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#1
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Dear all,
I have a problem with some statistic calculation. I have 2 series of measu the first column is the test series and the second column if the reference series. I have to calculate the standard 90% confidence intervals of the ratio test/reference (T/R) and I have to answer to this question: the products were considered bioequivalent if the difference between two compared parameters was found statistically insignificant (P < 0.05) and 90% confidence intervals for these parameters fell within 80-120%. I know the result for these series, but I have more to calculate: 90% CI : 88.1-115.6% Two one-sided t-test probability (Probability for T/R ratio (r) to be within 0.8 and 1.2) : 0.98 Someone can explain or suggest me an internet reference to solve this problem. Thanks a lot 5724.4 5204.49 1518.37 1172.63 1186.72 1720.13 805.7 2390.36 1155.86 762.54 2380.6 3050.71 1077.61 1044.55 1755.64 1760.97 973.11 1196.18 689.05 702.22 1323.86 1399.5 1367.1 232.78 3092.67 3354.46 1730.24 1475.54 1070.01 1381.2 1803.08 1454.36 2507.96 2160.75 2051.32 1192.23 1181.96 1220.58 1097.87 395.9 1339.46 1431.47 3610.07 2995.25 847.34 869.04 1655.91 1436.13 1285.73 3253.84 2044.78 1185.02 |
#2
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Something doesn't add up here. The average of the ratios (128.8%) is
not contained in your confidence interval. Also, either the data contain some outliers, or you must transform to approximate normality before using the t-distribution, since the distribution of sample ratios is skewed (e.g. 1367.1/232.78 = 587.3%) Jerry Rosario wrote: Dear all, I have a problem with some statistic calculation. I have 2 series of measu the first column is the test series and the second column if the reference series. I have to calculate the standard 90% confidence intervals of the ratio test/reference (T/R) and I have to answer to this question: the products were considered bioequivalent if the difference between two compared parameters was found statistically insignificant (P < 0.05) and 90% confidence intervals for these parameters fell within 80-120%. I know the result for these series, but I have more to calculate: 90% CI : 88.1-115.6% Two one-sided t-test probability (Probability for T/R ratio (r) to be within 0.8 and 1.2) : 0.98 Someone can explain or suggest me an internet reference to solve this problem. Thanks a lot 5724.4 5204.49 1518.37 1172.63 1186.72 1720.13 805.7 2390.36 1155.86 762.54 2380.6 3050.71 1077.61 1044.55 1755.64 1760.97 973.11 1196.18 689.05 702.22 1323.86 1399.5 1367.1 232.78 3092.67 3354.46 1730.24 1475.54 1070.01 1381.2 1803.08 1454.36 2507.96 2160.75 2051.32 1192.23 1181.96 1220.58 1097.87 395.9 1339.46 1431.47 3610.07 2995.25 847.34 869.04 1655.91 1436.13 1285.73 3253.84 2044.78 1185.02 |
#3
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Hi Jarry,
I have done the calculation after a log transformation, but my result doesn't match the one I have posted. Either my calculations or the result, I've already posted (I read it in a sientific article), are wrong... What do you think about? thanks a lot "Jerry W. Lewis" wrote in message ... Something doesn't add up here. The average of the ratios (128.8%) is not contained in your confidence interval. Also, either the data contain some outliers, or you must transform to approximate normality before using the t-distribution, since the distribution of sample ratios is skewed (e.g. 1367.1/232.78 = 587.3%) Jerry Rosario wrote: Dear all, I have a problem with some statistic calculation. I have 2 series of measu the first column is the test series and the second column if the reference series. I have to calculate the standard 90% confidence intervals of the ratio test/reference (T/R) and I have to answer to this question: the products were considered bioequivalent if the difference between two compared parameters was found statistically insignificant (P < 0.05) and 90% confidence intervals for these parameters fell within 80-120%. I know the result for these series, but I have more to calculate: 90% CI : 88.1-115.6% Two one-sided t-test probability (Probability for T/R ratio (r) to be within 0.8 and 1.2) : 0.98 Someone can explain or suggest me an internet reference to solve this problem. Thanks a lot 5724.4 5204.49 1518.37 1172.63 1186.72 1720.13 805.7 2390.36 1155.86 762.54 2380.6 3050.71 1077.61 1044.55 1755.64 1760.97 973.11 1196.18 689.05 702.22 1323.86 1399.5 1367.1 232.78 3092.67 3354.46 1730.24 1475.54 1070.01 1381.2 1803.08 1454.36 2507.96 2160.75 2051.32 1192.23 1181.96 1220.58 1097.87 395.9 1339.46 1431.47 3610.07 2995.25 847.34 869.04 1655.91 1436.13 1285.73 3253.84 2044.78 1185.02 |
#4
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What article?
