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In developing a generic drug, it is necessary for a manufacturer of biopharmaceutical products to show equivalence to the current product. The variable of interest is the absorption rate of the product. The current product has an absorption rate of 18 \(\mathrm{mg} / \mathrm{hr}\). If the new generic product has an absorption rate that is within \(0.50 \mathrm{mg} / \mathrm{hr}\) of this value, it will be considered equivalent. A random sample of 20 units of product is available, and the sample mean and standard deviation of absorption rate are \(18.22 \mathrm{mg} / \mathrm{hr}\) and \(0.92 \mathrm{mg} / \mathrm{hr},\) respectively. a. State the appropriate hypotheses that must be tested to demonstrate equivalence. b. What are your conclusions using \(\alpha=0.05 ?\)

Short Answer

Expert verified
The generic product is equivalent to the current product since it stays within the specified range.

Step by step solution

01

Understand the Problem

The task requires us to test if the absorption rate of the generic drug is equivalent to the current product, within a certain margin. The current absorption rate is 18 mg/hr and the new product is considered equivalent if it is within ±0.50 mg/hr.
02

Set Up Hypotheses

To test equivalence, the null hypothesis for each direction of difference must be tested against the alternatives that the mean is outside the tolerance interval. Formulate two one-sided hypotheses: \ 1. Test if the mean is greater than 18.5 mg/hr. \ 2. Test if the mean is less than 17.5 mg/hr. \ Formally, \( H_{0_1}: \mu \geq 18.5 \) and \( H_{a_1}: \mu < 18.5 \), \( H_{0_2}: \mu \leq 17.5 \) and \( H_{a_2}: \mu > 17.5 \).
03

Use T-Test for Both Hypotheses

With a significance level of \( \alpha = 0.05 \), split \(\alpha\) into two parts for the bi-directional nature of the hypothesis (\( \alpha = 0.025 \) for each). Given a sample size of 20, use a t-distribution with 19 degrees of freedom to calculate the test statistics. \[ t = \frac{\bar{x} - \mu_0}{\frac{s}{\sqrt{n}}} \] where \( \bar{x} = 18.22 \), \( s = 0.92 \), and \( n = 20 \).
04

Compute Test Statistics

Calculate \ \( t_{1} = \frac{18.22 - 18.5}{\frac{0.92}{\sqrt{20}}} \approx -1.37 \) \ and \ \( t_{2} = \frac{18.22 - 17.5}{\frac{0.92}{\sqrt{20}}} \approx 3.49 \).
05

Determine Critical Values and Conclusions

Consult a t-table for \( \alpha = 0.025 \) with 19 degrees of freedom; the critical value is approximately -2.093 and 2.093. Compare the test statistics to the critical values. For \( t_{1} = -1.37 \), since \(-1.37 > -2.093\), do not reject \( H_{0_1} \). For \( t_{2} = 3.49 \), since \(3.49 > 2.093\), reject \( H_{0_2} \). Since one null hypothesis \( H_{0_2} \) is rejected, the mean does not fall outside the tolerated interval.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Equivalence Testing
In the world of biopharmaceuticals and drug testing, the primary goal is often to demonstrate that a new product is equivalent to an established one. Equivalence testing is a statistical method used to determine if two products are similar enough in a specific aspect to be considered "equivalent." For example, in drug development, a generic product must show that its key properties, like absorption rate, do not significantly differ from the original brand product. To do this, a predefined margin is set to define what "difference" is acceptable without compromising therapeutic efficacy. In this exercise, the critical parameter under testing is the absorption rate.
The established product has an absorption rate of 18 mg/hr, and the new drug must deviate by no more than ±0.50 mg/hr to be considered equivalent. Equivalence testing involves setting two one-sided hypotheses instead of a typical two-sided hypothesis. - One test checks if the new drug's rate is significantly greater than the upper limit. - The other checks if it's significantly lower than the lower limit. These tests together help determine if the generic product's absorption rate is within the equivalent range.
T-Test
The t-test is a common statistical test used when you want to compare means, particularly in small sample sizes. It's used here to verify if the mean absorption rate of the new generic drug falls within the tolerated range of equivalence.The t-test assumes that the data follows a normal distribution and calculates a t-statistic, which reflects how far the sample mean deviates from the hypothesized mean under the null hypothesis, in terms of standard error.The formula for calculating the t-statistic is:\[ t = \frac{\bar{x} - \mu_0}{\frac{s}{\sqrt{n}}} \]where:
  • \( \bar{x} \) is the sample mean,
  • \( \mu_0 \) is the mean value under the null hypothesis,
  • \( s \) is the sample standard deviation, and
  • \( n \) is the sample size.
In this case, two t-tests are conducted, targeting each of the one-sided hypotheses needed for equivalence testing. The t-statistics then guide us in comparing against critical values to make conclusions about the hypotheses.
Significance Level
The significance level, denoted by \( \alpha \), is a probability threshold determining how confidently we reject the null hypothesis. It's a crucial part of hypothesis testing as it sets the criteria for statistical decision-making.Here, a significance level of \( \alpha = 0.05 \) is applied, indicating a 5% risk of concluding that there is an effect when there is none. For the equivalence test, since we have two separate hypotheses to test (one for each direction), the significance level is halved for each test,\( \alpha = 0.025 \) per test.
This splitting accounts for the bi-directional nature of the equivalence testing.Critical values derived from the t-distribution corresponding to 19 degrees of freedom are used to compare the calculated t-statistics. If a calculated t-statistic falls beyond these critical values, it suggests significant evidence against the null hypothesis.
Absorption Rate
Absorption rate is a critical factor in pharmacokinetics, describing how quickly a drug enters circulation after administration. In drug development, especially when creating generic alternatives, matching the absorption rate to the brand drug is crucial to ensure similar therapeutic outcomes. In this exercise, the absorption rate is a key metric for equivalence testing. The standard absorption rate for the existing drug is 18 mg/hr.
The generic drug must align closely with this rate, within a margin of ±0.50 mg/hr, to avoid significant divergence that may impact efficacy or safety. Analyzing and comparing the absorption rates involves not only looking at the sample mean but also understanding variability through the sample standard deviation, which is necessary for calculating the t-statistic. The measure gives insights into how dispersed the individual absorption rates might be relative to the mean, impacting the equivalence conclusions.

