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

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

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

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

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

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

                            =  0.03

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