P(every one of the seven intervals covers the true mean yield at its location) = 3.


 Bernard Dean
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1 1 Let = number of locations at which the computed confidence interval for that location hits the true value of the mean yield at its location has a binomial(7,095) (a) P(every one of the seven intervals covers the true mean yield at its location) = 3 The overall confidence level for the seven simultaneous statements is only 689% (b) 2 (a) (b) 3 Let Now, if and only if (a) Let has a under the alternative value (b) Let has a under the alternative value
2 (c) The power to detect the alternative mean value 295 is greater than for the alternative mean value 299 since it is easier to detect a value of µ which is further away from the null mean value (a) The boxplot suggests that the distribution of the data values is fairly symmetric No outliers are identified by the plot The value 110 litres/minute for the mean maximum voluntary ventilation value considered in part (c) is close to the center of the distribution of the data values as measured by the median (b) Let µ be the mean maximum voluntary ventilation value for this population of healthy college seniors 1116 litres/minute litres/minute litres/minute
3 For 95% confidence, A 95% confidence interval for µ is given by We are 95% confident that the mean maximum voluntary ventilation value for this population of healthy college seniors is between 8525 litres/minute and litres/minute (c) litres/minute litres/minute Since the value 110 is inside the 95% confidence interval in (b), these data are not statistically significant at the 5% level At the 5% level, one would accept At the 5% level, these data are consistent with the mean maximum voluntary ventilation value being 110 litres/minute (d) The analysis assumes the observations are independent and normally distributed Since our sample is a simple random sample, the observations are independent if the population size is much larger than the sample size Since there are no outliers and, even if the distribution of the maximum voluntary ventilation values is even somewhat skewed in the population, and hence not normally distributed, our confidence level will still be reasonably accurate In fact, the normal quantilequantile plot (or equivalently, normal probability plot) does not suggest that there is a significant problem with the normal assumption that would affect the analysis It could easily arise even with independent and normally distributed data with
4 5 Let µ be the mean diameter (in mm) of a skin test reaction for the population of adolescents who might participate in an immunologic study of this kind mm mm mm (a) mm mm Since we reject in favour of at the 1% level These data provide evidence at the 1% level that the mean diameter of the skin test reaction to the antigen is less than 30 mm (b) for 98% confidence mm A 98% confidence interval for µ is given by: We are 98% confident that the mean diameter of the skin test reaction for this population of adolescents is between 17 mm and 25 mm (c) The key assumption here for the analysis to be valid is that the diameters from different subjects be independent If different subjects respond independently to the skin test, this would be the case Since, the normal assumption is not necessary and the stated confidence level and significance levels will be reasonably accurate even if the distribution of diameter of the skin test reaction is strongly skewed in the population
5 6 It was expected that the blood stream concentration of a particular antibiotic one hour after its administration could vary substantially from individual to individual The study was therefore designed as a matched pairs study where the same individual was exposed to both antibiotics This allows a comparison between the two antibiotics while holding subject constant Person Penicillin Amoxicillin Difference (d=penamox) Let be the mean difference in blood stream concentration between penicillin and amoxicillin one hour after administration g/ml g/ml g/ml (a) Large values of provide evidence against in the direction of Using table C3, These data are consistent with the mean blood stream concentrations one hour after administration being the same for the two antibiotics In particular, these data are not even statistically significant at the 20% level (b) for 95% confidence A 95% confidence interval for is given by:
6 With 95% confidence, the mean blood stream concentration one hour after administration of pencillicin could be as much as 874 µg/ml less than amoxicillin or as much as 474 µg/ml more than amoxicillin according to our data (c) We assume that the differences are independent This will be the case if the subjects respond independently Independence of the two responses within subject is not necessary for this analysis We also assume that the differences are normally distributed It is important that this be based on past experience since these data alone possess very little information regarding this issue Moreover, since is very small, we cannot rely on the robustness properties of the onesample procedures associated with larger sample sizes The normal quantilequantile plot (or equivalently, normal probability plot) does not indicate a departure from normality 7 Let the mean protoporphyrin level for the population of adult male alcoholics with ring sideroblasts in the bone marrow that were sampled Let the standard deviation of the protoporphyrin level for the population of adult male alcoholics with ring sideroblasts in the bone marrow Let the mean protoporphyrin level for the population of apparently healthy nonalchoholic adult males that were sampled Let the standard deviation of the protoporphyrin level for the population of apparently healthy nonalchoholic adult males that were sampled
7 The ratio of the largest sample variance to the smallest sample variance which is much greater than a typical cutoff value of 3 (or the more conservative cutoff value 2) often used in practice to rule out the safe application of pooled procedures that assume equal population standard deviations when sample sizes are not nearly equal as is the case here If the protoporphyrin level in each population is normally distributed (not indicated here!), the two critical values for a test of : against : with are and Since we would as expected reject This test is neither necessary nor recommended in this situation It is highly nonrobust to departures from normality (a) Satterthwaite s degrees of freedom are: Large values of provide evidence against in the direction of Using Table C3, These data provide extremely strong evidence that the mean protoporphyrin level is higher in the represented alcoholic population than in the nonalcoholic population In particular, these data are statistically significant at the 05% level (b) To get the critical value for 99% confidence using table C3 we can round down to 50
8 The 99% confidence interval for is given by: 390) With 99% confidence, the mean protoporphyrin level is between 200 and 390 units greater in the represented alcoholic population than in the nonalcoholic population Using a computer, the value of without rounding is which yields the same interval in this case 8 (a) This is a completely randomized design (b) The distribution of the data values for the sensory deprivation treatment group is shifted towards lower values and has greater spread than that of the control group In particular, at least 75% of the data values in the control group are above the third quartile of the sensory deprivation treatment group There is a single outlier that was identified for the control group
9 (c) Let the mean alphawave frequency for a randomly selected individual when exposed to the sensory deprivation treatment Let the mean alphawave frequency for a randomly selected individual when treated as a control subject = Large values of provide evidence against in the direction of Using table C3, Hence, these data provide very strong evidence that the mean alphawave frequency under the deprived sensory treatment is different from that of the control treatment In particular, these data are statistically significant at the 05% level (d) for 95% confidence The 95% confidence interval for is given by:
10 We are 95% confident that the mean alphawave frequency under the deprived sensory treatment is between 030 and 130 units lower than under the control treatment (e) It seems reasonable to expect that the two samples are independent Provided exposure to a treatment for a subject occurred independently of other subjects subjected to the same treatment, responses within a specific treatment should also be independent The normal quantilequantile (or equivalently normal probability plot) for the deprived sensory treatment group does not indicate a departure from normality There is a value in the control group that is somewhat outlying With any outlier it is important to consider its potential cause Is it a recording error or an experimental error? Is it indicative of a different mechanism that is at play? How one proceeds depends on the reason for an outlier Here, it is probably chance variation associated with the underlying distribution among subjects subjected to the control treatment Even if this is not exactly normal for this group, since the sample sizes are equal, and moreover,, we can expect our pooled procedures to be quite robust to departures from equal variances, and moreover, normality in the absence of strong skewness Our plots do not indicate departures from the assumptions that would affect the validity of the analysis Normal quantilequantile plot for the sensory deprivation group:
11 Normal quantilequantile plot for the control group:
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