# Confidence Intervals

2086 words 9 pages
Confidence Intervals

Consider the following question: someone takes a sample from a population and finds both the sample mean and the sample standard deviation. What can he learn from this sample mean about the population mean?
This is an important problem and is addressed by the Central Limit Theorem. For now, let us not bother about what this theorem states but we will look at how it could help us in answering our question.

The Central Limit Theorem tells us that if we take very many samples the means of all these samples will lie in an interval around the population mean. Some sample means will be larger than the population mean, some will be smaller. The Central Limit Theorem goes on to state that 95% of the sample means will lie

Example 2:
The Roman Arches is an Italian restaurant. The manager wants to estimate the average amount a customer spends for lunch, Monday through Friday. A random sample of 115 customers' lunch tabs gives a mean of \$12.75 with a standard deviation of \$3.37. What is the 99% confidence interval for the mean of the lunch tabs for all customers?

Solution:
[pic]. Thus the manager can be 99% confident that the mean of all customers' lunch tabs is somewhere between \$11.94 and \$13.56.

Comment: If the population standard deviation is known, then we could use the formula for those cases where the sample size is less than 30. But since for most practical situations we will not know the population standard deviation, we will simply use the formula for those cases where the sample size is greater than 30.

Confidence Intervals when Sample Proportions are known

In situations where a sample proportion, denoted by[pic], is known, and we want to get information about the population proportion p, we will use the formula [pic].

Example 3:
A representative of a consumer organization took a random sample of 250 egg cartons from the dairy section of a very large supermarket and found that 80 cartons had at least one broken egg. Find a 90% confidence interval for the proportion of cartons in the population that had at least one broken egg in them.

Solution:
Use[pic][pic]. Thus we are 90% confident that the proportion of egg cartons in the population containing

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