1. Dependent and Independent Samples 1Section 10.1, Page 208

2. Problems 2Problems, Page 230

3. Problems 3Problems, Page 230

4. Confidence Interval – Dependent Samples Illustrative Problem sample mean = 6.33, Sx = 5.13, n = 6 Find the 95% confidence interval for the mean of the difference? (Sampling Distribution is t-distribution, 5df) C.I. = sample mean ± Margin of Error = sample mean ± critical value * standard error 6.33 ± 2.5706 * 6.33 ± 5.38 = (.95, 11.71) 4Section 10.2, Page 211

5. Confidence Interval – Dependent Samples Illustrative Problem TI-83 Black Box Program Find the 95% Confidence Interval d = B – A (We have to make the assumption that the data is taken from a normal population.) STAT – ENTER: Enter Brand A in L1 and B is L2 Highlight the title box for L3. =2 nd L2 – 2 nd L1 Enter (Difference set is now L3). STAT–TESTS- 8:TInterval Complete procedure as described in Section 9.1 Confidence Interval is (.96, 11.71) 5Section 10.2, Page 211

6. Hypothesis Test– Dependent Samples Illustrative Problem sample mean = 6.33, Sx = 5.13, n = 6 Test the hypotheses that the tread wear for Brand B is greater than for Brand A. H o : μ d = B – A =0 (No difference) H a : μ d = B – A > 0 (B is greater) μ d =0 p-value =.0146 p-value = PRGM – TDIST LOWER BOUND = 6.33 UPPER BOUND = 2 ND EE99 MEAN = 0 STANDARD ERROR = df = 5 Answer: p-value =.0147 (Reject Ho, B has greater tread wear.) 6Section 10.2, Page 211

7. Hypothesis Test– Dependent Samples Illustrative Problem – TI-83 Black Box Program Test the hypotheses that the tread wear for Brand B is greater than for Brand A. (d=Brand B – Brand A) H o : μ d = 0 (No difference in wear) H a : μ d > 0 (Brand B has more wear) STAT – ENTER: Enter Brand A in L1 and B in L2 Highlight L3 =2 nd L2 – 2 nd L3 Enter (L3 in now difference set) STAT-TESTS– 2:T-Test Complete the program as described in Section 9.2 with L3 as the difference set. P-Value =.0146 (Reject H 0, Brand B has greater tread wear.) 7Section 10.2, Page 211

8. Problems a.Test the hypotheses that the people increased their knowledge. Use α=.05 and assume normality. State the appropriate hypotheses. b.Find the p-value and state your conclusion. c.Find the 90% confidence interval for the mean estimate of the increase in test scores. 8Problems, Page 231

9. Problems a.State the appropriate hypotheses. b.Find the p-value and state your conclusion. c.Find the 98% confidence interval for the mean estimate of the increase in number of sit-ups. 9Problems, Page 232

10. Independent Samples Two Means: Unknown σ Suppose we want to compare the income of Shoreline students with UW students. From sample of 50 Shoreline Students: From sample of 60 UW students: Test the hypothesis that the populations incomes are not the same. To find a p-value we need a sampling distribution for 10Section 10.3, Page 215

11. Independent Samples Two Means: Unknown σ μ 1 σ 1 μ 2 σ 2 Population 1Population 2 Sampling Dist. 11Section 10.3, Page 215 A T-Distribution Applies

12. Independent Samples Two Means: Unknown σ Conditions for t-distribution model for sampling distribution: 1.The samples are independent of each other – two unrelated sets of sources are used, one from each population. 2.The populations are normal, or the sample size from each population is large, n ≥ 30 The correct degrees of freedom for the sampling distribution for the difference of two independent means is given by the following formula: The calculator will make this calculation. 12Section 10.3, Page 215

