Confidence Intervals and Hypothesis Testing of a Population Mean (Variance Known)

Transcription

1 Confidence Interval and Hypothei Teting of a Population Mean (Variance Known) Confidence Interval One-ided confidence level for lower bound, X l = X Z α One ided confidence interval for upper bound, X u = X + Z α Similarly, the two ided (1-α) confidence interval for the population mean would be given by, P (X Zα < μ < X + Zα ) = 1 α Therefore, you are (1-α)100% confident that the true mean lie between thee two value. Thi i a -ided confidence interval. Hypothei Tet The probability of erroneouly rejecting the null hypothei i given by, P ( Z > X μ 0 ) = p value If the p-value i le than α, which i the probability limit that you et for erroneouly rejecting H0, then you reject H0. Becaue, now the probability that you have made an error in rejecting H0 i lower than the limit you et for yourelf. Another way of doing the ame tet i, you can define Z calc = X μ 0 Similarly, you can find a Zcrit uch that,

2 P( Z > Z crit ) = α The only way for the p-value to be lower than α i for the Zcalc to be greater than Zcrit (ignore the ign). In ummary the hypothei tet procedure, 1) State H0 and H1 baed on the experiment ) Select a uitable value of α 3) Calculate p-value and/or Zcalc 4) Look up Zcrit from the table (ue α/ if ided tet and α if one ided tet) 5) Reject H0 i p value < α if 1 ided, α/ if ided or if Zcalc i greater than Zcrit. Confidence Interval and Hypothei Teting of a Population Mean (Variance Unknown) Thi time we do not know the variance. In uch a cae, it wa hown that the ample mean follow the T-ditribution, Confidence Interval X ~ T (μ, n ) Going by imilar procedure a earlier, the 1-ided (1-α) confidence interval to find a lower limit. μ l = X t α,n 1 In a imilar procedure, one can obtain the one ided confidence interval to find an upper limit to be, μ u = X + t α,n 1 Similarly, the two ided (1-α) confidence interval for the ample mean would be given by, But we are intereted in μ. So rearranging, P (μ tα,n 1 < X < μ + tα,n 1 ) = 1 α

3 μ (X ± tα,n 1 ) with (1 α) confidence Therefore, you are (1-α)100% confident that the true mean lie between thee two value. Thi i a -ided confidence interval. Hypothei Tet In the hypothei tet, intead of the Z-table we now ue the T-table. The probability of erroneouly rejecting the null hypothei i given by, P (T > X μ 0 ) = p value If the p-value i le than α, which i the probability limit that you et for erroneouly rejecting H0, then you reject H0. Becaue, now the probability that you have made an error in rejecting H0 i lower than the limit you et for yourelf. Another way of doing the ame tet i, you can define Similarly, you can find a tcrit uch that, t calc = X μ 0 P(t > t crit ) = α The only way for the p-value to be lower than α i for the tcalc to be greater than tcrit (ignore the ign). In ummary the hypothei tet procedure, 1) State H0 and H1 baed on the experiment ) Select a uitable value of α 3) Calculate p-value and/or tcalc 4) Look up tcrit from the table (ue α/ if ided tet and α if one ided tet) Reject H0 i p value < α if 1 ided, α/ if ided or i tcalc i greater than tcrit.

4 E 43 Spring 015 Recitation 9 April 10, 015 Example 1. A report give the 95% confidence interval for the mean concentration of benzene baed on 10 ample in a factory exit water i (7850, 7960). a. Find the ample mean and tandard deviation. b. The manufacturer claim that the exit water meet federal regulation with a mean of le than 7980 ppm of benzene. Tet the manufacturer claim at 0.5% ignificance.. Baed on 15 device, the lifetime of a biomedical device i found to have an average of hour with a tandard deviation of 6.1 hour. a. Can you conclude that the lifetime i greater than 5500 hour at the 5% ignificance level? Find the p-value. b. Contruct a 95% lower bound on the population mean. Practice Problem 3. A civil engineer i analyzing the compreive trength of concrete. Compreive trength i normally ditributed with a population variance of 1000 quare pi. A random ample of 1 pecimen ha a mean compreive trength of pi. a) Can you conclude that the mean compreive trength i not 3500 pi at the 5% ignificance level? Find the p-value. b) How many more ample are required to find the compreive trength to within 50 pi at 95% confidence? 4. A pot-mix beverage machine i adjuted to releae a certain amount of yrup into a chamber where it i mixed with carbonated water. A random ample of 5 beverage wa found to have a mean yrup content of ounce and a tandard deviation of ounce. a. Do the data upport the claim that the mean amount of yrup exceed 1 ounce at the 5% ignificance level? Give a range for the p-value. b. Find the 95% lower confidence bound of the yrup content.

5 E 43 Spring 015 Recitation 9 Solution April 10, a. μ (7850, 7960) 95% CI X + tα,n 1 = 7960 (1) X tα,n 1 = 7850 () X = X = 7905 From (1) b. H 0 : μ 7980 (7960 X ) = H 1 : μ < 7980 tα,n 1 = t crit = t 0.005,9 = 3.50 ( ) 10 t 0.05,9 [=.6] = t calc = X μ = = t calc < t crit Cannot reject H0. Cannot conclude manufacturer claim i true. X = n=15 =6.1 a. H 0 : μ 5500 H 1 : μ > < P value < 0.01

6 t calc = X μ = = t crit = t 0.05,14 = t calc > t crit Can reject H0. Lifetime i greater that 5500 hour. b. μ (X t α,n 1, ) 3. X = n=1 = X = < P value < (565.1 t 0.05,14 ) t0.05,14 = a. H 0 : μ = 3500 H 1 : μ 3500 Z crit = Z0.05 μ (55.30, ) Z calc = X μ = = = 1.96 Z calc < Z crit Cannot reject H0. Cannot conclude that compreive trength i not 3500 pi. b. μ (X ± Zα ) Z i.e. n=1537 (alway round up) P value = Zα = 50 = 50 n = ( ) = = 155 more ample are required.

7 n=5 =0.016 a. H 0 : μ 1 H 1 : μ > 1 α = 0.05 t calc = X μ 0 = = t crit = t 0.05,4 = t calc > t crit Can reject H0. Syrup content >1 P value < P value 0 b. μ (X t α,n 1, ) (1.098 t 0.05, , ) μ (1.093, )