Posted: December 21st, 2016

Two students are discussing the average family income of students on their campus.

Two students are discussing the average family income of students on their campus. They look up the average campus income for college students and it is $77,000 with a standard deviation of $20,000 (Pryor, 2008). They are curious if the average family income on their campus is above or below the national average. They go out late one afternoon and ask 10 students their family income (they volunteer to tell the truth) and it turns out to be $69.000. The two students conclude that, indeed, the average student family income on campus is truly below average. Why might their conclusion be strongly questioned?
1. the 10 students were not necessarily a random sample
2. they don t know how much random variability there might be from sample to sample
3. the sample size may be too small
4. all of the above
The two students from the above problem decide to redo their little study, and on the advice of their statistics professor, sample 40 students and ask them their family income. The mean of the 40 student families is 73,900. How much average error did they reduce by increasing from 10 to 40 in their sample? Hint: the average error is just the standard error of the mean, what proportion of this error was reduced by increasing from 10 to 40 students, (smean for N = 10 ) (smean for N = 40 ) ? What is the approximate ratio of these two values?
1. reduced the average error by about one sixth
2. reduced the average error by about one fifth
3. reduced the average error by about one half
4. reduced the average error by about one fourth
1 points Save

Given that the true mean of the average income on the two students campus was $77,000 but their results were simply due to sampling error, after all they didn t sample everyone just 40 randomly selected students, how likely is it to have gotten a mean income of 73,900 or less simply as a result of chance, sampling error? (Use the normal curve table for this with N = 40 and s = $20,000 and