Question

A 90% confidence interval for the mean height of students
is (60.128, 69.397). What is the value of the margin of error?
a) m = 129.525
b) m = 4.635
c) m = 64.763
d) m = 9.269

Answer 1

Answer:

$ME= \frac{69.397-60.128}{2}= 4.6345 \approx 4.635$

And the best answer on this case would be:

b) m = 4.635

Step-by-step explanation:

Let X the random variable of interest and we know that the confidence interval for the population mean $\mu$ is given by this formula:

$\bar X \pm t_{\alpha/2} \frac{s}{\sqrt{n}}$

The confidence level on this case is 0.9 and the significance $\alpha=1-0.9=0.1$

The confidence interval calculated on this case is $60.128 \leq \mu \leq 69.397$

The margin of error for this confidence interval is given by:

$ME =t_{\alpha/2} \frac{s}{\sqrt{n}}$

Since the confidence interval is symmetrical we can estimate the margin of error with the following formula:

$ME = \frac{Upper -Lower}{2}$

Where Upper and Lower represent the bounds for the confidence interval calculated and replacing we got:

$ME= \frac{69.397-60.128}{2}= 4.6345 \approx 4.635$

And the best answer on this case would be:

b) m = 4.635