Question

An insurance company regularly reviews official police reports of traffic accidents in a particular region of the country so that it can price its premiums in that region appropriately. A review of a random sample of 582 accident reports showed that in 91 accidents, a teenage driver was found to be at fault.

Required:
Report the upper bound of a 90% confidence interval for the true proportion of accidents in the area that are caused by teenagers, to three decimal places.

Answer 1

Answer and Step-by-step explanation:

Random sample = 582

No. of accidents = 91

P = 91 / 582 = 0.156

Confidence interval for proportion of accidents:

P ± z √(p(1-p)/n)

Where,

n = 582

p = 0.156

z = 1.645

Formula:

P ± z √(p(1-p)/n)

put these values in formula, we get:

= 0.156 ± 1.645 √(0.156(1-0.156)/582)

= 0.156 ± 1.645 √(0.1316/582)

= 0.156 ± 1.645 (0.01483)

= 0.156 ± 0.0244

= (0.1804, 0.1316)

Answer 2

The upper bound of the 90% confidence interval is 0.181

Explanation:

From the question we are told that

The sample size is n = 582

The number of accidents is k = 91

Generally the sample proportion is mathematically represented as

`\r p = ( k)/(n) = (582)/(91 )`

=>
`\r p = 0.1564`

Generally the confidence level is 90% , the level of significance is
`\alpha  =  (100-90)\%`

=>
`\alpha =  0.10`

From the normal distribution table the critical value of
`(\alpha )/(2)` is
`Z_{(\alpha )/(2) } =  1.645`

Generally the margin of error is mathematically represented as

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

=>
`E  =  1.645* \sqrt{( 0.1564 (1 - 0.1564))/(582) }`

=>
`E =  0.0248`

Generally the 90% confidence interval is mathematically represented as

`0.1564 - 0.0248  <  p  <  0.1564  -0.0248`

=>
`0.1316  <  p  <  0.1812`