Question

If a survey of a movie shows that 50% of all people who like the movie, enjoyed the official movie soundtrack too. If a total of 3850 people were asked, find the data of the survey. What is the sample size? If the margin of error for this survey is 1.6%, what is the 95% confidence interval using only 1 decimal?

Answer 1

Answer:

For the first question the sample size is n= 3850

$ME= z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}$

$\hat p \pm ME$

And replacing we got:

$0.5- 0.016 = 0.484$

$0.5+ 0.016 = 0.516$

We are confident at 95% that the true proportion of all people who like the movie is between 0.484 and 0.516.

Step-by-step explanation:

We have the following info given:

$\hat p = 0.5$ represent the estimated proportion of all people who like the movie

$n = 3850$ represent the sample size selected

For the first question the sample size is n= 3850

For the second part they gives to us the margin of error:

$ME = 0.016$

We know that the confidence interval for a population proportion $p$ of interest is given by:

$\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}$

And the margin of error is given by:

$ME= z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}$

So then the confidence interval can be found with this formula:

$\hat p \pm ME$

And replacing we got:

$0.5- 0.016 = 0.484$

$0.5+ 0.016 = 0.516$

We are confident at 95% that the true proportion of all people who like the movie is between 0.484 and 0.516.