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Confidence Intervals. W&W, Chapter 8. Confidence Intervals.

Confidence Intervals

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Confidence Intervals

W&W, Chapter 8

Although on average, M (the sample mean) is on target (or unbiased), the specific sample mean that we happen to observe is almost certain to be a bit high or a bit low. Accordingly, if we want to be reasonably confident that our inference is correct, we cannot claim that is precisely equal to M. Instead we must construct a confidence interval of the form:

= M +/- sampling error

The confidence interval gives us a range within which we are confident that the true falls.

Recall from Chapter 1 that we estimated a confidence interval for a proportion, which was:

= P 1.96 [P(1-P)/n]

The value 1.96 comes from a confidence level of 95%, where we assume that 95 out of every 100 samples we collect will contain the true population parameter.

Recall from the standard normal distribution that 95% of all values fall within 2 standard deviations of the mean. The exact Z-score for 95% is +/- 1.96.

= M +/- Z/2(S.E.)

where Z is the critical value based on = 1 - confidence level (as a proportion)

S.E. of the sampling distribution = /N

For 95%, = 1 - .95 = .05; /2 = .025; thus we go to the standard normal table and find the value that has a probability of .025 beyond it. This is 1.96!

Thus a 95% confidence interval for the mean is calculated as:

= M +/- 1.96(/N)

Note: As N increases, the size of the confidence interval shrinks, i.e., we are more confident that we are close to with a larger sample.

What if we wanted to be 99% confident that our confidence interval contains the true ? In other words, in the long run, 99 out of 100 samples would produce a confidence interval that contains .

= 1 - .99 = .01; /2 = .005

Z/2 = Z.005 = 2.58

= M +/- 2.58(/N)

So far we have assumed that the population standard deviation, , was known. Most of the time, this is not the case, so we will use the sample standard deviation, s, as a proxy for .

Confidence interval when is unknown:

= M +/- Z/2(s/N)

Example: We want to determine the mean number of years that FSU professors hold their jobs. We take a random sample of 100 professors and calculate the sample mean, which is 14.7 and the sample standard deviation, s = 2.5. What is the 95% confidence interval for the true average number of years FSU professors have held their jobs?

= M +/- Z/2(s/N)

= 14.7 +/- (1.96)(2.5/100)

= 14.7 +/- .49

Thus we are 95% confident that the mean number of years FSU professors have kept their jobs is between 14.2 and 15.2 years.

If we had collected this data from a sample of 500 professors, our interval estimate would shrink to 14.7 +/- .22 or 14.5 to 14.9 years, which demonstrates how the confidence interval is decreased by larger samples.

We have been using Z-scores from the standard normal distribution to construct our critical values. The use of Z-scores, however, is only appropriate if is known or if we have a reasonably sized sample (N100) when is unknown. In cases of small samples when is not known, we must use the student t distribution.

A man named William Gosset, who wrote under the name of Student, developed the t distribution to deal with the nature of small samples. He was working for the Guinness brewing company in Ireland in the early 1900's (article published in 1908). He wanted to construct a reliable test for examining the quality of a small sample of beer and from that, be able to conclude that the whole batch of beer was ok.

- Like the standard normal, it is symmetric with a mean of zero
- The larger tails imply that the proportion of area beyond a specific value of t is greater than the proportion of area beyond the corresponding Z

- The t critical points are presented relative to the degrees of freedom, which is n - # of things not free to vary. For the standard deviation, s,
S = (Xi - M)2

N - 1

Thus the degrees of freedom here is n - 1 (later we will see that df = n - k, where k is the number of parameters we are estimating).

The maker of a certain car model claims that the car averages 31 miles per gallon. A random sample of nine cars is selected and each car is driven on a tank of gas. The sample mean mpg is 29.5 and the sample standard deviation (s) is 3. Can we be 95% confident that the car actually gets 31mpg from this sample?

We must use t because 1) is unknown and 2) n is small (less than 100).

- = M +/- t/2(s/n)
/2 = .05/2 = .025

df = n - 1 = 8 (go to t table)

- = 29.5 +/- 2.31(3/9)
- = 29.5 +/- 2.31 = 27.2 to 31.8

Yes, we can be 95% confident that the car gets 31mpg.