Presentation on theme: "5.5 Confidence Intervals for a Population Mean ; t distributions t distributions t confidence intervals for a population mean Sample size required to estimate."— Presentation transcript:
5.5 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 n 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 dont know either. The sample standard deviation s provides an estimate of the population standard deviation 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 n Substitute s (sample standard deviation) for ssss s ss 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 is known, the sampling model for the mean x is N( /n). When is estimated from the sample standard deviation s, the sampling model for the mean x follows a t distribution t(, s/n) with degrees of freedom n 1. is the 1-sample t statistic
Confidence Interval Estimates n CONFIDENCE INTERVAL for n CONFIDENCE INTERVAL for n where: n t = Critical value from t- distribution with n-1 degrees of freedom n = Sample mean n s = Sample standard deviation n 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 n Very similar to z~N(0, 1) n Sometimes called Students t distribution; Gossett, brewery employee n Properties: i)symmetric around 0 (like z) ii)degrees of freedom
-3-20123 Z t 0123 -2-3 Students t Distribution Figure 11.3, Page 372
-3-20123 Z t1t1 0123 -2-3 Students t Distribution Figure 11.3, Page 372 Degrees of Freedom
-3-20123 Z t1t1 0123 -2-3 t7t7 Students t Distribution Figure 11.3, Page 372 Degrees of Freedom
13.07776.31412.70631.82163.657 21.88562.92004.30276.96459.9250............ 101.37221.81252.22812.76383.1693............ 1001.29011.66041.98402.36422.6259 1.2821.64491.96002.32632.5758 0.80 0.90 0.950.980.99 t-Table: text- inside back cover; Course Pack- inside back cover and following p. 15, lect. unit 5 n 90% confidence interval; df = n-1 = 10
0 1.8125 Students t Distribution P(t > 1.8125) =.05 -1.8125.05.90 t 10 P(t < -1.8125) =.05
Comparing t and z Critical Values Conf. leveln = 30 z = 1.64590%t = 1.6991 z = 1.9695%t = 2.0452 z = 2.3398%t = 2.4620 z = 2.5899%t = 2.7564
n 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 n Because cardiac deaths increase after heavy snowfalls, a study was conducted to measure the cardiac demands of shoveling snow by hand n The maximum heart rates for 10 adult males were recorded while shoveling snow. The sample mean and sample standard deviation were n Find a 95% CI for the population mean max. heart rate for those who shovel snow.
EXAMPLE: Consumer Protection Agency n Selected random sample of 16 packages of a product whose packages are marked as weighing 1 pound. n From the 16 packages: n a. find a 95% CI for the mean weight of the 1-pound packages n b. should the companys claim that the mean weight is 1 pound be challenged ?