Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
What if it is impossible or impractical to use a large sample? Apply the Student’s t distribution. Student’s t Variable x t s n The shape of the t distribution depends only only 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. Table 6 in Appendix II gives values of the variable t corresponding to the number of degrees of freedom (d.f.) Degrees of Freedom d.f. = n – 1 where n = sample size The t Distribution has a Shape Similar to that of the the Normal Distribution A Normal distribution A “t” distribution Find the critical value tc for a 95% confidence interval if n = 8. d.f.=7 c ' '' d.f. ... 6 7 8 ...0.90 ... 0.95 ... 0.98 ... 0.99 ... ...1.943 ...1.895 ...1.860 2.447 2.365 2.306 3.143 2.998 2.896 3.707 3.499 3.355 Confidence Interval for the Mean of Small Samples (n < 30) from Normal Populations xE xE where x Sample Mean s E tc n c = confidence level (0 < c < 1) tc = critical value for confidence level c, and degrees of freedom = n - 1 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. Mean = 15.7 ounces Standard deviation = 2.3 n = 8, so d.f. = nounces. –1=7 • • For c = 0.90, Table 6 in Appendix II gives t0.90 = 1.895. s 2.3 E tc 1.895 1.54 n 8 Mean = 15.7 ounces Standard deviation = 2.3 ounces. E = 1.54 The 90% confidence interval is: xE xE 15.7 - 1.54 < < 15.7 + 1.54 14.16 < < 17.24 The 90% Confidence Interval: 14.16 < < 17.24 We are 90% sure that the true mean weight of the fish in the lake is between 14.16 and 17.24 ounces.