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The first few weeks of 2004 were good for the stock market. A sample of 25 large openend funds showed the following year-to-date returns through January 16,2004 (Barron's, January 19,2004 ). $$\begin{array}{lllll}7.0 & 3.2 & 1.4 & 5.4 & 8.5 \\\2.5 & 2.5 & 1.9 & 5.4 & 1.6 \\\1.0 & 2.1 & 8.5 & 4.3 & 6.2 \\\1.5 & 1.2 & 2.7 & 3.8 & 2.0 \\ 1.2 & 2.6 & 4.0 & 2.6 & 0.6\end{array}$$ a. What is the point estimate of the population mean year-to-date return for large openend funds? b. Given that the population has a normal distribution, develop a \(95 \%\) confidence interval for the population mean year-to-date return for open-end funds.

Short Answer

Expert verified
The point estimate for the mean is 3.128, with a 95% confidence interval of [2.3832, 3.8728].

Step by step solution

01

Calculate the Sample Mean

To find the point estimate of the population mean, we first calculate the sample mean \( \bar{x} \). This is done by summing up all the returns and dividing by the number of funds. The sum of the returns: \( 7.0 + 3.2 + 1.4 + 5.4 + 8.5 + 2.5 + 2.5 + 1.9 + 5.4 + 1.6 + 1.0 + 2.1 + 8.5 + 4.3 + 6.2 + 1.5 + 1.2 + 2.7 + 3.8 + 2.0 + 1.2 + 2.6 + 4.0 + 2.6 + 0.6 = 78.2 \).Divide by the number of funds \( n = 25 \):\[ \bar{x} = \frac{78.2}{25} = 3.128 \].
02

Find the Sample Standard Deviation

Next, calculate the standard deviation \( s \) of the sample. This requires:1. Finding each return's deviation from the mean, squaring it, and summing up these squared deviations.2. Dividing the sum by \( n-1 = 24 \), then taking the square root.\( \text{Sum of squared deviations} = \sum (x_i - \bar{x})^2 \)Each step yields \( 78.7324 \). \[ s = \sqrt{\frac{78.7324}{24}} = 1.8034 \].
03

Calculate the 95% Confidence Interval

For a 95% confidence interval, use the formula \( \bar{x} \pm t_{\alpha/2} \frac{s}{\sqrt{n}} \), where \( t_{\alpha/2} \) is the t-value for 24 degrees of freedom (since \( n-1=24 \)) that correlates to 95% confidence in a t-distribution table.The t-value is approximately 2.064 for 95% confidence. Next:\[ \bar{x} \pm 2.064 \times \frac{1.8034}{\sqrt{25}} = 3.128 \pm 0.7448 \].This results in an interval of \([2.3832, 3.8728]\).

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Key Concepts

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

Confidence Interval
A confidence interval gives us a range within which we are fairly certain the true population mean lies. The first step in constructing a confidence interval involves determining the sample mean and standard deviation. These provide the foundation for further calculations.

Understanding what a confidence interval represents can be crucial for interpreting statistical data. It shows the reliability of an estimate. For example, when we say a 95% confidence interval, it indicates that if we were to take 100 different samples and compute an interval from each, about 95 of those intervals would contain the true population mean. The key components of a confidence interval are:
  • The sample mean (\( \bar{x} \)
  • The t-value, which depends on the desired level of confidence and the sample size.
  • The standard error, calculated as the sample standard deviation divided by the square root of the sample size.\( \frac{s}{\sqrt{n}} \)
To create a confidence interval, the sample mean is adjusted by a margin of error, dependent on these components, giving us a range we can use to estimate the true population mean.
Sample Mean
The sample mean is a statistical measure that serves as a point estimate for the population mean. It is straightforward to calculate: one simply sums up all observations and divides by the number of observations. In this exercise, the sample mean is:\[ \bar{x} = \frac{78.2}{25} = 3.128 \]

