Question

determine the margin of error for a 99% confidence interval to estimate the population mean when s = 38 for the sample sizes below.
a) n = 12
b) n = 35
c) n = 49
a) the margin of error for a 99% confidence interval when n = 12 is 34.07. (round to two decimal places as needed.)
b) the margin of error for a 99% confidence interval when n = 35 is . (round to two decimal places as needed.)

Explanation:

Step1: Identify the t - value

For a 99% confidence interval, the significance level $\alpha=1 - 0.99 = 0.01$, and $\frac{\alpha}{2}=0.005$. The degrees of freedom $df=n - 1$. We will use the t - distribution table or a calculator to find the t - value $t_{\frac{\alpha}{2},df}$.

Step2: Calculate margin of error formula

The formula for the margin of error $E = t_{\frac{\alpha}{2},df}\frac{s}{\sqrt{n}}$, where $s = 38$ is the sample standard - deviation.

For $n = 35$:

First, calculate the degrees of freedom $df=n - 1=35 - 1 = 34$. Using a t - distribution table or calculator, $t_{0.005,34}\approx 2.728$.
Then, calculate the margin of error $E = 2.728\times\frac{38}{\sqrt{35}}$
$E = 2.728\times\frac{38}{5.916}\approx 2.728\times6.423\approx 17.52$

For $n = 49$:

Calculate the degrees of freedom $df=n - 1=49 - 1 = 48$. Using a t - distribution table or calculator, $t_{0.005,48}\approx 2.682$.
Then, calculate the margin of error $E = 2.682\times\frac{38}{\sqrt{49}}$
$E = 2.682\times\frac{38}{7}=2.682\times5.429\approx 14.56$

Answer:

b) 17.52
c) 14.56