13. Independent Means – Illustrative Problem Confidence Interval – TI-83 Add-in Find the 95% confidence interval for the difference in the heights, μ m – μ f. C. I. = sample mean difference ± Margin of Error= sample mean difference ± critical value * standard error PRGM – STDERROR – 5: 2 MEANS Sx1 = 1.92; n1=30; Sx2 = 2.18; n2 = 20 Answer: SE =.6004; df = 37.2121 PRGM – CRITVAL – 2 C – LEVEL =.95; df (INTEGER) = 37 Answer: 2.0262 C.I. = 69.8-63.8 ± 2.0262 *.6004 = (4.78, 7.22) 13Section 10.3, Page 216

14. Independent Means – Illustrative Problem Confidence Interval – TI-83 Black Box Find the 95% confidence interval for the difference in the heights, μ m – μ f. STAT – TESTS – 0:2 – SampTInt Inpt: Stats : 69.8; Sx1: 1.92; n1: 30 : 63.8; Sx2: 2.18; n2: 20 C-Level:.95 Pooled: No Calculate Answer: (4.78, 7.22) 14Section 10.3, Page 216

15. Independent Means – Illustrative Problem Hypotheses Test– Add-In Programs Is this sufficient evidence to prove that male students are taller than female students. H o : μ m – μ f = 0  μ m = μ f Ha: μ m – μ f > 0  μ m > μ f 15Section 10.3, Page 216 PRGMS – TDIST LOWER BOUND = 6 UPPER BOUND = 2 ND EE99 MEAN = 0 STANDARD ERROR =.6004 (Slide 12) df = 37.2121 (Slide 13) Answer: p-value = 2.1930E-12 ≅ 0 (Reject Ho, there is sufficient evidence to prove males are taller.) 0 p-value ≅ 0

16. Independent Means – Illustrative Problem Hypotheses Test– TI-83 Black Box Is this sufficient evidence to prove that male students are taller than female students. H o : μ m – μ f = 0  μ m = μ f Ha: μ m – μ f > 0  μ m > μ f STAT – TESTS – 4: 2 – SampTTest Input: Stats :69.8; Sx1:1.92; n1:30 :63.8; Sx2:2.18; n2:20 μ1: >μ2 Pooled: No Calculate; Answer: p-value = 2.1947E- 12 ≅ 0 (Reject Ho, there is sufficient evidence to prove males are taller) 16Section 10.3, Page 216

17. Problems a.Write the necessary hypotheses. b.State the p-value and your decision. c.If you make an error, what type is it? d.Construct a 98% confidence interval for the difference mean selling prices (North of Cedar – Provo) 17Problems, Page 233

18. Problems a.Write the necessary hypotheses. b.State the p-value and your decision. c.If you make an error, what type is it? d.Construct a 98% confidence interval for the difference mean weight gained (Diet B – Diet A) 18Problems, Page 233

19. Problems Assume normality. a.Is there convincing evidence that nonorgan donors have higher anxiety about death than organ donors? Write the hypotheses and find the p-value. State your conclusion. b.Find the mean and standard error of the sampling distribution. c.Find the 98% confidence interval for the means; nondonor – donor. 19Problems, Page 232

20. Problems a.State the hypothesis (Assume Normality) b.Find the p-value, and state you conclusion. c.Find the mean and standard error of the sampling distribution d.Find the 95% confidence interval for the difference of the means; Gouda-Brie. 20Problems, Page 232

21. Independent Samples Two Proportions Suppose we want to compare the proportion of students getting financial aid at Shoreline and UW. From sample of 50 Shoreline Students: From sample of 60 UW students: Test the hypothesis that the population proportions are not the same. To find a p-value we need a sampling distribution for 21Section 10.3, Page 221

22. Independent Samples Two Proportions: Unknown σ p p1p1 p2p2 Population 1Population 2 Sampling Dist. p’ 1 Sampling Dist. p’ 2 Sampling Dist. p’ 1 – p’ 2 22Section 10.4, Page 221 The Normal Model Applies

23. Two Proportions Conditions for Normal Model The sampling distribution for the difference of two sample proportions is a normal model if: Samples are independent of each other Each sample size > 20 Each sample has more than 5 successes (np or np’) and 5 failures (nq or nq’). (q=1-p) Each sample is not more than 10% of its population 23Section 10.4, Page 221

