Question

The University College is interested in the average number of hours per week that freshmen spend going to parties. They took a random sample of 300 freshmen and calculated a mean of 6 hours a week going to parties and a 95% confidence interval for the population mean to be 4 to 8 hours a week. Based on this information, what was their standard error of the mean?

Answer 1

Answer:

And the confidence interval is given by:

$4 \leq \mu \leq 8$

We can estimate the margin of error like this:

$ME = \frac{8-4}{2}= 2$

And the margin of error is given by:

$ME=z_{\alpha/2} SE$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that $z_{\alpha/2}=1.96$

And solving for the standard error we got:

$SE= \frac{2}{1.96}= 1.02$

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:

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

And the confidence interval is given by:

$4 \leq \mu \leq 8$

We can estimate the margin of error like this:

$ME = \frac{8-4}{2}= 2$

And the margin of error is given by:

$ME=z_{\alpha/2} SE$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that $z_{\alpha/2}=1.96$

And solving for the standard error we got:

$SE= \frac{2}{1.96}= 1.02$