Question

A random sample of 1003 adult Americans was asked, "Do you think televisions are a necessity or a luxury you could do without?" Of the 1003 adults surveyed, 521 indicated that televisions are a luxury they could do without. Construct and interpret a 95% confidence interval for the population proportion of adult Americans who believe that televisions are a luxury they could do without out.

Answer 1

The 95% confidence interval is
`0.503 < p < 0.535`

The interpretation is that there is 95% confidence that the true population proportion lie within the confidence interval

Explanation:

From the question we are told that

The sample size is n = 1003

The number that indicated television are a luxury is k = 521

Generally the sample mean is mathematically represented as

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

`\r p = (521)/(1003)`

`\r p = 0.519`

Given the confidence level is 95% then the level of significance is mathematically evaluated as

`\alpha = 100 - 95`

`\alpha = 5\%`

`\alpha = 0.05`

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

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

The margin of error is mathematically represented as

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

=>
`E = 1.96 * \sqrt{ ( 0.519 (1- 0.519 ))/(1003) }`

=>
`E = 0.016`

The 95% confidence interval is mathematically represented as

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

=>
`0.519 - 0.016 < p < 0.519 + 0.016`

=>
`0.503 < p < 0.535`