24. Two Proportions – Confidence Interval Illustrative Problem - TI-83 Add-In A campaign manager wants to estimate the difference in support between women and men for his candidate. A sample of 1000 for each population was taken, and 459 women and 388 men favored his candidate. Find the 99% confidence interval for p w – p m. C.I. = sample proportion difference ± Margin of Error = sample proportion difference ± critical value * standard error PRGM - CRITVAL – 1:NORMAL DIST – CONF LEVEL =.99 Answer: 2.5758 PRGM – STDERROR- 2: 2 PROP: p1 = 459/1000; n1 = 1000; p2 = 388/1000; n2 = 1000 Answer: sample proportion difference =.0710; standard error =.0220 C.I. =.0710 ± 2.5758*.0220 = (.0143,.1277) We are 99% confident that the proportion of female support exceeds the true proportion of male support is between 1.43% and 12.77%. 24Section 10.4, Page 222

25. Two Proportions – Confidence Interval Illustrative Problem - TI-83 TI-83 Black Box A campaign manager wants to estimate the difference in support between men and women for his candidate. A sample of 1000 for each population was taken, and 459 women and 388 men favored his candidate. Find the 99% confidence interval for p w – p m. STAT – TESTS – B:2PropZInt x1 = 459 n1= 1000 x2 = 388 n2 = 1000 C-Level =.99 Calculate Answer: (.0142,.1278) 25Section 10.4, Page 222

26. Section 10.4, Page 22326 Difference between 2 Population Proportions For a hypothesis test, the null hypothesis is: H o : p 1 = p 2. If the two proportions are equal, then the standard deviation of the proportions must be also equal. Since p’ 1 seldom equals p’ 2, the σ (p’ 1 -p’ 2 ) does not reflect this fact. To get around this problem for hypothesis tests, we calculate the weighted average of p’ 1 and p’ 2, called p’ p, or pooled sample proportion, and substitute is value for both p’ 1 and p’ 2 in the standard error calculation.

27. Two Proportions – Hypotheses Test Illustrative Problem - TI-83 TI-83 Add-In A telephone salesman claims that his phones are better than the competition, and that no more of his phones are defective than the competition. A sample is taken. Can we reject the salesman’s claim? Use α =.05. H o : p s = p c H a : p s > p c PRGM – STDERROR – 3: 2-PROP POOLED p 1 = 15/150; n 1 = 150; p 2 = 6/150; n 2 = 150 Answer: p’ s – p’ c =.06; Standard Error =.0295 0.06 p-value =.0210 PRGM – NORMDIST – 1 LB =.06; UB = 2 nd EE99; MEAN = 0; SE( )=.0295 Answer: p-value =.0210. (Reject Null Hypotheses, and reject the salesman’s claim.) 27Section 10.4, Page 224

28. Two Proportions – Hypotheses Test Illustrative Problem - TI-83 TI-83 Black Box A telephone salesman claims that his phones are better than the competition, and that no more of his phones are defective than the competition. A sample is taken. Can we reject the salesman’s claim? Use α =.05. H o : p s = p c H a : p s > p c STAT – TESTS – 6:2PropZTest x1: 15 n1: 150 x2: 6 n2: 150 p1: >p2 Calculate Answer: p-value =.0208 (Reject the Null Hypotheses and the salesman’s claim.) 28Section 10.4, Page 224

29. Problems a.Test the hypotheses that the preference in city A is greater than the preference in city B. State the appropriate hypotheses. b.Find the p-value and state your conclusion. c.What model is used for the sampling distribution and what is the mean of the sampling distribution and its standard error? d.Find the 97% confidence interval for the difference in preferences, city A – city B. 29Problems, Page 235

30. Problems a.State the appropriate hypotheses. b.Find the p-value and state your conclusion. c.What model is used for the sampling distribution and what is the mean of the sampling distribution and its standard error? d.Find the 98% confidence interval for the difference in proportions, men – women. 30Problems, Page 234