Usually, we do not know the population mean and standard deviation. Our
goal is to estimate these numbers. The standard way to accomplish this is
to use the sample mean and standard deviation as a best guess for the true
population mean and standard deviation. We call this "best
guess" a A
x is a point estimate for m s is a point estimate for s
A point estimate is
We are not only interested in finding the point estimate for the mean, but also determining how accurate the point estimate is. The Central Limit Theorem plays a key role here. We assume that the sample standard deviation is close to the population standard deviation (which will almost always be true for large samples). Then the Central Limit Theorem tells us that the standard deviation of the sampling distribution is
We
will be interested in finding an interval around x such that there is a large
probability that the actual mean falls inside of this interval. This
interval is called a
Suppose that we check for clarity in 50 locations in Lake Tahoe and discover that the average depth of clarity of the lake is 14 feet. Suppose that we know that the standard deviation for the entire lake's depth is 2 feet. What can we conclude about the average clarity of the lake with a 95% confidence level?
We can use x to provide a point
estimate for m. How accurate
is x as a point
estimate? We construct a
We arrive at z = -1.96. Now we solve for x:
x - 14 x -
14 x - 14 = -0.55
We say that 0.55 is the We have that a 95% confidence interval for the mean clarity is (13.45,14.55) In other words there is a 95% chance that the mean clarity is between 13.45 and 14.55.
In general if z
When the population is normal or if the sample size is large,
then the sampling distribution will
also be normal, but the use of s to replace s
is not that accurate. The smaller the sample size the worse the
approximation will be. Hence we can expect that some adjustment will be
made based on the sample size. The adjustment we make is that we do not
use the normal curve for this approximation. Instead, we use the
Suppose that we conduct a survey of 19 millionaires to find out what percent of their income the average millionaire donates to charity. We discover that the mean percent is 15 with a standard deviation of 5 percent. Find a 95% confidence interval for the mean percent. Assume that the distribution of all charity percents is approximately normal.
We use the formula: (Notice the t instead of the z and s instead of s) We get
15
t Since n = 19, there are 18 degrees of freedom. Using the table in the back of the book, we have that
t Hence the margin of error is 2.10 (5) / = 2.4 We can conclude with 95% confidence that the millionaires donate between 12.6% and 17.4% of their income to charity.
Handout on finding the sample size needed for a confidence interval for a mean
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