Question

Hi, can you help me answer this question please, thank you!

Answer 1

Answer: ±2.55

Given:

n = 24

mean = 46

s = 5

To find the margin of error, we will use the following formula:

`E=\pm t_{(\alpha)/(2)}(\frac{s}{\sqrt[]{n}})`

First, let us find α/2.

Since we have a confidence level of 98%, this would mean that α = 0.02, and α/2 = 0.01

Next, we will find its degrees of freedom, by subtracting 1 to n.

degrees of freedom = n - 1 = 24 - 1 = 23

We will now look at the t-table to look at the distribution at 98% confidence level

From the table, we just need to find

the t distribution when:

df (degrees of freedom) = 23

Confidence level = 98%

That will give us:

t = 2.500

Now, substitute all of these on the formula:

`E=\pm t_{(\alpha)/(2)}(\frac{s}{\sqrt[]{n}})`
`E=\pm t_(0.01)(\frac{s}{\sqrt[]{n}})`
`E=\pm2.500(\frac{5}{\sqrt[]{24}})`
`E=\pm2.500(1.0206)`
`E=\pm2.55`

The margin of error would be at ±2.55