Download presentation

Presentation is loading. Please wait.

Published byPhyllis McKenzie Modified over 3 years ago

1
8.2 Estimating μ When σ is Unknown

2
What if it is impossible or impractical to use a large sample? Apply the Student ’ s t distribution: It is important to note that the t distribution uses a degrees of freedom (d.f.) = n – 1 The shape of the t distribution depends only on the sample size, n, if the basic variable x has a normal distribution. When using the t distribution, we will assume that the x distribution is normal.

3
Student’s t Distributions In order to use the normal distribution to find confidence intervals for a population mean μ we need to know the value of σ. However, much of the time, we don’t know either. In such cases, we use the sample standard deviation s to approximate σ. When we use s to approximate σ, the sampling distribution for x bar follows a distribution known as Student’s t distribution.

4
More about t distribution The graph is always symmetrical about its mean, which (as with z dist) is 0. t distribution is similar to normal z distribution except it has somewhat thicker tails. (as seen on next slide) As the degrees of freedom increase, the t distribution approaches the standard normal distribution. – As n increases… what happens to the distribution of any sample?

5
The t Distribution has a shape similar to that of the the Normal Distribution A Normal distribution A “ t ” distribution

6
Maximal Margin of Error, E Similar to what we saw in the previous section, we can determine the confidence interval by using the margin of error:

7
Example A company has a new process for manufacturing large artificial sapphires. In a trial run, 37 sapphires are produced. The mean weight for these 37 gems is 6.75 carats and σ = 0.33 carats. Let μ be the mean weight for distribution of all. Why is the t distribution the preferred method here? Find E for a 95% confidence interval. Then find the confidence interval. Interpret the confidence interval in the context of the problem.

8
The mean weight of eight fish caught in a local lake is 15.7 ounces with a standard deviation of 2.3 ounces. Construct a 90% confidence interval for the mean weight of the population of fish in the lake. Another Example

9
Mean = 15.7 ounces Standard deviation = 2.3 ounces. n = 8, so d.f. = n – 1 = 7 For c = 0.90, Table 6 in Appendix II gives t 0.90 = 1.895.

10
Mean = 15.7 ounces Standard deviation = 2.3 ounces. E = 1.54 The 90% confidence interval is: 15.7 - 1.54 < < 15.7 + 1.54 14.16 < < 17.24

11
Summary Examine problem Statement σ is known Use normal distribution with margin of error σ is unknown Use Student t’s distribution with margin of error (don’t forget about d.f.)

Similar presentations

OK

Point Estimates point estimate A point estimate is a single number determined from a sample that is used to estimate the corresponding population parameter.

Point Estimates point estimate A point estimate is a single number determined from a sample that is used to estimate the corresponding population parameter.

© 2018 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on surface chemistry Dot matrix display ppt online Ppt on bombay film industry Ppt on instrument landing system outer Ppt on retail marketing Synthesizing in reading ppt on ipad Ppt on magneto optical current transformer Ppt on summary writing graphic organizer Ppt on dual cycle Ppt on face recognition technology using matlab