/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 67 A manufacturer of college textbo... [FREE SOLUTION] | 91Ó°ÊÓ

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A manufacturer of college textbooks is interested in estimating the strength of the bindings produced by a particular binding machine. Strength can be measured by recording the force required to pull the pages from the binding. If this force is measured in pounds, how many books should be tested to estimate the mean force required to break the binding to within \(0.1\) pounds with \(95 \%\) conficence? Assume that \(\sigma\) is known to be \(0.8\) pound.

Short Answer

Expert verified
The manufacturer should test 246 books to estimate the mean force required to break the binding to within 0.1 pound with 95% confidence.

Step by step solution

01

Identify given values

From the problem, we know that \(\sigma = 0.8\) (pound), the confidence level is 95 percent and the margin of error \(E = 0.1\) (pound).
02

Compute Z

Now, we have to compute the corresponding z-score for 95 percent confidence. For 95 percent confidence level, the score is typically \(1.96\) (in a Z-table). Therefore, \(Z_{\alpha/2} = 1.96\).
03

Calculate Sample Size

Now that we have everything, we can compute for the sample size using the formula \( n = \left( \frac{Z_{\alpha/2} * \sigma}{E} \right)^{2} \). Substituting the previously computed values gives us \( n = \left( \frac{1.96 * 0.8}{0.1} \right)^{2} = 245.8624\). The sample size must be a whole number so we round up to the next higher whole number, so \( n = 246\) books to be tested.

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