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

The margin of error, at a 70% confidence level, is approximately 0.02.

Step-by-step explanation:

To find the margin of error, we can use the formula:

Margin of Error = Z * (standard deviation / sqrt(sample size))

Given that the sample size is 450, the mean GPA is 2.96, and the standard deviation is 0.21, we need to find the Z value for a 70% confidence level.

The Z value can be found using the Z-table or a statistical calculator. For a 70% confidence level, the Z value is approximately 1.04.

Plugging in the values into the formula:

Margin of Error = 1.04 * (0.21 / sqrt(450)) = 0.0243

Therefore, the margin of error to the nearest hundredth is approximately 0.02.

Answer 2

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 = (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(\sigma)/(√(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(\sigma)/(√(n))`

`M = 1.037(0.21)/(√(450))`

`M = 0.01`

The margin of error is of 0.01.