Question

450 high school freshmen were randomly selected for a national survey. Among survey participants, the mean grade-point average (GPA) was 2.96, and the standard deviation was 0.21. What is the margin of error, assuming a 70% confidence level, to the nearest hundredth

Answer 1

Answer:

The margin of error is of 0.01.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1 - 0.7}{2} = 0.15$

Now, we have to find z in the Z-table as such z has a p-value of $1 - \alpha$.

That is z with a pvalue of $1 - 0.15 = 0.85$, so Z = 1.037.

The margin of error is of:

$M = z\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

Standard deviation was 0.21.

This means that $\sigma = 0.21$

Sample of 450:

This means that $n = 450$

What is the margin of error, assuming a 70% confidence level, to the nearest hundredth?

$M = z\frac{\sigma}{\sqrt{n}}$

$M = 1.037\frac{0.21}{\sqrt{450}}$

$M = 0.01$

The margin of error is of 0.01.