Answer: ±2.55
Given:
n = 24
mean = 46
s = 5
To find the margin of error, we will use the following formula:
[tex]E=\pm t_{\frac{\alpha}{2}}(\frac{s}{\sqrt[]{n}})[/tex]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:
[tex]E=\pm t_{\frac{\alpha}{2}}(\frac{s}{\sqrt[]{n}})[/tex][tex]E=\pm t_{0.01}(\frac{s}{\sqrt[]{n}})[/tex][tex]E=\pm2.500(\frac{5}{\sqrt[]{24}})[/tex][tex]E=\pm2.500(1.0206)[/tex][tex]E=\pm2.55[/tex]The margin of error would be at ±2.55
