Question

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.

Answer 1

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).