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

The sample size for the survey is 7700 and the 95% confidence interval is (0.484, 0.516).

Step-by-step explanation:

To find the sample size, we can set up an equation using the information given. We know that 50% of the people who like the movie enjoyed the official movie soundtrack. This means that 50% of the people were part of the survey. Let's denote the sample size as 'x'. We can write the equation as:

0.5 * x = 3850

Solving for 'x', we get:

x = 3850 / 0.5

x = 7700

So, the sample size for the survey is 7700.

To calculate the 95% confidence interval using a margin of error of 1.6%, we can use the formula:

CI = p ± z * sqrt((p(1-p))/n)

Where:

CI = Confidence Interval

p = Proportion of people who enjoyed the movie soundtrack (0.5)

z = Z-score corresponding to a 95% confidence level (1.96)

n = Sample size (7700)

Plugging in the values, we get:

CI = 0.5 ± 1.96 * sqrt((0.5(1-0.5))/7700)

CI = 0.5 ± 1.96 * sqrt(0.25/7700)

CI = 0.5 ± 0.016

So, the 95% confidence interval is (0.484, 0.516).

Answer 2

For the first question the sample size is n= 3850

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

And the margin of error is given by:

`ME= z_(\alpha/2)\sqrt{(\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.