Download STAT 366 Spring 10 Quiz 5 3/31/10 (Hypothetical) A study was done

STAT 366 Spring 10 Quiz 5 3/31/10
(Hypothetical) A study was done at WMU to determine the number of times
(perhaps numerous times per day) a student visited a campus computer lab on
campus per week. A random sample of 81 students was taken, and the sample
average was 4 with a standard deviation of 2.7 (definitely not normally distributed,
or even mound shaped). For the following statements , determine if they are True or
False.(2pt each)
1) The average # of visits to a campus computer lab per week for all students at
WMU is 4, give or take 2.7.
FALSE: You could estimate this average by 4, but your error would not be give
or take 2.7, which is the SD of the sample. Your would need to calculate the SE.
2) About 50% of the students at WMU visit a campus computer lab 4 or more
times per week.
FALSE: Even if we estimate that the average to be 4, since we don’t know what
the actual distribution of the population is, i.e. if it is symmetric or not, we
cannot determine what % of the data lies above the average. (For example,
think of a left or right skewed distribution)
3) A 95% confidence interval for the average # of visits to a campus computer
lab per week for all students at WMU is 4 +- .6.
TRUE: The SE= 2.7/Sqrt(81)=.3. So a 95% C.I. = xbar +- 2*SE = 4 +-.6
4) If another random sample of size 81 is taken, there is a 95% chance that it
will also have a mean of 4.
FALSE: The sample mean, xbar, of a sample of size 81 is a random number
which is approx. normally distributed, with mean and SD unknown (since we
don’t know the population mean  or standard deviation . However, given
that it is a random number, and that it can be basically any number within a
specific range (e.g.  +- 2*, at least) then the chance that it takes on one
particular value in that range is essentially zero.
The students in the random sample above were also asked if they had visited the
recreation center at least once per week. Out of the random sample of 81, 27 said
they did visit the rec. center at least once per week. Find a 99% confidence interval
for the proportion of all WMU students who visit the rec. center at least once per
week. (4pt)
Ans: The sample proportion psamp = 27/81 = .33 and the sample standard deviation
SDsamp= Sqrt{.33*(1-.33)} =.47, which we will use as our estimate for population
proportion and SD. So the SE(p) = .47/Sqrt(81) =.47/9=.052, making the 99%
confidence interval for the population proportion= .33 +- 3*(.052) = .33 +- .156