Question

In a recent study of statistics students, a random sample of students were asked to provide the number of hours per week they spend studying for their statistics class. The results were used to compute confidence intervals for the population mean hours per week spent studying for statistics. The 95% confidence interval for the population mean hours per week that students spend studying for statistics was (6.55, 8.15). In this confidence interval, what is the margin of error (m)? Provide your answer as a number rounded to two decimal places.

Answer 1

Answer:

$ME=\frac{Upper-Lower}{2}= \frac{8.15-6.55}{2}= 0.80$

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X$ represent the sample mean for the sample  

$\mu$ population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size  

Solution to the problem

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$   (1)

For this case the confidence interval is given by (6.55,8.15). And we can estimate the margin of error with this formula since the confidence interval is symmetrical:

$ME=\frac{Upper-Lower}{2}= \frac{8.15-6.55}{2}= 0.80$