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Chapter 8: Problem 54

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 \(\hat{p}=.08\)

Short Answer

Expert verified
a) Yes, the sample size is large enough. b) No, the sample size is not large enough. c) Yes, the sample size is large enough. d) Yes, the sample size is large enough.

Step by step solution

01

Evaluate Case a

Check if both \(80 \cdot 0.85 \geq 10\) and \(80 \cdot (1 - 0.85) \geq 10\). If both conditions are met, the sample is 'large enough'.
02

Evaluate Case b

Check if both \(110 \cdot 0.98 \geq 10\) and \(110 \cdot (1 - 0.98) \geq 10\). If both conditions are met, the sample is 'large enough'.
03

Evaluate Case c

Check if both \(35 \cdot 0.40 \geq 10\) and \(35 \cdot (1 - 0.40) \geq 10\). If both conditions are met, the sample is 'large enough'.
04

Evaluate Case d

Check if both \(200 \cdot 0.08 \geq 10\) and \(200 \cdot (1 - 0.08) \geq 10\). If both conditions are met, the sample is 'large enough'.

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Most popular questions from this chapter

For each of the following, find the area in the appropriate tail of the \(t\) distribution. a. \(t=-1.302\) and \(d f=42\) b. \(t=2.797\) and \(n=25\) c. \(t=1.397\) and \(n=9\) d. \(t=-2.383\) and \(d f=67\)

What are the parameters of a normal distribution and a \(t\) distribution? Explain.

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For a population data set, \(\sigma=12.5\). How large a sample should be selected so that the margin of error of estimate for a \(99 \%\) confidence interval for \(\mu\) is \(2.50 ?\) 1\. How large a sample should be selected so that the margin of error of estimate for a \(96 \%\) confidence interval for \(\mu\) is \(3.20\) ?

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