Statistics 350 – Winter 2009: Solutions to Recommended Problems for HW 8 Textbook: Independent Samples t Procedures 9.72 a. Define 1 to be the mean time symptoms would last if everyone in the population were to take the placebos and 2 to be the mean time symptoms would last if everyone in the population were to take the zinc lozenges. The parameter of interest is 21 . In words, this is the mean of the additional time the symptoms would last for everyone in the population, if they were to take placebos instead of zinc lozenges. b. Define 1x to be the mean amount of time symptoms lasted for the 23 participants who took placebos and 2x to be the mean amount of time symptoms lasted for the 25 participants who took zinc lozenges. The sample statistic is 21 xx . c. Because the sample sizes are small the populations of times symptoms would last under the two conditions must both be bell-shaped. 9.73 a. The mean is 21 = 8 – 4.5 = 3.5 days. b. The standard deviation is 514.0 232255.1 2 2 . c. The figure is centered at 3.5 and has a standard deviation of 0.514. 11.41 a. Conclude that there is a difference between the population means. The value 0 would occur if there was no difference between population means. The interval encompasses the plausible values for the population mean difference. So if the interval does not cover 0 it says that a mean difference of 0 is not plausible. b. We cannot conclude that there is a difference between the population means. The value 0 would occur if there was no difference between population means. So, if the interval covers 0 we cannot reject a statement (hypothesis) of no difference. We can only conclude that the true difference is likely to be somewhere in the range covered by the interval. 11.43 a. Females = 12.55; Males = 13.65; difference = 1.10 (−1.10 if computed as Females − Males as in output). b. −3.24 to 1.04. With 95% confidence, we can say that the difference (Females − Males) in number of letters printed in the populations of females and males is between −3.24 and 1.04. c. We cannot say there is a difference because the confidence interval includes the value 0. d. 0 6 7.177.074.0][][ 222221 SESE Equivalently, compute 3446.42901.4 22 . e. 2.00 11.45 a. 21 xx = 57.5 – 55.3 = 2.2 cm. b. s.e.( 21 xx )= 5.0 368.1364.2 22222121 nsns cm. c. Approximate 95% confidence interval for the difference in population means is sample estimate ± 2× Standard error, which is 2.2 ± (2 × 0.5), or 1.2 to 3.2 cm. 13.35 a. H0: 21 = 0, or equivalently 1 = 2 versus Ha: 21 0, or equivalently, 1 2 b. −2.16 c. 17.23 8 6 8.56 4.1 2 2 79.8081 3 4 0)4.4 2 93 0 7( 22 (Differs from reported t = −2.16 due to rounding.) 13.37 Step 1: H0: 21 = 0, or equivalently 1 = 2 versus Ha: 21 0, or equivalently, 1 2 , 1 = mean reported fastest ever speed for population of college women 2 = mean reported fastest ever speed for population of college men Step 2: The sample sizes are sufficiently large to proceed. We assume, for the question of fastest ever driving speed, that the sample represents a larger population of college students. For the unpooled procedure, we will use the conservative df = 86, the smaller of 11n and 12n , and the test statistic is 09.835.2 19 87 4.17 102 4.14 4.1074.880)( e r r o r s t a n d a r d N u l l v a l u eN u l l-s t a t i st i c S a m p l e 22 2 22 1 21 21 n s n s xxt 13.43 a. Step 1: H0: 21 = 0, or equivalently 1 = 2 versus Ha: 21 >0, or equivalently, 1 > 2 , (mean number of hangover symptoms is higher if parents have problems) where 1 = mean number of hangover symptoms in the population of college students whose parents have alcohol problems and 2 =mean number of hangover symptoms in the population of college students whose parents do not have alcohol problems Step 2: We must assume the samples represent random samples from the larger populations of students of both types. The sample sizes are large enough to proceed with showing any graphical displays. Using the unpooled procedure, the test statistic is 15.4 24.0 01e r r o r st a n d a r d N u l l v a l u eN u l l-st a t i st i c S a m p l e t . Details are: Sample statistic is 19.49.521 xx symptom and the standard error = 24.0 94540.328260.3 22222121 nsns Step 3: Using the Table A.3 and the conservative df = smaller of sample sizes – 1, df = 282 – 1 = 281, all we can say is that the p-value is less than 0.002. Steps 4 & 5: The null hypothesis can be rejected and concluded that the mean number of hangover symptoms for students whose parents have alcohol problems is higher than for students whose parents do not. b. Step 1 and the “checking the conditions” part of Step 2 are the same as in part (a). For the pooled procedure, 28.42339.0 01 t . Details are: 447.388.112945282 4.3)1945(6.3 )1282( 22 ps , and pooled standard error is 2 3 3 9.09 4 512 8 214 4 7.311).(. 2121 nnsxxes p Step 3: df = 282 + 945 – 2 = 1225. From Minitab, p-value = .000. Using Table A.3 with df = 1000, the p-value range is less than .001. Steps 4 & 5: The null hypothesis can be rejected and concluded that the mean number of hangover symptoms for students whose parents have alcohol problems is higher than for students whose parents do not. c. The results of the two procedures are similar, with test statistics of 4.15 and 4.28. The sample standard deviations are very close (3.4 and 3.6) so it is not unreasonable to use the pooled procedure. 13.44 a. No. The null value is covered by the 95% confidence interval. b. Yes. The null value is not covered by the 95% confidence interval. jonksky Recommended Problems