We are asked
to create the 95% confidence interval for the mean number of nights that people stayed for
vacation. We are told that the sample size was 1500, and the sample mean is 7.5. We are also
told to assume that the population standard deviation is 0.8.
When creating
the interval about a mean realize that what we are doing is taking some point estimate and
adding/subtracting a term encompassing possible errors. There are two primary components of this
error term: one factor involves how confident we want to be, and the other factor is the
standard error of the mean.
- The point estimate: a point estimate
is a "guess" as to the actual value. The sample mean makes a good point estimate as it
will vary much less than other measures of the center if we repeat the survey.
`bar(x)=7.5` - We are asked for a confidence level of 95%. Then `alpha=.05` .
Thus we would expect the actual mean to fall out of the interval we create 5% of the time.
Assuming a fairly "normal" distribution (not highly skewed) the mean will fall below
the interval 2.5% of the time and above 2.5%.
Converting the
data distribution to a standard normal distribution, we can find the z-score that corresponds to
2.5% outside the interval. We call this value `z_(alpha/2)` .
Looking in a
standard normal table or using technology we find the corresponding z-value to be
1.96.
The standard error of the mean is defined to be `sigma/sqrt(n)` . We
can use this as we know the population standard deviation. If we did not know the population
standard deviation we would have to use another technique.
The
interval we seek is given by:
`bar(x)-z_(alpha/2)(sigma/sqrt(n)) `7.5-(1.96)(0.8/sqrt(1500)) `7.4595 (You will need to consult your text or
instructor as to their requirements for rounding.)
No comments:
Post a Comment