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?

Since the Confidence is 0.95 or 95%, the value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], 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 [tex]z_{\alpha/2}=1.96[/tex]

And solving for the standard error we got:

[tex] SE= \frac{2}{1.96}= 1.02[/tex]

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".

[tex]\bar X[/tex] represent the sample mean for the sample

[tex]\mu[/tex] 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:

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

And the confidence interval is given by:

[tex]4 \leq \mu \leq 8[/tex]

We can estimate the margin of error like this:

[tex] ME = \frac{8-4}{2}= 2[/tex]

And the margin of error is given by:

[tex] ME=z_{\alpha/2} SE[/tex]

Since the Confidence is 0.95 or 95%, the value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], 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 [tex]z_{\alpha/2}=1.96[/tex]

And solving for the standard error we got:

[tex] SE= \frac{2}{1.96}= 1.02[/tex]