Statistical Analysis of Two Population Means: Hypothesis Testing and Confidence Intervals , Assignments of Mathematical Statistics

Download Statistical Analysis of Two Population Means: Hypothesis Testing and Confidence Intervals and more Assignments Mathematical Statistics in PDF only on Docsity! HW1 7.28 8.42 We must assume that the distribution of the LOS's for all patients is normal. a. For confidence coefficient .90, a = Appendix B, with df=n—1=20- interval is: ~ 90 = .10 and a/2 = 10/2 =.05. From Table VI, 19, fos = 1.729. The 90% confidence i. Fl pe 38 1.72942. 538+ 464 (3.336, 4.264) v20 We are 90% confident that the mean LOS in 2003 is between 3.336 and 4.264 days. “90% confidence” means that if repeated samples of size n are selected from a Population 90% confidence intervals are constructed, 90% of all intervals thus constructed will contai the population mean. The p-value is p= P(z> 1.29}4P(z < -1.29)=(.5-.4015)+(.5 -.4015)=.1970. (Using Table IV, Appendix B.) The p-value is p= P(z 21.29)= (.5~.4015) =.0985. (Using Table IV, Appendix B.) 52.3-51 ot S088 10.4/-J/50 The p-valuc is p= P(z> 0.88}4+P(z < -0.88) = (.5-.3 106) 1 (.5 -.3106) =.3788 . (Using Table IV, Appendix B.) In part a, in order to reject Hy, @ would have to be greater than .1970. In part b, in order to reject Ho, @ would have to be greater than .0985. In part c, in order to reject Ho, a would have to be greater than .3788. For a two-tailed test, a/2 = 01/2 = .005. From Table TV, Appendix B, 2,9; = 2.58. a2 5759-523) 9 59 § -59.3- 51-5 3649513 9 -3.56 Oo; s/¥so 50 For a one-tailed test, @=.01. From Table IV, Appendix B, zo, = 2.33. x Ae 5 2,33 = 23S 9 938 593-51=5.39959 13-55 =3.95 oO; s/V50 50 8.45 a)H 0 : u=16.5 vs Ha: u>16.5 , where n=200, mean=17.05 , S=5.21 b) P-value=.068 , at ά =.05, we reject null hypothesis, at ά =.10, we accept null hypothesis. c) A two-tailed test is inappropriate because the alternative hypothesis is greater than 16.5, not unequal to 16.5. 9.23 Compare means of two populations, we use 2-sample t test H 0 : u1= u2 vs Ha: u1≠ u2 , at ά =.05 Pooled standard deviation: S2 =60.526, Use formula at text page 491, we get t=.29< t ά =.025 , v Note: v is degree of freedom calculated from text page 492 Conclusion: we can’t reject null hypothesis, no trend exists. 9.87 a) 90% confidence interval is (3.59,4.21) b) H 0 : u1= u2 vs Ha: u1≠ u2 Use formula at text page 491, t=20.6> t ά =.005 , v Note: v is degree of freedom calculated from text page 492 Conclusion: we reject null hypothesis, two sample means are not equal. c) Assume n1=n2=n, we use formula at text page 522, n=346 Yi & bh 38% Lowes Verabie 8 mn Stbev 36 Kean Bound 2 Lifeexf 108 70.15 19.57 LoL 67.60 5.C9 P-value is less than @, statistic Z >critical Z, thus we reject Hp. Repeat parts d)-f), but now test for more than 70 years with a=.01. Report the p- value and interpret. Is your conclusion different than in part f)? Explain why He: u=70 Ta: w> 70 We use Z-test One-Sample Z: lifeext Test of ma 70 va > The asstwe? slaaxard deviation = 1¢. 99% Lower Variabl N Mean StDev SB Year Becpd Pp rey o BI80 0.13 9,438 ritical Z, thus we don’t reject He liteex! 1c3 70. P-value is greater than a, statistic Z < Suppose we wish to examine whether the mean life expectancy for women is Tonger in couniries where daily the calorie intake is high (2500). First ereate the bwo samples of countries based on calorie intake and then test the claim at a=.05 Report below appropriate mull and alternative hypotheses, the value of the test statistic and its p-value. Do you reject the clainy? Hint: Follow the steps in the workbook py 15 Ho: portti=0 Haz wotr>0 Two-Sample T-Test and Cl: lifeexf, calhigh Tyo-sample T “or Litecet calhigh X Bean SzDev sh Mean a 3287.80 9,03 17 1 76.6 b.1G 0.77 Difference - ma (9) - wa ti mate tor diffure: 954 ord T-Tect of di! 4, -15.07) not =) Yelue = 5.9938 We would reject He, hecause P value is less than a. Now examine whether the mean life expectancy for women is higher than that for men, Tirst obtain a 95% Cl for the dilTerenes between average life expectancies ‘Test the claim at level =.05, and discuss the results. Why is a paired t-test appropriate in this case’ Paired T-Test and Cl: lifeexf, lifeexm  i, Paired Y Ser 1ifeaxf - lifeexr N Mean St¥ev SK yean lifee: 109 ts Vireaxm ie a Difterenze 1D¥ boa, s net — 0): T-Value = 24.11 P 55% CT for meaa Dest of meen 2.900 95% C'] shows that the difference between mean life expectancy is positive. So we are 95% confident to say that mean life expectancy for women is higher than the one for men. AL.05 a level . we reject the Ho because p value is less than a. Because we are examining (he same variables using different samples Finally, produce a scatter plot belween the two Life expectaneies. Is there a linear trend? What does it mean in terms of the variables involved? Oblain the Pearson cortelation coefficient to help you interpret these results. Seatterplot of lifeesf vs fifeexm Trisa linear trend. It means when everything else does not change, life expectancy for women increases when for men increases. Correlations: lifeexf, lifeexm ca ~ 0.982 aassen correlation Pa¥elae ~ 9,000 Correlation of 0.982 means there is a strong linear relationship between life expeetaney for women and that for men, 8.108 a. Since the standard deviation is almost the same as the mean, and we know that fal intake cannot be negative, the distribution of fat intake per day is skewed to the right. b. To determine if the mean fat intake for middle-age men on weight-control programs exceeds 30 grams, we test: He: w= 30 Hy p> 30 The wet statistic isz= 24 37-90. _ 95 oO, 32/J64 The rejection statistics requires @= .10 in the upper tail of the z-distribution. From Table TV, Appendix B, z19 = 1.28. The rejection region is z > 1.28 Since the observed value of the test statistic falls in the rejection region (z = 1.75 > 1.28), Ho is rejected. There is sufficient evidence to indicate the mcan fat intake for middle-age men on weight-control programs exceeds 30 grams at a= .10. e. Kor @= 05, the rejection region requires @= .05 in the upper tail of the z-distribution. From Table [V, Appendix B, zs = 1.645. The rejection region is z> 1.645. Since the observed value of the test statistic falls in the rejection region (z = 1.75 > 1.645), Ho is rejected. The conclusion is the same. Kor a= .01, the rejection region requires a = .01 in the upper tail of the z-distribution. From Table IV, Appendix B, zo, = 2.33. The rejection region is z > 2.33. Since the observed value of the test statistic does not fall in the rejection region (z= 1.75 # 2.33), Ap is not rejected. The conclusion is now different. esponses for all business students who have access to 9.20 a. The first population is the set of re Jecture noles and the second population is the set of responses for all business students not having access to lecture notes.