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%.

Step-by-step explanation:

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

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

$E=z^*\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$ , 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: $E=(1.96)\sqrt{\dfrac{0.65(1-0.65)}{18000}}$

$E=(1.96)\sqrt{0.0000126388888889}$

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

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%.