Question

Tv ratings in 7000 households. Nielsen estimates that 18,000 people live here. Suppose Nielsen reports American idol had 65% of tv audience. Interpret results with 95% confidence based on sample size of18,000 people.

Margin of error?


Confidence intervals is from ———% to ———%

Answer 1

Answer: Margin of error = 0.07

Confidence intervals is from 64.3% to 65.7%.

Explanation:

Let p be the population proportion of tv audience American idol had.

The formula to find the margin of error is given by :-

`E=z^*\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}` , where z* = Critical value ,
`\hat{p}` = Sample proportion and n = sample size.

We are given that :

n= 18000

`\hat{p}=0.65`

By z-table , Critical value for 95% confidence interval : z*= 1.96

Margin of error:
`E=(1.96)\sqrt{(0.65(1-0.65))/(18000)}`

`E=(1.96)√(0.0000126388888889)`

`E=(1.96)(0.00355512150128)\\\\ E=0.00696803814252\approx0.007`

i.e. Margin of error = 0.07

Confidence interval for population proportion =
`(\hat{p}-E,\ \hat{p}+E)`

=
`(0.65-0.007,\ 0.65+0.007)=(0.643,\ 0.657)=(64.3\%,\ 65.7\%)`

Hence, Confidence intervals is from 64.3% to 65.7%.