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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 average force required to break the binding to within .1 lb with 95% confidence? Assume that s is known to be .8.

Short Answer

Expert verified

The sampler size is \(n = 246\).

Step by step solution

01

Step 1:Sample size.

In order for confidence interval\(\left( {\overline x - {Z_{\alpha /2}} \times \frac{\sigma }{{\sqrt n }},\overline x + {Z_{\alpha /2}} \times \frac{\sigma }{{\sqrt n }}} \right)\) to have width \(w\), the necessary sample size \(n\) is,\(n = {\left( {2{Z_{\alpha /2}} \times \frac{\sigma }{w}} \right)^2}\)

02

To find the sample size.

The average force required to break the binding to be within \(0.1\) means that the width of the interval is \(2(0.1) = 0.2\).

In order for confidence interval\(\left( {\overline x - {Z_{\alpha /2}} \times \frac{\sigma }{{\sqrt n }},\overline x + {Z_{\alpha /2}} \times \frac{\sigma }{{\sqrt n }}} \right)\) to have width \(w\), the necessary sample size \(n\) is,

\(n = {\left( {2{Z_{\alpha /2}} \times \frac{\sigma }{w}} \right)^2}\)

The smaller width \(w\), the larger \(n\) must be.

Using the formula for \(\sigma = 0.8\) and \({Z_{\alpha /2}} = 1.96\), the sample size is

\(\begin{aligned}{}n &= {\left( {2{Z_{\alpha /2}} \times \frac{\sigma }{w}} \right)^2}\\ &= {\left( {2(1.96) \times \frac{{0.8}}{{0.2}}} \right)^2}\\ &= 245.86\end{aligned}\)

But the sampler size needs to be an integer, therefore round it up to \(n = 246\).

Hence, The sampler size is\(n = 246\).

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Most popular questions from this chapter

Example 1.11 introduced the accompanying observations on bond strength.

11.5 12.1 9.9 9.3 7.8 6.2 6.6 7.0

13.4 17.1 9.3 5.6 5.7 5.4 5.2 5.1

4.9 10.7 15.2 8.5 4.2 4.0 3.9 3.8

3.6 3.4 20.6 25.5 13.8 12.6 13.1 8.9

8.2 10.7 14.2 7.6 5.2 5.5 5.1 5.0

5.2 4.8 4.1 3.8 3.7 3.6 3.6 3.6

a.Estimate true average bond strength in a way that conveys information about precision and reliability.

(Hint: \(\sum {{{\bf{x}}_{\bf{i}}}} {\bf{ = 387}}{\bf{.8}}\) and \(\sum {{{\bf{x}}^{\bf{2}}}_{\bf{i}}} {\bf{ = 4247}}{\bf{.08}}\).)

b. Calculate a 95% CI for the proportion of all such bonds whose strength values would exceed 10.

The article โ€œDistributions of Compressive Strength Obtained from Various Diameter Coresโ€ (ACI Materials J., 2012: 597โ€“606) described a study in which compressive strengths were determined for concrete specimens of various types, core diameters, and length -to-diameter ratios. For one particular type, diameter, and l/d ratio, the 18 tested specimens resulted in a sample mean compressive strength of 64.41 MPa and a sample standard deviation of 10.32 MPa. Normality of the compressive strength distribution was judged to be quite plausible.

a.Calculate a confidence interval with confidence level 98% for the true average compressive strength under these circumstances.

b.Calculate a 98% lower prediction bound for the compressive strength of a single future specimen tested under the given circumstances. (Hint: t.02,17 = 2.224.)

