4. Example A sample of weights of 34 male freshman students was obtained. 185 161 174 175 202 178 202 139 177 170 151 176 197 214 283 184 189 168 188 170 207 180 167 177 166 231 176 184 179 155 148 180 194 176 If one wanted to estimate the true mean of all male freshman students, you might use the sample mean as a point estimate for the true mean.
5. Example After looking at a histogram and boxplot of the data (below) you might notice that the data seems reasonably symmetric with a outlier, so you might use either the sample median or a sample trimmed mean as a point estimate. Calculated using Minitab 260 220 180 140
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10. Recall * n 10 and n (1- ) 10 Specifically when n is large*, the statistic p has a sampling distribution that is approximately normal with mean and standard deviation . For the sampling distribution of p, p = and for large* n The sampling distribution of p is approximately normal.
15. Example - continued So np = 182(0.4121) = 75 >10 and n(1-p)=182(0.5879) = 107 >10 we can use the formulas given on the previous slide to find a 95% confidence interval for . The 95% confidence interval for is (0.341, 0.484).
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18. Some Common Critical Values Confidence level z critical value 80% 1.28 90% 1.645 95% 1.96 98% 2.33 99% 2.58 99.8% 3.09 99.9% 3.29
19. Terminology The standard error of a statistic is the estimated standard deviation of the statistic.
20. Terminology The bound on error of estimation, B , associated with a 95% confidence interval is (1.96) ·(standard error of the statistic). The bound on error of estimation, B , associated with a confidence interval is (z critical value) ·(standard error of the statistic) .
21. Sample Size The sample size required to estimate a population proportion to within an amount B with 95% confidence is The value of may be estimated by prior information. If no prior information is available, use = 0.5 in the formula to obtain a conservatively large value for n. Generally one rounds the result up to the nearest integer.
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25. One-Sample z Confidence Interval for 2. The sample size n is large (generally n 30), and 3. , the population standard deviation, is known then the general formula for a confidence interval for a population mean is given by
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27. Example Find a 90% confidence interval estimate for the true mean fills of catsup from this machine. A certain filling machine has a true population standard deviation = 0.228 ounces when used to fill catsup bottles. A random sample of 36 “6 ounce” bottles of catsup was selected from the output from this machine and the sample mean was 6.018 ounces.
28. Example I (continued) The z critical value is 1.645 90% Confidence Interval (5.955, 6.081)
29. Unknown - Small Size Samples [All Size Samples] An Irish mathematician/statistician, W. S. Gosset developed the techniques and derived the Student’s t distributions that describe the behavior of .
30. t Distributions If X is a normally distributed random variable, the statistic follows a t distribution with df = n-1 (degrees of freedom).
31. t Distributions This statistic is fairly robust and the results are reasonable for moderate sample sizes (15 and up) if x is just reasonable centrally weighted. It is also quite reasonable for large sample sizes for distributional patterns (of x) that are not extremely skewed.
35. One-Sample t Procedures Suppose that a SRS of size n is drawn from a population having unknown mean . The general confidence limits are and the general confidence interval for is
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37. Confidence Interval Example We find the critical t value of 1.833 by looking on the t table in the row corresponding to df = 9, in the column with bottom label 90%. Computing the confidence interval for is Calculating the sample mean and standard deviation we have n = 10, = 86, s = 11.842
38. Confidence Interval Example To calculate the confidence interval, we had to make the assumption that the distribution of weekly viewing times was normally distributed. Consider the normal plot of the 10 data points produced with Minitab that is given on the next slide.
39. Confidence Interval Example Notice that the normal plot looks reasonably linear so it is reasonable to assume that the number of hours of television watched per week by shut-ins is normally distributed. Typically if the p-value is more than 0.05 we assume that the distribution is normal P-Value: 0.753 A-Squared: 0.226 Anderson-Darling Normality Test