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Chapter 8: Problem 105

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.

Short Answer

Expert verified
Based on our calculations, we can be 95% confident that the mean relief time for all patients after taking the drug is between 21.92 and 26.08 minutes.

Step by step solution

01

Identify the provided values

The sample mean (\(\bar{X}\)) is 24 minutes, the standard deviation (\(蟽\)) is 4.5 minutes, the sample size (\(n\)) is 18 patients, and our level of confidence is 0.95.
02

Find the critical z value

For a 95% confidence interval, we need to find the z value that cuts off the central 95% of the distribution. This is typically denoted \(Z_{0.025}\), as we're looking at both tails of the distribution. From the standard z table, or from a z calculator, we find that \(Z_{0.025} = 1.96\).
03

Plug the values into the confidence interval formula

We now substitute our values into the confidence interval formula. This gives us: \(24 \pm 1.96 * \frac{4.5}{\sqrt{18}}\).
04

Evaluate the expression

We evaluate the expression which gives us: \(24 \pm 2.08\). This is an interval between \(21.92\) and \(26.08\) minutes.

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Most popular questions from this chapter

Inside the Box Corporation makes corrugated cardboard boxes. One type of these boxes states that the breaking capacity of this box is 75 pounds. Fifty-five randomly selected such boxes were loaded until they broke. The average breaking capacity of these boxes was found to be \(78.52\) pounds. Suppose that the standard deviation of the breaking capacities of all such boxes is \(2.63\) pounds. Calculate a \(99 \%\) confidence interval for the average breaking capacity of all boxes of this type.

a. A sample of 100 observations taken from a population produced a sample mean equal to \(55.32\) and a standard deviation equal to \(8.4\). Make a \(90 \%\) confidence interval for \(\mu\). b. Another sample of 100 observations taken from the same population produced a sample mean equal to \(57.40\) and a standard deviation equal to \(7.5\). Make a \(90 \%\) confidence interval for \(\mu\). c. A third sample of 100 observations taken from the same population produced a sample mean equal to \(56.25\) and a standard deviation equal to \(7.9\). Make a \(90 \%\) confidence interval for \(\mu\). d. The true population mean for this population is \(55.80 .\) Which of the confidence intervals constructed in parts a through \(\mathrm{c}\) cover this population mean and which do not?

Determine the sample size for the estimate of \(\mu\) for the following. a. \(E=2.3, \quad \sigma=15.40\), confidence level \(=99 \%\) b. \(E=4.1, \quad \sigma=23.45\), confidence level \(=95 \%\) c. \(E=25.9, \quad \sigma=122.25, \quad\) confidence level \(=90 \%\)

A department store manager wants to estimate at a \(90 \%\) 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?

Determine the sample size for the estimate of \(\mu\) for the following. a. \(E=.17, \quad \sigma=.90\), confidence level \(=99 \%\) b. \(E=1.45, \quad \sigma=5.82\), confidence level \(=95 \%\) c. \(E=5.65, \quad \sigma=18.20\), confidence level \(=90 \%\)
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