Let X1, X2,โ€ฆ, Xn be a random sample from a continuous probability distribution having median \(\widetilde {\bf{\mu }}\) (so that \({\bf{P(}}{{\bf{X}}_{\bf{i}}} \le \widetilde {\bf{\mu }}{\bf{) = P(}}{{\bf{X}}_{\bf{i}}} \le \widetilde {\bf{\mu }}{\bf{) = }}{\bf{.5}}\)).

a. Show that

\({\bf{P(min(}}{{\bf{X}}_{\bf{i}}}{\bf{) < }}\widetilde {\bf{\mu }}{\bf{ < max(}}{{\bf{X}}_{\bf{i}}}{\bf{)) = 1 - }}{\left( {\frac{{\bf{1}}}{{\bf{2}}}} \right)^{{\bf{n - 1}}}}\)

So that \({\bf{(min(}}{{\bf{x}}_{\bf{i}}}{\bf{),max(}}{{\bf{x}}_{\bf{i}}}{\bf{))}}\)is a \({\bf{100(1 - \alpha )\% }}\) confidence interval for \(\widetilde {\bf{\mu }}\) with (\({\bf{\alpha = 1 - }}{\left( {\frac{{\bf{1}}}{{\bf{2}}}} \right)^{{\bf{n - 1}}}}\).Hint :The complement of the event \(\left\{ {{\bf{min(}}{{\bf{X}}_{\bf{i}}}{\bf{) < \mu < max(}}{{\bf{X}}_{\bf{i}}}{\bf{)}}} \right\}\)is \({\bf{\{ max(}}{{\bf{X}}_{\bf{i}}}{\bf{)}} \le \widetilde {\bf{\mu }}{\bf{\} }} \cup {\bf{\{ min(}}{{\bf{X}}_{\bf{i}}}{\bf{)}} \ge \widetilde {\bf{\mu }}{\bf{\} }}\). But \({\bf{max(}}{{\bf{X}}_{\bf{i}}}{\bf{)}} \le \widetilde {\bf{\mu }}\) iff \({\bf{(}}{{\bf{X}}_{\bf{i}}}{\bf{)}} \le \widetilde {\bf{\mu }}\)for all i.

b.For each of six normal male infants, the amount of the amino acid alanine (mg/100 mL) was determined while the infants were on an isoleucine-free diet,

resulting in the following data:

2.84 3.54 2.80 1.44 2.94 2.70

Compute a 97% CI for the true median amount of alanine for infants on such a diet(โ€œThe Essential Amino Acid Requirements of Infants,โ€ Amer. J. Of Nutrition, 1964: 322โ€“330).

c.Let x(2)denote the second smallest of the xiโ€™s and x(n-1) denote the second largest of the xiโ€™s. What is the confidence level of the interval(x(2), x(n-1)) for \(\widetilde \mu \)?

Assume that the helium porosity (in percentage) of coal samples taken from any particular seam is normally distributed with true standard deviation .75.

a. Compute a 95% CI for the true average porosity of a certain seam if the average porosity for 20 specimens from the seam was 4.85.

b. Compute a 98% CI for true average porosity of another seam based on 16 specimens with a sample average porosity of 4.56.

c. How large a sample size is necessary if the width of the 95% interval is to be .40?

d. What sample size is necessary to estimate true average porosity to within .2 with 99% confidence?

Let\({{\rm{\alpha }}_{\rm{1}}}{\rm{ > 0,}}{{\rm{\alpha }}_{\rm{2}}}{\rm{ > 0,}}\)with\({{\rm{\alpha }}_{\rm{1}}}{\rm{ + }}{{\rm{\alpha }}_{\rm{2}}}{\rm{ = 0}}\)

\({\rm{P}}\left( {{\rm{ - }}{{\rm{z}}_{{\alpha _{\rm{1}}}}}{\rm{ < }}\frac{{{\rm{\bar X - m}}}}{{{\rm{s/}}\sqrt {\rm{n}} }}{\rm{ < }}{{\rm{z}}_{{\alpha _{\rm{2}}}}}} \right){\rm{ = 1 - }}\alpha \)

a. Use this equation to derive a more general expression for a \({\rm{100(1 - \alpha )\% }}\)CI for \({\rm{\mu }}\)of which the interval (7.5) is a special case.

b. Let \({\rm{\alpha = 0}}{\rm{.5}}\)and \({{\rm{\alpha }}_{\rm{1}}}{\rm{ = \alpha /4,}}\)\({{\rm{\alpha }}_{\rm{2}}}{\rm{ = 3\alpha /4}}{\rm{.}}\)Does this result in a narrower or wider interval than the interval (7.5)?

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