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Most popular questions from this chapter

A 1992 article in the Journal of the American Medical Association ("A Critical Appraisal of 98.6 Degrees F, the Upper Limit of the Normal Body Temperature, and Other Legacies of Carl Reinhold August Wunderlich") reported body temperature, gender, and heart rate for a number of subjects. The body temperatures for 25 female subjects follow: 97.8,97.2 97.4,97.6,97.8,97.9,98.0,98.0,98.0,98.1,98.2,98.3,98.3 98.4,98.4,98.4,98.5,98.6,98.6,98.7,98.8,98.8,98.9,98.9 and 99.0 . a. Test the hypothesis \(H_{0}: \mu=98.6\) versus \(H_{1}: \mu \neq 98.6\) using \(\alpha=0.05 .\) Find the \(P\) -value. b. Check the assumption that female body temperature is normally distributed. c. Compute the power of the test if the true mean female body temperature is as low as 98.0 . d. What sample size would be required to detect a true mean female body temperature as low as 98.2 if you wanted the power of the test to be at least \(0.9 ?\) e. Explain how the question in part (a) could be answered by constructing a two-sided confidence interval on the mean female body temperature.

A manufacturer produces crankshafts for an automobile engine. The crankshafts wear after 100,000 miles \((0.0001\) inch \()\) is of interest because it is likely to have an impact on warranty claims. A random sample of \(n=15\) shafts is tested and \(\bar{x}=2.78\). It is known that \(\sigma=0.9\) and that wear is normally distributed. a. Test \(H_{0}: \mu=3\) versus \(H_{1}: \mu \neq 3\) using \(\alpha=0.05\). b. What is the power of this test if \(\mu=3.25 ?\) c. What sample size would be required to detect a true mean of 3.75 if we wanted the power to be at least \(0.9 ?\)

In a little over a month, from June 5,1879 , to July 2 , 1879 , Albert Michelson measured the velocity of light in air 100 times (Stigler, Annals of Statistics, 1977 ). Today we know that the true value is \(299,734.5 \mathrm{~km} / \mathrm{sec} .\) Michelson's data have a mean of \(299,852.4 \mathrm{~km} / \mathrm{sec}\) with a standard deviation of \(79.01 .\) a. Find a two-sided \(95 \%\) confidence interval for the true mean (the true value of the speed of light). b. What does the confidence interval say about the accuracy of Michelson's measurements?

Did survival rate for passengers on the Titanic really depend on the type of ticket they had? Following are the data for the 2201 people on board listed by whether they survived and what type of ticket they had. Does survival appear to be independent of ticket class? (Test the hypothesis at \(\alpha=0.05 .)\) What is the \(P\) -value of the test statistic? $$\begin{array}{lccccr} & \text { Crew } & \text { First } & \text { Second } & \text { Third } & \text { Total } \\\\\text { Alive } & 212 & 202 &118 & 178 & 710 \\\\\text { Dead } & 673 & 123 & 167 & 528 & 1491 \\\\\text { Total } & 885 & 325 & 285 & 706 & 2201\end{array}$$

Consider the test of \(H_{0}: \sigma^{2}=7\) against \(H_{1}: \sigma^{2} \neq 7\) What are the critical values for the test statistic \(\chi_{0}^{2}\) for the following significance levels and sample sizes? a. \(\alpha=0.01\) and \(n=20\) b. \(\alpha=0.05\) and \(n=12\) c. \(\alpha=0.10\) and \(n=15\)

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