Question

In a survey of 1,000 television viewers, 40% said they watch network news programs. For a 90% confidence level, the margin of error for this estimate is 2.5%. If we want to be 95% confident, how will the margin of error change

Answer 1

Increase due to a higher confidence interval

Explanation:

Answer 2

Answer: The margin of error =
`3\%`

Step-by-step explanation:

Given

Sample size (n) = 1000

Population proportion = 0.4

`\alpha` = 1 - confidence level

= 1 - 0.95

= 0.05

`margin\; of\; error = z_{(\alpha )/(2)}\sqrt{\frac{{\widehat{p}}{(1 -\widehat{p})}}{n}}`

`margin\; of\; error = z_{(0.05 )/(2)\sqrt{\frac{{(0.4)}{(1 -0.4)}}{1000}}`

`= z_(0.025)\sqrt{\frac{{(0.4)}{(0.6)}}{1000}}`

`= 1.96\sqrt{\frac{{(0.4)}{(0.6)}}{1000}}`

= 0.03

The margin of error change to
`2.5\%` to
`3\%`