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Chapter 8: Problem 32

A department store manager wants to estimate at a \(98 \%\) 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?

Short Answer

Expert verified
The department store manager should survey a sample size of 191 customers.

Step by step solution

01

Identify Known Values

The manager knows that the standard deviation (\(\sigma\)) of amounts spent is \$31, the maximum acceptable error (E) is \$3 and the confidence level is \(98\%\).
02

Determine the Z-score

The z-score corresponding to a \(98\%\) confidence level (which leaves \(2\%\) split between two tails of the distribution) is \(2.33\) (This value can be found in a z-table or obtained from statistical software).
03

Substitute values into Sample Size Formula

Substitute the known values of \(Z_{\alpha/2} = 2.33\), \(\sigma = 31\), and \(E = 3\) into the sample size formula: \(n = \left(\frac{2.33 \cdot 31}{3}\right)^2\).
04

Calculate and Round Up

Conduct the calculation to find \(n \approx 190.77\). Because we can't have a fraction of a customer, we round up to get the next whole number. So, a sample size of 191 customers is needed.

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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 expect the true population parameter to lie, with a certain level of certainty. In this case, the department store manager wants a 98% confidence interval. This means that if the same population is sampled multiple times, approximately 98% of those intervals will contain the population mean. A greater confidence level means that you want more certainty that your interval will encompass the true population mean. As the confidence level increases, the interval becomes wider because you're allowing for more possible variation. But this also means you might need a larger sample size to achieve the precision you desire.
Standard Deviation
The standard deviation is a measure of how spread out the numbers are in a data set. In other words, it shows us the amount of variation or dispersion from the average. In our exercise, the standard deviation of amounts spent by all customers is given as $31.
This value represents the average deviation of each amount from the mean. It's essential for calculating the sample size because it directly influences the width of the confidence interval.
  • If the standard deviation is high, the data points are spread out widely.
  • If it's low, the data points are closer to the mean.
Understanding standard deviation helps us grasp the stability or variability of what is being measured - in this case, customer spending.
Z-score
A z-score tells us how many standard deviations an element is from the mean. For confidence intervals, it helps in determining the critical value needed to calculate the interval itself. In this case, a confidence level of 98% corresponds to a z-score of 2.33.
This value is found using a z-table or specialized statistical software. The z-score indicates how unlikely we want our sample mean to be when compared to the population mean. When the z-score is set higher, it represents a higher level of confidence. This means we require our sample mean to be more concentrated around the true mean, thus affecting the needed sample size.
Population Mean
The population mean is the average of all values in the population. It is the ultimate goal of the estimation in our exercise. The department store manager wants to find something very close to this true average amount spent by all customers.
Directly calculating this mean would require gathering data from every customer, which is often impractical. Instead, a sample mean, calculated from just a part of the entire population, serves as an estimate.
  • By increasing the sample size, the estimate of the population mean becomes more accurate.
  • A large enough sample size reduces the margin of error, bringing the sample mean closer to the population mean.
The manager's primary goal is to ensure this estimation is precise within $3 of the true mean, guiding decisions with reliable data.

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Most popular questions from this chapter

An economist wants to find a \(90 \%\) confidence interval for the mean sale price of houses in a state. How large a sample should she select so that the estimate is within \(\$ 3500\) of the population mean? Assume that the standard deviation for the sale prices of all houses in this state is \(\$ 31,500\).

A drug that provides relief from headaches was tried on 18 randomly selected patients. The experiment showed that the mean time to get relief from headaches for these patients after taking this drug was 24 minutes with a standard deviation of \(4.5\) minutes. Assuming that the time taken to get relief from a headache after taking this drug is (approximately) normally distributed, determine a \(95 \%\) confidence interval for the mean relief time for this drug for all patients.

A local gasoline dealership in a small town wants to estimate the average amount of gasoline that people in that town use in a 1-week period. The dealer asked 44 randomly selected customers to keep a diary of their gasoline usage, and this information produced the following data on gas used (in gallons) by these people during a 1-week period. The population standard deviation is not known. \(\begin{array}{rrrrrrrrrrr}23.1 & 13.6 & 25.8 & 10.0 & 7.6 & 18.9 & 26.6 & 23.8 & 12.3 & 15.8 & 21.0 \\ 26.9 & 22.9 & 18.3 & 23.5 & 21.6 & 15.5 & 23.5 & 11.8 & 15.3 & 11.9 & 19.2 \\ 14.5 & 9.6 & 12.1 & 18.0 & 20.6 & 14.2 & 7.1 & 13.2 & 5.3 & 13.1 & 10.9 \\ 10.5 & 5.1 & 5.2 & 6.5 & 8.3 & 10.5 & 7.4 & 7.4 & 5.3 & 10.6 & 13.0\end{array}\) Construct a \(95 \%\) confidence interval for the average weekly gas usage by people in this town.

You are interested in estimating the mean age of cars owned by all people in the United States. Briefly explain the procedure you will follow to conduct this study. Collect the required data on a sample of 30 or more cars and then estimate the population mean at a \(95 \%\) confidence level. Assume that the population standard deviation is \(2.4\) years.

Check if the sample size is large enough to use the normal distribution to make a confidence interval for \(p\) for each of the following cases. a. \(n=80\) and \(\hat{p}=.85\) b. \(n=110 \quad\) and \(\quad \hat{p}=.98\) c. \(n=35\) and \(\hat{p}=.40\) d. \(n=200\) and \(\hat{p}=.08\)
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