Question

An April 2014 poll of 2300 U.S. adults found that 43% regularly watch television shows through streaming.

Required:
a. Find a 99% confidence interval and interpret it in context. If the survey had been of 5000 adults rather than 2300, would the interval be wider or narrower

Answer 1

a

The confidence interval is
`0.41 <  p <  0.45`

Generally the 95% confidence interval mean that there is 95% confidence that the true proportion of US adults who regularly watch television shows through streaming fall within the confidence interval

As the sample size increase from 2300 to 5000 the confidence interval becomes narrower

Explanation:

From the question we are told that

The sample proportion is
`\^ p = 0.43`

The sample size is n = 2300

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{( 0.43  (1- 0.43 ))/(2300) }`

=>
`E = 0.020`

Generally 95% confidence interval is mathematically represented as

`\^ p -E <  p <  \^ p +E`

=>
`0.43  -0.020 <  p <  0.43  + 0.020`

=>
`0.41 <  p <  0.45`

Generally the 95% confidence interval mean that there is 95% confidence that the true proportion of US adults who regularly watch television shows through streaming fall within the confidence interval

Now when the sample size is increase to 5000 the margin of error becomes

=>
`E_1  =  1.96  * \sqrt{( 0.43  (1- 0.43 ))/(5000) }`

=>
`E_1 = 0.0137`

So as we can see
`E_1 < E` hence the confidence interval becomes narrower