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Chapter 9: Problem 16

A random sample of 12 graduates of a certain secretarial school typed an average of 79.3 words per minute with a standard deviation of 7.8 words per minute. Assuming a normal distribution for the number of words typed per minute, find a \(95 \%\) confidence interval for the average number of words typed by all graduates of this school.

Short Answer

Expert verified
The \(95\%\) confidence interval is given by \((\bar{x} - E, \bar{x} + E)\). Upon calculation, we will find the confidence interval.

Step by step solution

01

Declare the Known Values

Here, the sample mean \(\bar{x}\) is \(79.3\) words per minute, the sample standard deviation \(s\) is \(7.8\) words per minute and the sample size \(n\) is \(12\). The value of \(z\) for \(95\%\) confidence level is \(1.96\).
02

Find the Standard Error

The standard error is given by the formula \(SE = \frac{s}{\sqrt{n}}\) . Substituting the given values, the standard error is \(SE = \frac{7.8}{\sqrt{12}}\).
03

Calculate the Margin of Error

The margin of error is given by \(E = z * SE\) Substituting the known values, the margin of error \(E = 1.96 * SE\).
04

Calculate the Confidence Interval

The average number of words typed per minute, \( \mu \), lies within the interval \(\bar{x} \pm E\). After substituting the values \(\bar{x}\) and \(E\), we get the confidence interval.

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Key Concepts

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

Sample Statistics
Sample statistics are important because they help us make conclusions about a whole population based on a smaller sample. Let's take a closer look at some key components:
  • Sample Mean (\(\bar{x}\)): This is the average value obtained from your sample. In our example, 12 graduates typed, on average, 79.3 words per minute.
  • Sample Standard Deviation (s): This measures how much the words per minute vary among the graduates. A standard deviation of 7.8 words per minute tells us about the spread of the data.
  • Sample Size (n): The number of observations in the sample, which affects the reliability of our estimates. Here, n is 12.
By examining these components, we gain insights into the entire group of secretarial school graduates using just the sample.
Standard Error
The standard error (SE) helps us understand how far our sample mean (\(\bar{x}\)) is likely to be from the true population mean. It's a critical component when calculating confidence intervals. We compute the standard error using the formula:
  • Standard Error: \( SE = \frac{s}{\sqrt{n}} \)
In the current example:
  • Sample standard deviation (s) is 7.8, and
  • Sample size (n) is 12.
When you substitute these numbers into the formula, you find out how much the sample mean of 79.3 words per minute could differ from the population mean. A smaller SE indicates a more precise estimate.
Margin of Error
The margin of error tells us the range of values within which the true population average likely falls. It depends on the standard error and the confidence level we choose, which is typically expressed with a z-score:
  • Margin of Error (E): \( E = z * SE \)
For a 95% confidence level, the z-score is 1.96. This means we're pretty confident our range captures the true mean. In our example, multiply 1.96 by the calculated standard error to find how wide the confidence interval will be around our sample mean of 79.3 words per minute. This gives us the cushion around that mean where we expect the true average to fall.
Normal Distribution
Understanding the normal distribution is key when calculating confidence intervals because it describes how data points are spread around the mean. It's characterized by:
  • Symmetry: The data is evenly distributed around the average.
  • Bell Shape: Most values cluster around a central peak and taper off equally on both sides.
  • Mean, Median, Mode: These measures of central tendency are all the same in a perfectly normal distribution.
In our exercise, we assume the number of words typed follows this distribution. This allows us to use z-scores to find the 95% confidence interval and ensures that our sample data reliably predicts the population characteristics. A normal distribution simplifies the process of making probabilistic statements about how data behaves.

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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?

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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