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Chapter 9: Problem 13

Potency of an Antibiotic A drug manufacturer claimed that the mean potency of one of its antibiotics was \(80 \%\). A random sample of \(n=100\) capsules were tested and produced a sample mean of \(\bar{x}=79.7 \%\) with a standard deviation of \(s=.8 \% .\) Do the data present sufficient evidence to refute the manufacturer's claim? Let \(\alpha=.05 .\) a. State the null hypothesis to be tested. b. State the alternative hypothesis. c. Conduct a statistical test of the null hypothesis and state your conclusion.

Short Answer

Expert verified
Answer: Yes, there is sufficient evidence to refute the manufacturer's claim that the mean potency of their antibiotic is 80%.

Step by step solution

01

a. State the null hypothesis to be tested.

The null hypothesis, denoted as H_{0}, is a statement that there is no significant difference between observed and expected data. In this case, the null hypothesis is that the true mean potency of the antibiotic is equal to the manufacturer's claim of 80%. Mathematically, it can be written as: H_{0}: \mu = 80\%.
02

b. State the alternative hypothesis.

The alternative hypothesis, denoted as H_{1}, is a statement that contradicts the null hypothesis. In this case, it suggests that the true mean potency of the antibiotic is not equal to 80%. Mathematically, it can be written as: H_{1}: \mu \neq 80\%.
03

c. Conduct a statistical test of the null hypothesis and state your conclusion.

To conduct a statistical test, we will use a t-test, as we are dealing with a small sample size and do not know the population standard deviation. First, we need to calculate the t-score, which is given by the formula: t = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}}. Using the given data, we can calculate the t-score: t = \frac{79.7 - 80}{\frac{0.8}{\sqrt{100}}} = \frac{-0.3}{0.08} \approx -3.75. Next, we need to determine the critical t-value for a two-tailed test at a significance level of 伪 = 0.05 and using the degrees of freedom, df = n - 1 = 100 - 1 = 99. By referring to a t-table or using a calculator, we find the critical t-values to be approximately 卤2.626. Since -3.75 lies to the left of the lower critical value (-2.626), we reject the null hypothesis. Thus, the data provides sufficient evidence to refute the manufacturer's claim that the mean potency of their antibiotic is 80%.

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Key Concepts

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

Null Hypothesis
In hypothesis testing, the null hypothesis (\( H_0 \)) is a foundational starting point. It's essentially a statement suggesting that there is no effect or no difference. Here, it is used to assert that the mean potency of the antibiotic is \( 80\% \), as claimed by the manufacturer. This hypothesis serves as a benchmark against which we compare our sample data. The idea is to maintain this hypothesis unless there is strong evidence to reject it. In statistical terms, the null hypothesis for this exercise can be written as:
  • \( H_{0}: \mu = 80\% \)
Where:
  • \( \mu \) is the population mean potency of the antibiotic.
Alternative Hypothesis
The alternative hypothesis (\( H_1 \)) represents the statement that contrasts with the null hypothesis. It suggests a new perspective, one that will be taken if there is strong evidence against the null hypothesis. For this exercise, the alternative hypothesis claims that the mean potency is not equal to what the manufacturer asserts, i.e., \( 80\% \). Mathematically, it is expressed as:
  • \( H_{1}: \mu eq 80\% \)
This is a two-tailed hypothesis, implying that the actual mean could be either less than or greater than \( 80\% \). Such a hypothesis is useful when any significant deviation, either upwards or downwards, needs to be detected.
t-test
The t-test is a statistical tool used to determine if there's a significant difference between the sample mean and the known population mean. In this scenario, the t-test helps us understand if the sample mean potency of \( 79.7\% \) significantly differs from the claimed \( 80\% \).Here's how the t-test is applied:
  • Calculate the t-score using the formula: \[ t = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}} \]
Where:
  • \( \bar{x} \) is the sample mean (\( 79.7\% \))
  • \( \mu \) is the population mean (\( 80\% \))
  • \( s \) is the sample standard deviation (\( 0.8 \% \))
  • \( n \) is the sample size (\( 100 \))
For the given data, the t-score was calculated as approximately \( -3.75 \). We then compare this score with critical t-values from a t-distribution table at a significance level (\( \alpha \)) of \( 0.05 \) for \( 99 \) degrees of freedom. Because \( -3.75 \) falls outside the critical range (\( \pm 2.626 \)), we reject the null hypothesis. This indicates that there is enough evidence to say the potency of the antibiotic differs from \( 80\% \).

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Most popular questions from this chapter

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