Question

determine the margin of error for a 90% confidence interval to estimate the population mean with $sigma = 49$ for the following sample sizes.
a) $n = 33$
b) $n = 46$
c) $n = 63$
click the icon to view the cumulative probabilities for the standard normal distribution.
a) when $n = 33$, the margin of error for a 90% confidence interval is . (round to two decimal places as needed.)

Explanation:

Step1: Find the z - score for 90% confidence interval

For a 90% confidence interval, the significance level $\alpha=1 - 0.90 = 0.10$, and $\alpha/2=0.05$. Looking up in the standard - normal distribution table, the z - score $z_{\alpha/2}=z_{0.05}\approx1.645$.

Step2: Recall the formula for the margin of error

The formula for the margin of error $E$ when the population standard deviation $\sigma$ is known is $E = z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$.

a) When $n = 33$

Substitute $\sigma = 49$, $z_{\alpha/2}=1.645$, and $n = 33$ into the formula:
$E=1.645\times\frac{49}{\sqrt{33}}$
$E = 1.645\times\frac{49}{5.7446}$
$E=1.645\times8.53$
$E\approx14.03$

b) When $n = 46$

Substitute $\sigma = 49$, $z_{\alpha/2}=1.645$, and $n = 46$ into the formula:
$E = 1.645\times\frac{49}{\sqrt{46}}$
$E=1.645\times\frac{49}{6.7823}$
$E=1.645\times7.225$
$E\approx11.88$

c) When $n = 63$

Substitute $\sigma = 49$, $z_{\alpha/2}=1.645$, and $n = 63$ into the formula:
$E = 1.645\times\frac{49}{\sqrt{63}}$
$E=1.645\times\frac{49}{7.9373}$
$E=1.645\times6.173$
$E\approx10.15$

Answer:

a) 14.03
b) 11.88
c) 10.15