3 Inference comparing two population means

1、STATISTICS AND THE TI-84Lesson #11 Inferential Statistics: Two populations1. Inference: comparing two population meansExercise 1. There are two populations. A sample of size 120 from one of the populations gave a mean of 15 and a standard deviation of 1.3. A sample of size 88 from the other populati

2、on gave a mean of 13.5 and a standard deviation of 1.5. Find a 98% confidence interval for the difference between the population means.Sample information: · STAT TESTS 9 (2-Samp Z Int) Stats ENTER· (enter the values) s1 = 1.3 s2 = 1.5 C-Level: .98· Calculate ENTERResult: (1.0368, 1.96

3、32)Exercise 2: The average length of twenty trout caught in a lake was 10.8 inches with standard deviation of 2.3 inches, and the average length of fifteen trout caught in another lake was 9.7 inches with standard deviation of 1.5 inches. Construct a 90 percent confidence interval for the difference

4、 in the true mean lengths of trout in the two lakes.· STAT TESTS 0 (2-Samp t Int) Stats ENTER· Calculate ENTER Results: pooled: (-.0561, 2.2561) df=33 Sxp=2.00007576· STAT TESTS 0 (2-Samp t Int) Stats ENTER· Calculate ENTER Results: non-pooled: (.00993, 2.1901) , df=32.48284419 -

5、99-Exercise 3. A farmer tried Feed A on 256 cattle and Feed B on 144 cattle. The mean weight of cattle given Feed A was found to be 1350 pounds with a standard deviation of 180 pounds. On the other hand, the mean weight of the cattle given Feed B was found to be 1430 pounds with a standard deviation

6、 of 210 pounds. At the 5 percent level of significance, is Feed B significantly better than Feed A? Find the p-value.Reject the null hypothesis is p-value<0.05· STAT TESTS 3 Stats ENTER·· Calculate ENTERConclusion: z=3.845 p-value = 6.0203004E-5 <0.05Reject the null, we have sta

7、tistical evidence to believe that Feed B is significantly better than Feed AExercise 4. At a certain university twelve voters were picked at random from those who are in favor of impeachment of the president, and ten were selected at random from those who are against. The following table give their

8、ages.In favor 27 34 28 30 29 50 30 44 29 32 41 35Against 31 36 43 40 32 48 30 29 42 49 H0: m1-m2=0 H1: m1-m20At a 10% level of significance, is it true that the age of those in favor of impeachment significantly differs from the age of those against. Use a two sample t-test with pooled variance.Deci

9、sion Rule: Reject H0 if p-value <.05· 27, 34, 28, 30, 29, 50, 30, 44, 29, 32, 41, 35 STOÞ L1 ENTER· 31, 36, 43, 40, 32, 48, 30, 29, 42, 49 STOÞ L2 ENTER· STAT TESTS 4 DATA ENTER·· List 1: L1 List 2: L2 Freq 1: 1 Freq 2: 1· m1: m2 Pooled: Yes ENTER· Cal

10、culate ENTER Conclusion: t=-1.246579 p-value =.2269 > 0.10Do not Reject the null hypothesis, we have statistical evidence to believe that there is no significant difference between the ages of those in favor of impeachment and those against.-100- Exercise 5; Paired t-test. Dr. Williams claims tha

11、t the special diet that he recommends significantly reduces weight. A sample of eight persons was selected and they were put on the diet for a period of 6 weeks. The table below shows the weights (in pounds) of those eight person before and after dieting.Before182180195178177221198208After1681831871

12、69161204194196a) Construct a 99% confidence interval for the mean difference md in weight before and after using the dieting recommended by Dr. Williams. Use a paired difference .· 182,180, 195,178, 177, 221, 198, 208 STOÞ L1 ENTER· 168,183, 187,169, 161, 204, 194, 196 STOÞ L2 EN

13、TER · L1-L2 STOÞ D ENTER· STAT TESTS 8 (T Interval) · DATA ENTER List: D Freq: 1 C-Level: .99 · Calculate ENTERanswer: (1.3401, 17.91) b) Using a 1% level of significance, can you conclude that the mean weight loss for all persons due to this special diet is greater than zer

14、o?· STAT TESTS 2 (T-test) · DATA ENTER · mo: 0 List: D Freq: 1 >mo ENTER· Calculate ENTER· t=4.0655 p-value=.002388076 Conclusion: Reject H0 because p-value < a. We have statistical evidence at a 0.01 significance level to believe that the special diet recommended by D

15、r. Williams is effective.-101- 2. Inference: comparing two population proportionsExercise 6. In a study to estimate the proportion of residences in a certain city and its suburbs that subscribe to a certain magazine, it is found that 63 of 120 urban residences subscribe while only 34 of 125 suburban

16、 residences subscribe. Find a 90% confidence interval for the difference in the proportion of urban and suburban residences that subscribe to writer's digest.· STAT TESTS B (2-PropZInt.)· X1: 63 N1: 120 X2: 34 N2: 125 C-Level: 0.90 · Calculate ENTERanswer: (.15346, .35254)Problem

17、7. A jar containing 130 mosquitoes was sprayed with an insecticide of Brand A and it was found that 98 of them were killed. When another jar containing 150 mosquitoes of the same type was sprayed with Brand B. 120 of them were killed. At the 2 percent level of significance, do the two brands differ

18、in their effectiveness?Decision rule: Reject the null hypothesis, if p-value<.02· STAT TESTS 6 X1: 98 N1: 130 X2: 120 N2:150 p1: p2 · Calculate ENTER answer: z= -0.9276414111 p-value =0.3535935629Do not reject the null. We dont have statistical evidence to believe that the brands differ in effectiveness, p-value >.02Problem 8. During the first exam, the average grade of 7 men in a statistic exam was 78 points with a standard deviation of 5.3. For 11 women in the same class the average grade was 83 with a standard deviation of 6.2. At the 5% level of significance can w