Question

The manufacturer wants to estimate the proportion of their cars that get over 100 mpg. Their sample of 30 indicates that 28% can obtain over 100 mpg. Construct a 95% confidence interval for the population proportion.

Answer 1

The 95% confidence interval is
`0.1193 < p <0.4407`

Explanation:

From the question we are told that

The sample size is
`m = 30`

The sample proportion is
`\r p = 0.28`

Given that the confidence interval is 95% then the level of significance is mathematically represented as

`\alpha = 100 - 95`

`\alpha = 5\%`

`\alpha = 0.05`

Next we obtain the critical value of the
`(\alpha )/(2)` from the normal distribution table, the value is

`Z_{( \alpha )/(2) } = 1.96`

Generally the margin of error is mathematically represented as

`E = Z_{( \alpha )/(2) } * \sqrt{( ( \r p (1 - \r p )))/(n) }`

=>
`E = 1.96 * \sqrt{( (0.28 (1 - 0.28 )))/( 30) }`

=>
`E = 0.1607`

The 95% confidence interval is

`\r p - E < p < \r p + E`

=>
`0.28 - 0.1607 < p < 0.28 + 0.1607`

=>
`0.1193 < p <0.4407`