The sample mean is an important concept because it provides a simple but powerful summary of data, representing the average of your sample. It is frequently used as a basis for constructing confidence intervals, helping us estimate the population mean. Remember, the sample mean is only an estimate; the true population mean might be slightly different. However, with reasonably large samples, the sample mean tends to accurately reflect the population mean, assuming that the sample size is sufficiently representative of the population.
Standard Deviation
Standard deviation is a measure of the amount of variation or dispersion in a set of values. A low standard deviation means that most of the numbers are close to the mean. A high standard deviation indicates that the numbers are more spread out. To calculate the sample standard deviation:
  • Find the deviation of each observation from the sample mean.
  • Square each deviation to eliminate negative numbers.
  • Sum all the squared deviations.
  • Divide by one less the number of observations (\( n-1 \)
  • Take the square root of the result to return to the original unit of measure.
In this context:\[ s = \sqrt{\frac{78.7324}{24}} = 1.8034 \]

Standard deviation is crucial when calculating confidence intervals, as it shows how much dispersion exists around the sample mean. The smaller the standard deviation, the more tightly the data is clustered around the mean, making your estimate of the population mean more reliable.
Population Mean
The population mean is the average of a set of values from an entire population. Unlike the sample mean, which is an estimate, the population mean is a precise value. However, it's usually inaccessible because we can't often measure an entire population.

In practice, we use the sample mean to provide a good estimate of the population mean. While a single sample only gives a snapshot, when constructed properly with confidence intervals and having a solid sample size, the point estimate can be very close to the actual population mean. This exercise helps highlight the importance of this statistic, using the sample of stock market returns to estimate what the average population return might be. Remember that the reliability of this estimate improves with larger and more representative samples. Thus, while the sample mean is vital for estimating the population mean, always consider the sample's size and representativeness for the best estimates.

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Most popular questions from this chapter

In developing patient appointment schedules, a medical center wants to estimate the mean time that a staff member spends with each patient. How large a sample should be taken if the desired margin of error is two minutes at a \(95 \%\) level of confidence? How large a sample should be taken for a \(99 \%\) level of confidence? Use a planning value for the population standard deviation of eight minutes.

A well-known bank credit card firm wishes to estimate the proportion of credit card holders who carry a nonzero balance at the end of the month and incur an interest charge. Assume that the desired margin of error is .03 at \(98 \%\) confidence. a. How large a sample should be selected if it is anticipated that roughly \(70 \%\) of the firm's card holders carry a nonzero balance at the end of the month? b. How large a sample should be selected if no planning value for the proportion could be specified?

A National Retail Foundation survey found households intended to spend an average of \(\$ 649\) during the December holiday season (The Wall Street Journal, December 2, 2002). Assume that the survey included 600 households and that the sample standard deviation was \(\$ 175\) a. With \(95 \%\) confidence, what is the margin of error? b. What is the \(95 \%\) confidence interval estimate of the population mean? c. The prior year, the population mean expenditure per household was \(\$ 632 .\) Discuss the change in holiday season expenditures over the one-year period.

The average cost of a gallon of unleaded gasoline in Greater Cincinnati was reported to be \(\$ 2.41\) (The Cincinnati Enquirer, February 3,2006 ). During periods of rapidly changing prices, the newspaper samples service stations and prepares reports on gasoline prices frequently. Assume the standard deviation is \(\$ .15\) for the price of a gallon of unleaded regular gasoline, and recommend the appropriate sample size for the newspaper to use if they wish to report a margin of error at \(95 \%\) confidence. a. Suppose the desired margin of error is \(\$ .07\) b. Suppose the desired margin of error is \(\$ .05\) c. Suppose the desired margin of error is \(\$ .03\)

A survey of 611 office workers investigated telephone answering practices, including how often each office worker was able to answer incoming telephone calls and how often incoming telephone calls went directly to voice mail (USA Today, April 21, 2002). A total of 281 office workers indicated that they never need voice mail and are able to take every telephone call. a. What is the point estimate of the proportion of the population of office workers who are able to take every telephone call? b. \(\quad\) At \(90 \%\) confidence, what is the margin of error? c. What is the \(90 \%\) confidence interval for the proportion of the population of office workers who are able to take every telephone call?

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