1 / 21

210 likes | 293 Views

Chapter 7 Confidence Intervals for a Population Mean ; t distributions. t distributions t confidence intervals for a population mean Sample size required to estimate . The Importance of the Central Limit Theorem.

Download Presentation
## Chapter 7 Confidence Intervals for a Population Mean ; t distributions

**An Image/Link below is provided (as is) to download presentation**
Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.
Content is provided to you AS IS for your information and personal use only.
Download presentation by click this link.
While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.
During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

**Chapter 7Confidence Intervals for a Population Mean ; t**distributions t distributions t confidence intervals for a population mean Sample size required to estimate **The Importance of the Central Limit Theorem**• When we select simple random samples of size n, the sample means we find will vary from sample to sample. We can model the distribution of these sample means with a probability model that is**Since the sampling model for x is the normal model, when we**standardize x we get the standard normal z**If is unknown, we probably don’t know either.**The sample standard deviation s provides an estimate of the population standard deviation s For a sample of size n,the sample standard deviation s is: n − 1 is the “degrees of freedom.” The value s/√n is called the standard error of x , denoted SE(x).**Standardize using s for **• Substitute s (sample standard deviation) for s s s s s s s s Note quite correct Not knowing means using z is no longer correct**t-distributions**Suppose that a Simple Random Sample of size n is drawn from a population whose distribution can be approximated by a N(µ, σ) model. When s is known, the sampling model for the mean x is N(m, s/√n). When s is estimated from the sample standard deviation s, the sampling model for the mean x follows at distribution t(m, s/√n) with degrees of freedom n − 1. is the 1-sample t statistic**Confidence Interval Estimates**• CONFIDENCE INTERVAL for • where: • t = Critical value from t-distribution with n-1 degrees of freedom • = Sample mean • s = Sample standard deviation • n = Sample size • For very small samples (n < 15), the data should follow a Normal model very closely. • For moderate sample sizes (n between 15 and 40), t methods will work well as long as the data are unimodal and reasonably symmetric. • For sample sizes larger than 40, t methods are safe to use unless the data are extremely skewed. If outliers are present, analyses can be performed twice, with the outliers and without.**t distributions**• Very similar to z~N(0, 1) • Sometimes called Student’s t distribution; Gossett, brewery employee • Properties: i) symmetric around 0 (like z) ii) degrees of freedom **Z**-3 -2 -1 0 1 2 3 -3 -2 -1 0 1 2 3 Student’s t Distribution**Z**t -3 -2 -1 0 1 2 3 -3 -2 -1 0 1 2 3 Student’s t Distribution Figure 11.3, Page 372**Degrees of Freedom**Z t1 -3 -2 -1 0 1 2 3 -3 -2 -1 0 1 2 3 Student’s t Distribution Figure 11.3, Page 372**Degrees of Freedom**Z t1 t7 -3 -2 -1 0 1 2 3 -3 -2 -1 0 1 2 3 Student’s t Distribution Figure 11.3, Page 372**t-Table: text- inside back cover**• 90% confidence interval; df = n-1 = 10 0.80 0.95 0.98 0.99 0.90 Degrees of Freedom 1 3.0777 6.314 12.706 31.821 63.657 2 1.8856 2.9200 4.3027 6.9645 9.9250 . . . . . . . . . . . . 10 1.3722 1.8125 2.2281 2.7638 3.1693 . . . . . . . . . . . . 100 1.2901 1.6604 1.9840 2.3642 2.6259 1.282 1.6449 1.9600 2.3263 2.5758**Student’s t Distribution**P(t > 1.8125) = .05 P(t < -1.8125) = .05 .90 .05 .05 t10 0 -1.8125 1.8125**Comparing t and z Critical Values**Conf. level n = 30 z = 1.645 90% t = 1.6991 z = 1.96 95% t = 2.0452 z = 2.33 98% t = 2.4620 z = 2.58 99% t = 2.7564**Example**• An investor is trying to estimate the return on investment in companies that won quality awards last year. • A random sample of 41 such companies is selected, and the return on investment is recorded for each company. The data for the 41 companies have • Construct a 95% confidence interval for the mean return.**Example**• Because cardiac deaths increase after heavy snowfalls, a study was conducted to measure the cardiac demands of shoveling snow by hand • The maximum heart rates for 10 adult males were recorded while shoveling snow. The sample mean and sample standard deviation were • Find a 90% CI for the population mean max. heart rate for those who shovel snow.**EXAMPLE: Consumer Protection Agency**• Selected random sample of 16 packages of a product whose packages are marked as weighing 1 pound. • From the 16 packages: • a.find a 95% CI for the mean weight of the 1-pound packages • b. should the company’s claim that the mean weight is 1 pound be challenged ?

More Related