Question

A poll asked 1057 American adults if they believe there was a conspiracy in the assassination of President Kennedy, and found that 614 believe there was a conspiracy. Estimate the proportion of Americans who believe in this conspiracy using a 95% confidence interval. Round to three decimal places.

Answer 1

The 95% confidence interval of the proportion of Americans who believe in the conspiracy is
`0.551<  p <  0.610`

Explanation:

From the question we are told that

The sample size is n =1057

The number that believe there was a conspiracy is k = 614

Generally the sample proportion is mathematically represented as

`\^ p = (614)/(1057 )`

=>
`\^ p = 0.5809`

From the question we are told the confidence level is 95% , hence the level of significance is

`\alpha = (100 - 95 ) \%`

=>
`\alpha = 0.05`

Generally from the normal distribution table the critical value of
`(\alpha )/(2)` is

`Z_{(\alpha )/(2) } =  1.96`

Generally the margin of error is mathematically represented as

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

=>
`E = 1.96 * \sqrt{(0.5809 (1- 0.5809))/(1057) }`

=>
`E = 0.02975`

Generally 95% confidence interval is mathematically represented as

`\^ p -E <  p <  \^ p +E`

=>
`0.5809 - 0.02975<  p <  0.5809 + 0.02975`

=>
`0.551<  p <  0.610`