Question.3172 - 1. Confidence interval for μ when σ is known Suppose n = 9 people are selected at random from a large population. Assume the heights of the people in this population are normal, with mean μ = 68.71 inches and σ = 3 inches. Simulate the results of this selection 20 times and in each case find a 90% confidence interval for μ. The following commands may be used: MTB > random 9 c1-c20; SUBC> normal 68.71 3. MTB > zinterval 0.90 3 c1-c20 a. [2] How many of your intervals contain μ? b. [2] What is the probability that 100 (not 20) such intervals would contain μ? c. Do all the intervals have the same width? [1] ........ Why [1] (what is the theoretical width)?———————————— d. [1] Suppose you constructed 80% intervals instead of 90%. Would they be narrower or wider?—————– e. [1] How many of your intervals contained the value 71?———————— f. Suppose you took samples of size n = 4 instead of n = 9. Would you expect more or fewer intervals to contain 71? [1] ...... What about 68.71? [1] ........ What about the width of the intervals for n = 4: Would they be narrower or wider than for n = 9? [1] ........ 1 2. Confidence interval for μ when σ is NOT known Repeat the simulation of Question 1 but now assume σ is unknown and use the tintervals command to get the 20 90% intervals: MTB > random 9 c1-c20; SUBC > normal 68.71 3. MTB > tinterval 0.90 c1-c20 a. [2] How many of your intervals contain μ?———————————- b. Would you expect all 20 of the intervals to contain μ? [0.5]........ Why? [1.5]————— c. Do all the intervals have the same width? [1]......... Why (what is the theoretical width)? [1]—————— d. [1] Suppose you took 95% intervals instead of 90%. Would they be narrower or wider?— ————————– e. [2] How many of your intervals contain the value 71?—————————- f. Suppose you took samples of size n = 64 instead of n = 9. Would you expect more or fewer intervals to contain 71? [1] ...... What about 68.71? [1]——————— What about the width of the intervals for n = 64: Would they be narrower or wider than for n = 9? [1] ......... 3. Hypothesis testing for μ when σ is known Imagine choosing n = 16 women at random from a large population and measuring their heights. Assume that the heights of the women in this population are normal with μ = 63.8 inches and σ = 3 inches. Suppose you then test the null hypothesis H0 : μ = 63.8 versus the alternative that Ha : μ 6= 63.8, using α = 0.10. Assume σ is known. Simulate the results of doing this test 30 times as follows: MTB > random 16 c1-c30; SUBC > normal 63.8 3. MTB > ztest 63.8 3 c1-c30 a. [2] In how many tests did you reject H0. That is, how many times did you make an “incorrect decision”? ........ 2 b. [2] Are the p-values all the same for the 30 tests? ...... c. Suppose you used α = 0.001 instead of α = 0.10. Does this change any of your decisions to reject or not?[1]........... In general, should the number of rejections increase or decrease if α = 0.001 is used instead of α = 0.10?[2]............. d. Now assume that the population really has a mean of μ = 63, instead of 63.8, and carry out the above 30 simulations, (thus, use the above minitab commands with ’normal 63.8 3’ changed to ’normal 63 3’. Once again, using α = 0.10 and assuming σ known, in how many tests did you reject H0?[2] ............. A rejection of H0 in part (a) is a “correct decision”. True or False? [0.5] ......... A rejection of H0 in part (d) is a “correct decision”. True or False?[0.5] .......... 4. Hypothesis testing for μ when σ is NOT known Repeat Question 3, using ttest instead of ztest, and answer parts (a), (b), and (c) again. (Thus ‘ztest 63.8 3 c1-c30’ changes to ‘ttest 63.8 c1-c30’) a. [2] In how many tests did you reject H0. That is, how many times did you make an “incorrect decision”? .......... b. [1] Are the p-values all the same for the 30 tests? ....... c. Suppose you used α = 0.00008 instead of α = 0.10. Does this change any of your decisions to reject or not? [2] ........... In general, should the number of rejections increase or decrease if α = 0.00008 is used instead of α = 0.10? [2] ............ Part II Comprehension questions 5. A fast food franchiser is considering building a restaurant at a certain location. According to a financial analysis, a site is acceptable only if the number of pedestrians passing the location averages more than 100 per hour. A random sample of 50 hours produced ?x = 110 and s = 12 pedestrians per hour. (a) [6] Do these data provide sufficient evidence to establish that the site is acceptable? Use α = 0.05. (b) [2] What are the consequences of Type I and Type II errors? Which error is more expensive to make? (c) [4] Considering your answer in part (b), should you select α to be large or small? Explain. 3 (d) [3] What assumptions about the number of pedestrians passing the location in an hour are necessary for your hypothesis test to be valid? 6. An experiment was conducted to test the effect of a new drug on a viral infection. The infection was induced in 100 mice, and the mice were randomly split into two groups of 50. The first group, the control group, received no treatment for the infection. The second group received the drug. After a 30-day period, the proportions of survivors, ˆp1 and ˆp2, in the two groups were found to be 0.36 and 0.60, respectively. (a) [6] Is there sufficient evidence to indicate that the drug is effective in treating the viral infection? Test at 5% significance level. (Make sure to state your null and alternative hypotheses.) (b) [6] Use a 95% confidence interval to estimate the actual difference in the cure rates, i.e. p1 −p2, for the treatment versus the control groups. [3] Based on this confidence interval can you conclude that the drug is effective? Why? 7. In an investigation of pregnancy-induced hypertension, one group of women with this disor- der was treated with low-dose aspirin, and a second group was given a placebo. A sample consisting of 23 women who received aspirin has mean arterial blood pressure 111 mm Hg and standard deviation 8 mm Hg; a sample of 24 women who were given the placebo has mean blood pressure 109 mm Hg and standard deviation 8 mm Hg. (a) [6] At the 0.01 level of significance, test the null hypothesis that the two populations of women have the same mean arterial blood pressure. Justify any procedure you use. (b) [5] Construct a 99% confidence interval for the true difference in population means. [1] Does this interval contain the value 0? [3] Based on this confidence interval, what is you conclusion regarding the effect of the two treatments on the blood pressure of pregnant women? 8. [7] A company is interested in offering its employees one of two employee benefit packages. A random sample of the company’s employees is collected, and each person in the sample is asked to rate each of the two packages on an overall preference scale of 0 to 100. Results were 4 Employee Program A Program B 1 45 56 2 67 70 3 63 60 4 59 45 5 77 85 6 69 79 7 45 50 8 39 46 9 52 50 10 58 60 11 70 82 Do you believe that the employees of this company prefer, on the average, one package over the other? Explain. 9. In an attempt to compare the starting salaries for university graduates who majored in education and the social sciences, random samples of 100 recent university graduates were selected from each major and the following sample information was obtained: Major Mean St. Dev. Education $50,554 $2225 Social Science $48,348 $2375 Conduct an appropriate hypothesis test at the 5% level of significance to determine if there is a difference in the average starting salaries for all university graduates who majored in education and the social sciences. Conduct an appropriate hypothesis test at the 5% level of significance to determine if there is a difference in the average starting salaries for all university graduates who majored in education and the social sciences. Conduct this test using (i) [3] the p-value method , (ii) [3] the critical value method , and (iii) [3] the confidence interval method.
