1. Inference about a Normal Population Mean: Variance Unknown 2. Inference about a Normal Population Variance 3. Inference about a Population Proportion Based on Large Samples Section 6.1 Inference about a Normal Population Mean: Variance Unknown Assumption The population follows 2 N(, ). Let X be the sample mean of a sample taken from the population, S be the sample standard deviation and n be the sample size. It can be X follows the t distribution proved that S/ n with n 1 degrees of freedom. BS06 1

This graph shows 0.4 the probability N(0, 1) density functions of 0.3 N(0, 1) and some t(20) t(n), the t distribution with n 0.2 degrees of freedom where n is a 0.1 t(5) positive integer. t(1) 3 2 1 0 1 2 3

Interval estimation The 100(1 )% confidence interval for the population mean is s s , x t / 2, n1 x t / 2, n1 n n where the value t, n can be obtained from the t distribution table. Comparing with the case with known , here we use s instead of and t2,n1 instead of z2. BS06 2

Density function of the t distribution with n degrees of freedom Area

t,n Example 1 A paint manufacturer wants to determine the average drying time of a new brand of interior wall paint. If for 12 test areas of equal size he obtained a mean drying time of 66.3 minutes and a standard deviation of 8.4 minutes, construct a 95% confidence interval for the true population mean assuming normality. [Solution] n 12, x 66.3, s 8.4, 1 0.95 0.05 and t2,n1 t0.025,11 2.201. The 95% confidence interval for is 8.4 8.4 , 66.3 2.201 66.3 2.201 , 12 12 that is, (61.0, 71.6). BS06 3

Hypothesis test When the null hypothesis is 0, 0 or 0, the test statistic is chosen to be X 0 . S/ n Let be the level of significance. Reject H0 if and Test H0 H1 only if x 0 Left- 0 t,n1 0 tailed or 0 s/ n x 0 Right- 0 t,n1 0 tailed or 0 s/ n x 0 Two t2,n1 0 0 tailed s/ n For a two-tailed test, the followings are equivalent (for each particular sample): 1. do not reject H0: 0 at the level of significance; 2. the 100(1 )% confidence interval for contains 0. BS06 4

Example 2 A psychologist claims that the mean age at which children start walking is 12.5 months. Carol wanted to check if this claim is true. She took a random sample of 18 children and found that the mean age at which these children started walking was 12.9 months and the standard deviation was 0.80 months. Using the 1% significance level, can you conclude that the mean age at which all children start walking is different from 12.5 months? [Solution] H0: 12.5, H1: 12.5, 0.01. The value of the test statistic is x 0 12.9 12.5 2.121. s / n 0.80 / 18 Since |2.121| 2.898 t0.005,17, we do not reject H0. Thus we cannot conclude that the mean age at which all children start walking is different from 12.5 months. Example 3 According to the last census in a city, the mean family annual income was 316 BS06 5

thousand dollars. A random sample of 900 families taken this year produced a mean family annual income of 313 thousand dollars and a standard deviation of 70 thousand dollars. At the 2.5% significance level, can we conclude that the mean family annual income has declined since the last census? [Solution] H0: 316, H1: 316, 0.025. The value of the test statistic is x 0 313 316 1.286. s / n 70 / 900 By Excel, t0.025,899 1.963. Since 1.286 1.963 t0.025,899, we do not reject H0. Thus we cannot conclude that the mean family annual income has declined since the last census. Example 4 A federal agency responsible for enforcing laws governing weights and measures routinely inspects packages to determine whether the weight of the contents is at least as great as that advertised on the BS06 6

package. A random sample of 18 containers whose...