Question

In 1980 more than 35% of cars purchased had a manual transmission (i.e. stick shift). By 2007 the proportion had decreased to 7.7%. A random sample of college students who owned cars revealed the following: out of 133 cars, 31 had stick shifts. Estimate the proportion of college students who drive sticks with 90% confidence. Use a graphing calculator and round the answers to at least three decimal places. <

Answer 1

The 90% confidence interval for the proportion of college students who drive sticks is (0.173, 0.293).

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)`.

A random sample of college students who owned cars revealed the following: out of 133 cars, 31 had stick shifts.

This means that
`n = 133, \pi = (31)/(133) = 0.233`

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)} = 0.233 - 1.645\sqrt{(0.233*0.767)/(133)} = 0.173`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.233 + 1.645\sqrt{(0.233*0.767)/(133)} = 0.293`

The 90% confidence interval for the proportion of college students who drive sticks is (0.173, 0.293).