Answer Below:
1a) 18 of my intervals contains µ but anything above 14 is also acceptable. 1b) 0.90 1c) Yes. The theoretical width is 2*z 0.05 (3/) =2*z 0.05 (3/3) =2*z 0.05 1 =2*1.645 =3.29. 1d) Narrower. 1e) 9 of my intervals contained the value 71, but any number in [5, 15] is acceptable. 1f)More. More. Wider. 2a) 16 of my intervals contain µ but anything above 14 is also acceptable. 2b) No. Expected is 20*0.9=18. 2c) 2*t 0.05 (s/), s varies from sample to sample. 2d) Wider. 2e) 9, but any number in [5, 15] is acceptable. 2f) Fewer. Fewer. Narrower. 3a) 4 p- values less than 0.10 but any number ≤9 is acceptable. 3b) No 3c) Yes, but some may report No. 3d) 11 p-values 0.10 but any number in [3,22] is acceptable. 4a) 3, but any ≤9 is acceptable. 4b) Yes, but some may report No. Decrease. 5a) H 0 : µ = 100 vs. H α : µ > 100, since n is large enough, test statistic is z =(x- µ 0 )/(s/) =(110-100)/(12/) = 5.89. Null hypothesis is rejected since 5.89 > 1.645 = z 0.05 5b) Type I error is rejecting H 0 when it is true and the result of this error is to construct the site when there are not enough pedestrians passing the location (bankruptcy). Type II error is to accept H 0 when it is not true, so the site does not construct in location more than 100 pedestrians (missing a good location). Type I error is more expensive. (It is possible type II error be more expensive. Accept if it supports with good explanation) 5c) According to part b, α should be a small number so as to reduce the errors. Greater the value of α, the value of Z will be larger and it will accept the null hypothesis. 5d) Pedestrians are selected randomly. 6a) H 0 : (p1−p2) = D 0 = 0 Hα : (p1−p2) < D 0 = 0 (Our test is a one-tailed test.) At the 5% signi?cance level, α = 0.05. For rejection region, z = 1.645 Area = 0.05 For the sake of being thorough, note that n1p1= 18; n1q1= 32; n2p2= 30; n2q2=20. Since all of these are greater than or equal to 5, we can assume normality. 6b) First note, that since the hypothesis is centered about zero, we will use a pooled estimator for p, when computing s p1−p2 . p = (x1+x2)/(n1+n2) =(18+30)/(50+50) = 0.48 σ p1−p2 ≈ s p1−p2 = = =0.09992 Our test statistic is z =((p1−p2)−D 0 )/σ p1-p2 =((0.36−0.60)−0)/0.09992 = −2.4019 Since z = −2.4019 is in the rejection region, we reject H 0 and accept H α . There is su?cient evidence to indicate that the drug is e?ective in treating the viral infection. 7a) Level of significance, α=0.01. Degrees of freedom, d.f. =n 1 +n 2 -2 =8+8-2 =14 From t-table, since our alternative hypothesis is 2-tailed, the critical values t=±2.977. Decision Rule: Reject H 0 if t<-2.977 or if t>2.977, else null hypothesis holds valid. Test statistic: t=(x1-x2)/(s p () S p = n1=23 n2=24 s1=s2=8 Putting the values in the above equation, s p =8 t=(111-109)/(8() t≈0.857 Since t<2.977, therefore null hypothesis is accepted. So there is no statistical evidence that the two populations of women have the same mean arterial blood pressure. 7b) Sample n x-bar s A 23 111 8 B 24 109 8 8,8 %99,24,23,109,111 21 2121 ?? ????? ss Cnnxx df = min {23 – 1, 24 – 1} = 22 ? t* = 2.82 The critical t-value from the table using two tailed is 2.82 Confidence interval = ???? 24 8 23 8 82.2)109111( 22 (-4.58, 8.58) Yes the interval contains the value 0. Based on this confidence interval, we are 99% sure that the effect of the two treatments on the blood pressure of pregnant women is (-4.58, 8.58). So there is no statistical evidence that the two populations of women have the same mean arterial blood pressure. 8) Assuming each data set comes from a normal distribution and taking α= 0.05 H 0 : µ 1 - µ 2 =0 vs. H α : µ 1 - µ 2 ≠0 Or H 0 : µ d vs. H α : µ d ≠0 It is paired data where d i : -11, -12, 3, 5, 8, 21, -3, -2, 6, -1, -12. d =∑d i /n = -14/11 = -1.27 and s d 2 : (998-17.82)/10 = 98.02 Test statistic t= (d-0)/(s d /) =(-1.27-0)/(9.9/) =-0.43 For n-1 = 10 degrees of freedom, t 0.025 = 2.228, so H 0 is not rejected. 9) The hypotheses are: H 0 : µ 1 - µ 2 =0 vs. H α : µ 1 - µ 2 ≠0 Since our sample sizes are greater than 30, the test statistic is Z= {(x 1 -x 2 )-D 0 )}/() = {(50554 48348) -0}/ =6.78 At α = 0.05, the critical values are z = 1.96. H 0 is rejected since 6.78 > 1.96 Calculate the p-value p-value = 2P(Z > 6.78) ≈ 0 Compute 95% confidence interval as follows (x1-x2)±z 0.025 ) = (50554-48348) 1.96 ± =(1568, 2844)More Articles From Statistics
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