assignment#4

assignment#4 - 7.40 a) 95% confidence interval would give...

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Unformatted text preview: 7.40 a) 95% confidence interval would give us the formula so therefore b) This means that at 95% confidence interval, the true mean range would be from 303.3105 to 413.5895. c) The two conditions required for the inference of part b is that the random sample is selected from the target population and that the population has a relative frequency distribution that is approximately normally distributed. Therefore these conditions are both satistfied so we can infer part b. d) No, we don’t agree since we know that the mean is 358.45 and that the median is 367.5. 7.82 a) At 90% confidence interval with a sampling error at 0.1 and a standard deviation of 2mm, we can find our sample size n required to estimate the population mean to a 90% confidence at 0.1 mm. b) If n = 100, then the confidence interval will be wider than the confidence interval of part a because with a higher n, the width of the confidence interval will decrease. c) For a maximum confidence level, we would prefer the 99% confidence interval in order to maximize our confidence level. 7.104 a) Our 95% confidence interval for the correlation values would be from 0.06716 to 0.48956. 7.112 a) We can infer that there a high risk of fatality of leaving the airbag turned on when a child is in the passenger seat b) If it turns out that 24 of the 35 children weren’t even wearing seatbelts, then our inference doesn’t work because we can no longer say that it was because of the airbags that they died. 8.32 We can find that the mean of the data is 11066.4328 and that the standard deviation is 1583.0125. a) The null hypothesis would tell us that µ = 10 000 at H0. The alternative hypotehsis would tell us that µ > 10 000 at Ha. Also we can know that our level of significance would be from 0.05. Then we know that our critical region would be from anything higher than 1.645. So then we can reject H0 since 5.5143 is higher than 1.645 and is within the critical region. b) A type I error would be to conclude that the mean heat rate of gas turbines isn’t 10 000 when the mean equals to 10 000. A type II error would be to conclude that the mean heat rate of gast turbines is 10 000 when the mean isn’t equal to 10 000. 8.68 Our rejection region is from any t value lower than ‐1.345 (0.10 level of significance) and because our t value is lower than what is required to be in the rejection region, we can conclude that the mean repellence of the new mosquisto is less than 95%. b) The two conditions that are necessary for our inferences is that a random sample is selected from the target population and that the population from where the sample is selected has an approximately normal distribution. 8.86 The rejection region for this problem is Z < ‐1.645 so therefore, our value is in the acceptance region. This means that the verb buy that are DOD’s is significant tot eh value 0.33 with a level of significance of 0.05. 8.152 H0: µ = 26 Ha : µ doesn’t’ equal 26 Our rejection region lies from < ‐2.0076 to > 2.0076. Therefore because our value is within the acceptance region, we do not reject our Ho hypothesis. b) 1 vs 16 2 vs 15 3. vs 14 4 vs 13 Our rejection regions lies at 1> 1.6753 Therefore 1 vs 16 and 2 vs 15 should have victories of more than 10 points since they lie within the rejection region but 3 vs 14 and 4 vs 13 shouldn’t since their t value lies within the acceptance region. c) In winning by an average of less than 5 points, Our rejection region lies at t < ‐1.6753 (51 degrees of freedom and 0.05 level of significance.) Therefore we can infer that 8 vs 9 will win by an average of less than 5 points in the first round of games since it lies within the rejection region. e) Therefore since the z value lies in the acceptance region (between ‐1.96 and 1.96), we can say that the average is a good predictor of the victory margin in the NCAA tournament games. 9.22 the rejection region is from < ‐2.012 and >2.012. therefore because our t value lies within the acceptance region, we conclude that there is no significant difference between the population mean numbers of high vfrequency vocal responses by the two methods. 9.48 The conditions required for valid small‐sample inferences are that it is a random sample of differences selected from the target population and that the population of differences has an approximately normal distribution. Both conditions are satisfied in this problem. Therefore, since the t value is in the rejection region of t> 1.729, we reject hypothesis H0. This means that there is no significant difference between the mean homophone confusion errors during time 1 and time 2. 9.70 Since our rejection region lies within < ‐ 2.576 and >2.576, our value is within the rejected region and we can conclude that there is a significant difference between the proportions of boys who gambled weekly or daily in 1992 and 1998. Yes I agree with the Professor that the interpretaions of the differences should include a judgment about the magnitude of the difference and its significance. Therefore we will use the confidence interval estimation of the difference. With this, we have a 99% confidence that the interval between that range of ‐ 0/014938 and ‐0.007052 contains the difference between the proportions of boys who gambled weekly/daily on any game in 1992 and 1998. Also because the interval is less than 0, we know that the proportion of boys who gamebled in 1992 is less than that in 1998. We can infer with 99% confidence that there’s between 1.49% and 0.7% less gamblers in 1992 than in 1998. 9.114 Therefore the 9% confidence interval for the difference between the mean spillage amount caused by collision and that caused by fires is (‐43.1889, 45.3555). b) Since the rejection region lies in > ‐2.069 ad > 2.069, our test statistic lies in the acceptance region meaning there is no significant difference between the spillages caused by grounding and that caused by hull failure. c) The conditions required for valid small sample inference is that the two samples are randomly selected in an independent manner from the two target populations, both sampled populations have approximately normal distribution and the population variances are equal. All conditions are met in the problem. 9.134 This means that we will need to sample at least 31 male students to estimate the difference between the population mean attitude towards fathers and mothers with a 90% confidence interval and a sampling error of 0.3. ...
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This note was uploaded on 10/12/2011 for the course MATH 203 taught by Professor Dr.josecorrea during the Spring '08 term at McGill.

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