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Chapter 9: Problem 10

An efficiency expert wishes to determine the average time that it takes to drill three holes in a certain metal clamp. How large a sample will he need to be \(95 \%\) confident that his sample mean will be within 15 seconds of the true mean? Assume that it is known from previous studies that \(\sigma=40\) seconds.

Short Answer

Expert verified
The sample size needed for 95% confidence that the sample mean will be within 15 seconds of the true mean is 18.

Step by step solution

01

Identify the values

We first identify the necessary values from the problem statement. We know the standard deviation (蟽) is 40 seconds, the margin of error (E) is 15 seconds and the Z-value for a 95% confidence level is 1.96.
02

Apply the formula for sample size

Next, we plug these values into the formula for determining sample size in a confidence interval, which is \(n = (Z*蟽/E)^2\). Substituting the known values into the equation, we get: \(n = (1.96*40/15)^2\).
03

Calculate the sample size

Solve this equation to find the value of n. After performing the calculation, you find that \(n \approx 17.2\). Since we cannot have a fractional sample, we round up to the nearest whole number, which is 18.

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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 helps us understand the range in which we expect the true population parameter to lie, such as a mean, with a certain level of confidence. In this case, the efficiency expert wants to know how confident they can be about the average drilling time. When we say "95% confident," it means if the study were repeated lots of times, 95% of those intervals would contain the true mean time.
  • The confidence level indicates how sure you want to be; common values are 90%, 95%, and 99%.
  • The interval is calculated using the sample data, considering the variability and desired confidence level.
  • This interval provides a buffer around the sample mean to account for variability.
Understanding this concept is crucial because it expresses the reliability of the prediction based on the sample taken.
Standard Deviation
Standard deviation is a measure that indicates the amount of variability or spread in a set of data. In the context of the drilling time, it tells us how much the times vary from the average time. A lower standard deviation means that the data points tend to be close to the mean, while a higher standard deviation indicates more spread out data.
  • In our example, \(\sigma = 40\) seconds. This suggests there is considerable variation in drilling times.
  • It is a critical input when calculating confidence intervals, as it affects the width of these intervals.
  • Precision in measurement can be understood by how tightly our data is clustered around the mean.
Grasping standard deviation helps in making more informed decisions about the reliability of the data measurements.
Margin of Error
The margin of error represents how much you expect your sample estimates to differ from the true population value. Here it signifies how far your estimated average drilling time might be from the true average. It鈥檚 influenced by the sample size and the variability in the data (standard deviation).
  • In the exercise, the margin of error is 15 seconds; it tells us how precise the estimate of the mean time will be.
  • The larger the sample size, the smaller the margin of error, leading to a more precise estimate.
  • Properly choosing this value helps ensure that the confidence interval is meaningful.
Knowing how to set an acceptable margin of error allows for better planning and resource allocation in research.
Z-Value
The Z-value, or Z-score, is a statistical measurement that describes a data point's relationship to the mean of a group of data. In context, it helps to scale the data and establish confidence intervals. For a 95% confidence level, the Z-value is commonly 1.96, which corresponds to how far the sample mean can deviate from the true mean while still falling within that confidence level.
  • This Z-value comes from the standard normal distribution, which is a bell-shaped curve.
  • It is crucial in sample size determination, as reflected in the formula used for calculating necessary sample size: \(n = (Z\sigma/E)^2\).
  • The Z-value adjusts the margin of error to ensure that the confidence interval estimates reliably cover the true parameter in 95% of the cases.
A clear comprehension of the Z-value is essential for evaluating how well a sample performs against the overall data distribution.

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Most popular questions from this chapter

A random sample of 100 automobile owners shows that, in the state of Virginia, an automobile is driven on the average 23,500 kilometers per year with a standard deviation of 3900 kilometers. Assume the distribution of measurements to be approximately normal. (a) Construct a \(99 \%\) confidence interval for the average number of kilometers an automobile is driven annually in Virginia. (b) What can we assert with \(99 \%\) confidence about the possible size of our error if we estimate the average number of kilometers driven by car owners in Virginia to be 23.500 kilometers per year?

A study is to be made to estimate the proportion of residents in a certain city and its suburbs who favor the construction of a nuclear power plant. How large a sample is needed if one wishes to be at least \(95 \%\) confident that the estimate is within 0.04 of the true proportion of residents in this city and its suburbs that favor the construction of the nuclear power plant?

According to USA Today (March 17. \(\lfloor 997\) ), women made up \(33.7 \%\) of the editorial staff at local TV stations in 1990 and \(36.2 \%\) in \(1994 .\) Assume 20 new employees were: hired as editorial staff. (a) Estimate the number that would have been women in each year, respectively, (b) Compute a \(95 \%\) confidence interval to see if there is evidence that the proportion of women hired as editorial staff in 1994 was higher than the proportion hired in 1990 .

(a) A random sample of 200 voters is selected and 114 are found to support an annexation suit. Find the \(96 \%\) confidence interval for the fraction of the voting population favoring the suit. (b) What can we assert with \(96 \%\) confidence about the possible size of our error if we estimate the fraction of voters favoring the annexation suit to be \(0.57 ?\)

Suppose that there are \(n\) trials \(x_{1}, x_{2}, x_{n}\) from a Bernoulli process with parameter \(p\), the probability of a success. That is, the probability of \(r\) successes is given by \(\left(\begin{array}{l}n \\\ r\end{array}\right) p^{r}(1-p)^{n-F} .\) Work out the maximum likelihood estimator for the parameter \(p\).
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