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Twenty eight concrete blocks were sampled and tested for crushing strength in order to estimate the proportion that were sufficiently strong for a certain application. Twenty six of the 28 blocks were sufficiently strong. Use the small-sample method to construct a 90% confidence interval for the proportion of blocks that are sufficiently strong. Round the answers to at least three decimal places.

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Answer:

The 90% confidence interval for the proportion of blocks that are sufficiently strong is (0.849, 1).

Explanation:

In a sample with a number n of people surveyed with a probability of a success of
\pi, and a confidence level of
1-\alpha, we have the following confidence interval of proportions.


\pi \pm z\sqrt{(\pi(1-\pi))/(n)}

In which

z is the zscore that has a pvalue of
1 - (\alpha)/(2).

Twenty six of the 28 blocks were sufficiently strong.

This means that
n = 28, \pi = (26)/(28) = 0.929

90% confidence level

So
\alpha = 0.1, z is the value of Z that has a pvalue of
1 - (0.1)/(2) = 0.95, so
Z = 1.645.

The lower limit of this interval is:


\pi - z\sqrt{(\pi(1-\pi))/(n)} = 0.929 - 1.645\sqrt{(0.929*0.071)/(28)} = 0.849

The upper limit of this interval is:


\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.929 + 1.645\sqrt{(0.929*0.071)/(28)} = 1.009

Since a proportion cannot be above 1, the upper limit is 1.

The 90% confidence interval for the proportion of blocks that are sufficiently strong is (0.849, 1).

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