Jerry Rosario wrote: Hi Jarry, I have done the calculation after a log transformation, but my result doesn't match the one I have posted. Either my calculations or the result, I've already posted (I read it in a sientific article), are wrong... What do you think about? thanks a lot "Jerry W. Lewis" wrote in message ... Something doesn't add up here. The average of the ratios (128.8%) is not contained in your confidence interval. Also, either the data contain some outliers, or you must transform to approximate normality before using the t-distribution, since the distribution of sample ratios is skewed (e.g. 1367.1/232.78 = 587.3%) Jerry Rosario wrote: Dear all, I have a problem with some statistic calculation. I have 2 series of measu the first column is the test series and the second column if the reference series. I have to calculate the standard 90% confidence intervals of the ratio test/reference (T/R) and I have to answer to this question: the products were considered bioequivalent if the difference between two compared parameters was found statistically insignificant (P < 0.05) and 90% confidence intervals for these parameters fell within 80-120%. I know the result for these series, but I have more to calculate: 90% CI : 88.1-115.6% Two one-sided t-test probability (Probability for T/R ratio (r) to be within 0.8 and 1.2) : 0.98 Someone can explain or suggest me an internet reference to solve this problem. Thanks a lot 5724.4 5204.49 1518.37 1172.63 1186.72 1720.13 805.7 2390.36 1155.86 762.54 2380.6 3050.71 1077.61 1044.55 1755.64 1760.97 973.11 1196.18 689.05 702.22 1323.86 1399.5 1367.1 232.78 3092.67 3354.46 1730.24 1475.54 1070.01 1381.2 1803.08 1454.36 2507.96 2160.75 2051.32 1192.23 1181.96 1220.58 1097.87 395.9 1339.46 1431.47 3610.07 2995.25 847.34 869.04 1655.91 1436.13 1285.73 3253.84 2044.78 1185.02 |
#5
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Rosario,
There are a couple of ways to approach this problem. I start by specifying two one side hypothesis tests H01: Mt/Mc <= 0.8 or H02: Mt/Mc = 1.2 versus HA1: Mt/Mc 0.8 and HA2: Mt/Mc < 1.2 where Mt is the mean value of the test and Mc is the mean value of the control. The combined alternative is what we are trying to assert as bioequivalence HA: 0.8< Mt/Mc < 1.2. This scenario of testing is called the Intersection-Union test since the null hypothesis is a union hypothesis and the alternative is a intersection hypothesis. The testing strategy is to accept HA if both H01 rejects at alpha=0.05 and H02 rejects at alpha=0.05. It can be shown that this stratagy has an overall type 1 error rate of 0.05 if both tests are level 0.05 tests. They need not be independent to prove this either. I won't go into the proof here. Now, there are a couple of common ways to rewrite H01 and H02 to get linear hyptheses and then do t-tests. You can compare arithmetic means: H01: Mt-0.8Mc <= 0 or H02: Mt-1.2Mc = 0 or you can compare geometric means H01: log(Mt)-log(Mc) <= log(0.8) or H02: log(Mt)-log(Mc) = log(1.2) A lot of people compare geometric means since these kinds of tests are often done on blood plasma concentrations of drugs, which tend to have skewed distributions. The log transformation pulls the outliers in, and makes the data more normally distributed. In this case you just have to show that a two sided 90% confidence interval falls withing log(0.8) and log(1.2). Personally, I find this hard to interpret since I geometric means are appropriate (have some physical meaning) for averaging angles or rates. I prefer using the arithmetic means since they have a normal distribution for most data sets because of the central limit theorem. I just rewrite the hypotheses as linear tests and do the individual tests. For instance, H01 states in English that the test mean is less than 80% of the control mean. The standard errors of the difference in means has to be modified to incorportate the linear coefficients in the test. You can also look up Feillers theorem on the ratio of means. For this data set, you can conclude bioequivelence if the Feiller interval falls within 0.8 to 1.2. Now it appears that the data you have are paired observations from individuals, so the means are correlated between tests and controls. This adds a difficulty since it becomes a repeated measures experiment. Bioequivalence studies are often crossover studies, which makes things even more complicated. Anyway, I have attached a SAS program that test takes both the arithmetic means and the geometric means approach assuming paired observations. My analysis finds bioequivalence in the arithmetic hypothesis tests, but does not conclude bioequivalence in the geometric means. I can email you an output listing if you wish. Mark Von Tress data one; input test control; sample = _n_; cards; 5724.4 5204.49 1518.37 1172.63 1186.72 1720.13 805.7 2390.36 1155.86 762.54 2380.6 3050.71 1077.61 1044.55 1755.64 1760.97 973.11 1196.18 689.05 702.22 1323.86 1399.5 1367.1 232.78 3092.67 3354.46 1730.24 1475.54 1070.01 1381.2 1803.08 1454.36 2507.96 2160.75 2051.32 1192.23 1181.96 1220.58 1097.87 395.9 1339.46 1431.47 3610.07 2995.25 847.34 869.04 1655.91 1436.13 1285.73 3253.84 2044.78 1185.02 ; proc univariate data=one plot normal; var test control; run; proc sort data=one; by sample; proc transpose data=one out=onet; by sample; var test control; run; data onet; set onet; log_c = log10(col1); rename col1=conc _name_=group; run; *raw data - test equivalence of arithmetic means; proc mixed data=onet ratio covtest; class group sample; model conc = group / ddfm=satterth solution noint; random sample; estimate 'test 1 H01: -0.8control + 1.0 test 0' group -0.8 1 / upper; estimate 'test 2 H02: -1.2control + 1.0 test < 0' group -1.2 1 / lower; * reject both H01 and H02, so conclude bioequivalence; run; *log data - test equivalence of geometric means; proc mixed data=onet ratio covtest; class group sample; model log_c = group / ddfm=satterth solution noint; random sample; estimate 'difference in geometric means' group -1 1 / cl alpha=0.1; * test rule: accept bioequivalence at alpha = 0.05 if 2 sided 90% confidence interval (i.e. 2 one sided 95% ci's) on difference in mean log_c falls within log10(0.8)=-.1 and log10(1.2)=0.08; * Using results: do not accept bioequivalence since lci=-0.04297 -0.1, but uci=0.1149 0.08; run; |
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