Question

A simple random sample of 1100 males aged 12 to 17 int he United States were asked whether they played massive multiplayer online role-playing games (MMORPGs); 775 said that they did. Construct a 95% confidence interval to estimate the proportion of all U.S. males aged 12 to 17 who play MMORPGs. Using Scenario 2, if you wanted to have a margin of error of LaTeX: \pm 2% with 99% confidence, would your sample have to be larger, smaller, or the same size as the sample in the previous question

Answer 1

Answer:

$n' = 2212$ (larger)

Step-by-step explanation:

From the question we are told that

Initial sample size n=1100

Sample range 5

Sample Proportion$\hat P= x/n=>775/1100=>0.704$

99% confidence

Margin for error $E=\pm2 \%=>0.02$

Generally at confidence level  99%

Level of significance $\alpha =1-0.99=>0.01$

$\alpha /2 =\frac{0.01}{2} =0.005$

Therefore

$Z_\alpha_ /_2=Z_0_._0_0_5=2.576$

$Z_\alpha_ /_2=2.576$

Generally sample size is mathematically given by

$n' = (Z_\alpha_/_2 / E )^2 * \hat P * (1 - \hat P )$

$n' = (2.576) / 0.025)^2 * 0.704 * (1 - 0.704)$

$n' = 2212.463274$

$n' = 2212$

Therefore the new sample will be larger