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Chapter 9: Problem 37

Samples of two different models of cars were selected, and the actual speed for each car was determined when the speedometer registered \(50 \mathrm{mph}\). The resulting \(95 \%\) confidence intervals for mean actual speed were \((51.3,52.7)\) and \((49.4,50.6)\). Assuming that the two sample standard deviations are equal, which confidence interval is based on the larger sample size? Explain your reasoning.

Short Answer

Expert verified
The confidence interval with the larger sample size is \( (49.4, 50.6) \). This is because its width is smaller than the other given confidence interval, suggesting a larger sample size given equal standard deviations.

Step by step solution

01

Identify Confidence Intervals

Identify the two sets of confidence intervals, which are: \( (51.3, 52.7) \) and \( (49.4, 50.6) \).
02

Calculate the Width of Confidence Intervals

Calculate the width of each confidence interval by subtracting the lower limit from the upper limit. For the first confidence interval, the width is \( 52.7 - 51.3 = 1.4 \). And for the second, it is \( 50.6 - 49.4 = 1.2 \).
03

Compare the Widths

Compare the widths of these confidence intervals. Because the confidence interval will be narrower with larger sample size, given constant standard deviation, the narrower interval suggests a larger sample size.

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Most popular questions from this chapter

Acrylic bone cement is sometimes used in hip and knee replacements to fix an artificial joint in place. The force required to break an acrylic bone cement bond was measured for six specimens under specified conditions, and the resulting mean and standard deviation were \(306.09\) Newtons and \(41.97\) Newtons, respectively. Assuming that it is reasonable to believe that breaking force under these conditions has a distribution that is approximately normal, estimate the mean breaking force for acrylic bone cement under the specified conditions using a \(95 \%\) confidence interval.

Discuss how each of the following factors affects the width of the confidence interval for \(p\) : a. The confidence level b. The sample size c. The value of \(\hat{p}\)

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The formula used to compute a confidence interval for the mean of a normal population when \(n\) is small is $$\bar{x} \pm(t \text { critical value }) \frac{s}{\sqrt{n}}$$ What is the appropriate \(t\) critical value for each of the following confidence levels and sample sizes? a. \(95 \%\) confidence, \(n=17\) b. \(90 \%\) confidence, \(n=12\) c. \(99 \%\) confidence, \(n=24\) d. \(90 \%\) confidence, \(n=25\) e. \(90 \%\) confidence, \(n=13\) f. \(95 \%\) confidence, \(n=10\)

The formula used to compute a large-sample confidence interval for \(p\) is $$\hat{p} \pm(z \text { critical value }) \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$$ What is the appropriate \(z\) critical value for each of the following confidence levels? a. \(95 \%\) d. \(80 \%\) b. \(90 \%\) e. \(85 \%\) c. \(99 \%\)
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