)) to determine how wide the margin around our point estimate should be:\[20 \pm 1.96 \times 0.7143\]This resulted in a confidence interval of \( (18.6, 21.4) \). Understanding the Z-score is essential for constructing confidence intervals and helps convey the level of certainty or uncertainty we have in our sample estimates.

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Most popular questions from this chapter

Pharmaceutical companies promote their prescription drugs using television advertising. In a survey of 80 randomly sampled television viewers, 10 indicated that they asked their physician about using a prescription drug they saw advertised on TV. Develop a \(95 \%\) confidence interval for the proportion of viewers who discussed a drug seen on TV with their physician. Is it reasonable to conclude that \(25 \%\) of the viewers discuss an advertised drug with their physician?

A sample of 81 observations is taken from a normal population with a standard deviation of \(5 .\) The sample mean is \(40 .\) Determine the \(95 \%\) confidence interval for the population mean.

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A recent survey of 50 executives who were laid off during a recent recession revealed it took a mean of 26 weeks for them to find another position. The standard deviation of the sample was 6.2 weeks. Construct a \(95 \%\) confidence interval for the population mean. Is it reasonable that the population mean is 28 weeks? Justify your answer.

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