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 . (round to two decimal places as needed.)

Explanation:

Step1: Find degrees of freedom and t - value

For a 99% confidence interval, the significance level $\alpha=1 - 0.99 = 0.01$, and $\frac{\alpha}{2}=0.005$.

• When $n = 12$, degrees of freedom $df=n - 1=12 - 1 = 11$. Looking up in the t - distribution table, $t_{\frac{\alpha}{2},df}=t_{0.005,11}=3.106$.

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

Step2: Calculate margin of error

Substitute $t_{0.005,11}=3.106$, $s = 38$, and $n = 12$ into the formula:
$E=3.106\times\frac{38}{\sqrt{12}}\approx3.106\times\frac{38}{3.464}\approx33.97$.

• When $n = 35$, degrees of freedom $df=n - 1=35 - 1 = 34$. Since the t - distribution table may not have 34 exactly, we can use the closest value or use a statistical software. Using a software or approximation, $t_{0.005,34}\approx2.728$.

$E = 2.728\times\frac{38}{\sqrt{35}}\approx2.728\times\frac{38}{5.916}\approx17.60$.

• When $n = 49$, degrees of freedom $df=n - 1=49 - 1 = 48$. Using a software or approximation, $t_{0.005,48}\approx2.682$.

$E=2.682\times\frac{38}{\sqrt{49}}=2.682\times\frac{38}{7}\approx14.63$.

Answer:

a) 33.97
b) 17.60
c) 14.63