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The standard deviation for a population is \(\sigma=15.3\). A sample of 36 observations selected from this population gave a mean equal to \(74.8\). a. Make a \(90 \%\) confidence interval for \(\mu\). b. Construct a \(95 \%\) confidence interval for \(\mu\). c. Determine a \(99 \%\) confidence interval for \(\mu\). d. Does the width of the confidence intervals constructed in parts a through \(\mathrm{c}\) increase as the confidence level increases? Explain your answer.

Short Answer

Expert verified
a) The 90% confidence interval for \(\mu\) is [70.608, 78.992],\nb) The 95% confidence interval for \(\mu\) is [69.805, 79.795],\nc) The 99% confidence interval for \(\mu\) is [68.231, 81.369],\nd) Yes, the width of the confidence intervals increases as the confidence level increases.

Step by step solution

01

Calculate the Standard Error

The standard error for the mean (SE) is given by \(\sigma /\sqrt{n}\), so, it is calculated as \(15.3/ \sqrt{36}=2.55\).
02

Find Z-Scores

For the confidence levels \(90\%, 95\%, 99%\), the associated z-scores that we get from the standard normal distribution table are \(1.645, 1.96, 2.576\) respectively.
03

Calculate the Confidence Intervals

A confidence interval is constructed as \(\overline{x} \pm \) (z-score * SE). Thus}\na. The 90% confidence interval for \(\mu\) is \(74.8 \pm 1.645 * 2.55\) or \(74.8 \pm 4.192\), thus the interval is \([70.608, 78.992]\).\nb. The 95% confidence interval for \(\mu\) is \(74.8 \pm 1.96 * 2.55 = 74.8 \pm 4.995\), thus the interval is \([69.805, 79.795]\).\nc. The 99% confidence interval for \(\mu\) is \(74.8 \pm 2.576 * 2.55 = 74.8 \pm 6.569\), thus the interval is \([68.231, 81.369]\).
04

Analyze the Results

As the level of confidence increases, we can see that the interval width also increases. This is because as we want to be more certain that we capture the population mean, we need a broader interval. So, the answer to part d is YES.

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Key Concepts

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

Understanding Standard Deviation
Standard deviation, often symbolized as \( \sigma \) for a population, is a key concept in statistics. It measures the spread or dispersion of a set of values. A low standard deviation means most of the numbers are close to the mean, while a high standard deviation indicates the numbers are more spread out. In our exercise, the population standard deviation is \(15.3\).
To visualize this, imagine a set of data points plotted on a graph. If they cluster closely around the average (the mean), the standard deviation is small. Conversely, if they widely vary and spread from the mean, the standard deviation is larger. A practical use of the standard deviation is in constructing confidence intervals, which are intervals that estimate the range where a population parameter lies with a certain probability. It helps quantify the amount of variation or dispersion in a set of data values.
Decoding Standard Error
The standard error (SE) of the mean is an indicator of how much variability we can expect in the sample mean compared to the population mean. It’s calculated using the formula \( \sigma / \sqrt{n} \), where \( \sigma \) is the standard deviation of the population and \( n \) is the sample size. In the exercise example, with a \( \sigma \) of 15.3 and a sample size of 36, the standard error is \( 2.55 \).
The smaller the standard error, the more representative the sample mean is of the population mean. It essentially tells us how accurate our sample mean is likely to be. This statistic becomes crucial when constructing confidence intervals because it determines how "wide" or "narrow" these intervals will be.
  • A smaller standard error indicates that the sample mean is more likely to be close to the population mean.
  • A larger sample size results in a smaller standard error, given the same population standard deviation, reflecting more confidence in the sample's representation of the population.
Exploring Z-Scores
Z-scores are a way to express data points relative to the mean of a group of values, measured in standard deviations. It tells us how many standard deviations an element is from the mean. In the context of confidence intervals, z-scores are used to determine how far from the mean we should look when creating an interval.
For different confidence levels, we have different z-scores:
  • For a 90% confidence interval, the z-score is \( 1.645 \).
  • For a 95% confidence interval, the z-score is \( 1.96 \).
  • For a 99% confidence interval, the z-score is \( 2.576 \).
The larger the z-score, the more coverage we require to account for more certainty in capturing the true population mean. As a result, a higher confidence level results in a wider interval. Understanding z-scores helps to predict where data points will fall, assuming a normal distribution.
Grasping Population Mean
In statistics, the population mean is the average of all individual measurements in a population. It is a critical measure because it gives a single value that’s representative of the entire group. In our example, we seek to estimate the population mean using the sample mean.
The sample mean of 74.8 was taken from a group of 36 observations. To make inferences about the population mean, statisticians create confidence intervals using this sample mean. These intervals provide a range wherein the population mean is likely to fall, with a specified level of certainty (e.g., 90%, 95%, or 99%). Understanding the population mean helps in assessing whether an intervention or change has a significant effect when comparing to another group or a target value. It's foundational to hypothesis testing and effective decision making in research and practical applications.

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Most popular questions from this chapter

A random sample of 36 mid-sized cars tested for fuel consumption gave a mean of \(26.4\) miles per gallon with a standard deviation of \(2.3\) miles per gallon. a. Find a \(99 \%\) confidence interval for the population mean, \(\mu\). b. Suppose the confidence interval obtained in part a is too wide. How can the width of this interval be reduced? Describe all possible alternatives. Which alternative is the best and why?

You are interested in estimating the mean commuting time from home to school for all commuter students at your school. Briefly explain the procedure you will follow to conduct this study. Collect the required data from a sample of 30 or more such students and then estimate the population mean at a \(99 \%\) confidence level. Assume that the population standard deviation for such times is \(5.5\) minutes.

You are working for a supermarket. The manager has asked you to estimate the mean time taken by a cashier to serve customers at this supermarket. Briefly explain how you will conduct this study. Collect data on the time taken by any supermarket cashier to serve 40 customers. Then estimate the population mean. Choose your own confidence level.

A marketing researcher wants to find a \(95 \%\) confidence interval for the mean amount that visitors to a theme park spend per person per day. She knows that the standard deviation of the amounts spent per person per day by all visitors to this park is \(\$ 11\). How large a sample should the researcher select so that the estimate will be within \(\$ 2\) of the population mean?

The express check-out lanes at Wally's Supermarket are limited to customers purchasing 12 or fewer items. Cashiers at this supermarket have complained that many customers who use the express lanes have more than 12 items. A recently taken random sample of 200 customers entering express lanes at this supermarket found that 74 of them had more than 12 items. a. Construct a \(98 \%\) confidence interval for the percentage of all customers at this supermarket who enter express lanes with more than 12 items. b. Suppose the confidence interval obtained in part a is too wide. How can the width of this interval be reduced? Discuss all possible alternatives. Which alternative is the best?

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