/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 13 Potency of an Antibiotic A drug ... [FREE SOLUTION] | 91Ó°ÊÓ

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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
Based on the one-sample t-test with a significance level of 0.05, we reject the null hypothesis, which states that the mean potency of the antibiotic is 80%. Our calculated t-value (-3.75) is beyond the critical t-value (±1.984), indicating there is sufficient evidence to refute the manufacturer's claim that the mean potency of the antibiotic is 80%.

Step by step solution

01

State the null hypothesis

The null hypothesis \((H_0)\) is the drug manufacturer's claim that the mean potency of the antibiotic is 80%: $$H_0: \mu = 80\%$$
02

State the alternative hypothesis

The alternative hypothesis \((H_1)\) is that the mean potency of the antibiotic is different from 80%: $$H_1: \mu \neq 80\%$$
03

Conduct a statistical test of the null hypothesis

Given that we have a sample of 100 capsules, we can use the one-sample t-test to test the null hypothesis. We are given sample mean (\(\bar{x}\) = 79.7), sample standard deviation (\(s\) = 0.8), and the sample size (\(n\) = 100). The test statistic (t-value) can be calculated using the following formula: $$t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}}$$ Plugging in the values: $$t = \frac{79.7 - 80}{0.8 / \sqrt{100}} = -3.75$$ With \(\alpha = 0.05\), we will use a two-tailed test because our alternative hypothesis is a not equal to sign. We can find the critical t-value corresponding to this significance level and degrees of freedom (n-1=99) using a t-distribution table or calculator: The critical t-value for a two-tailed test with α = 0.05 and 99 degrees of freedom is approximately ±1.984.
04

State your conclusion

We can now compare the calculated t-value (-3.75) with the critical t-value (±1.984). Since the calculated t-value is beyond the critical t-value (-3.75 < -1.984), we reject the null hypothesis. Therefore, there is sufficient evidence to refute the manufacturer's claim that the mean potency of the 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
Understanding the null hypothesis is crucial in the process of hypothesis testing. It is essentially a statement or a theoretical concept asserting that there is no effect or no difference, and it serves as the starting point for statistical analysis. When a drug manufacturer claims that their antibiotic has a mean potency of 80%, this claim is considered the null hypothesis. In our specific scenario, the null hypothesis is denoted as: \( H_0: \text{The mean potency, } \text{μ, is equal to } 80\text{%} \). It represents the status quo that will be challenged by the test's results. If the evidence gathered through the test is sufficiently strong to disprove the null hypothesis, only then can it be rejected in favor of the alternative hypothesis.
Alternative Hypothesis
The alternative hypothesis contrasts the null hypothesis by proposing what we might begin to believe if the null hypothesis is rejected. It suggests that there is indeed an effect, or there is a difference. It is not the opposite of the null hypothesis but rather what we suspect might be the actual state of affairs. For our antibiotic potency example, the alternative hypothesis is set to envision a different mean potency other than 80%: \( H_1: \text{The mean potency, } \text{μ, is not equal to } 80\text{%} \). In hypothesis testing, we search for evidence strong enough to support this alternative hypothesis.
T-test
A t-test is a statistical procedure used to determine if there is a significant difference between the means of two groups or if a sample mean significantly differs from a known or hypothesized population mean. It is often used when the sample size is small and the population standard deviation is unknown. The t-test takes into account the sample mean, the sample standard deviation, and the sample size to calculate a t-value. This value is then compared to a critical value from the t-distribution to decide whether to reject the null hypothesis. In the antibiotic potency case, we conducted a one-sample t-test because we compared a sample mean to a claimed population mean.
Sample Mean
The sample mean, denoted by \( \bar{x} \), is the average value of a sample. It is calculated by adding up all the numbers in the sample and then dividing by the number of observations. The sample mean provides an estimate of the population mean and is a fundamental component in calculating the t-value for the t-test. For the present exercise, the sample mean of the antibiotic potencies is \( \bar{x} = 79.7\text{%} \), which will be pivotal in determining if the actual mean potency differs from the claimed 80%.
Standard Deviation
The standard deviation is a measure of how spread out the values in a dataset are around the mean. A low standard deviation indicates that the values tend to be close to the mean, while a high standard deviation indicates that the values are spread out over a larger range. In the context of hypothesis testing, particularly the t-test, the sample standard deviation (denoted by \( s \)) is used to understand the variability of the sample. It is a key part of the formula that calculates the t-value. In our antibiotic example, the standard deviation is \( s = 0.8\text{%} \).
Significance Level
The significance level, often denoted by \( \alpha \), is a threshold that determines the probability of rejecting a true null hypothesis, also known as a Type I error. It's a measure of tolerance for making such an error and is predetermined by the researcher. The lower the significance level, the less likely you are to incorrectly reject the null hypothesis. Commonly used levels are 0.05 or 5%, indicating a 5% risk of concluding a difference exists when there is none. In the antibiotic potency investigation, an \( \alpha = 0.05 \) was used, meaning there's a 5% chance the test could lead to a false rejection of the true null hypothesis.
Degrees of Freedom
Degrees of freedom are a concept in statistics that refers to the number of independent values or quantities which can be assigned to a statistical distribution. In the case of the t-test, the degrees of freedom (often abbreviated as df) are calculated as the number of observations in a sample minus one (\( n - 1 \)). They are used to determine the critical values of the t-distribution, which are crucial in deciding whether to reject the null hypothesis. For our example, we have 100 capsules tested, thus the degrees of freedom are 99 (\( 100 - 1 = 99 \)).
T-distribution
The t-distribution, also known as Student's t-distribution, is a probability distribution that arises while estimating the mean of a normally distributed population in situations where the sample size is small, and the population standard deviation is unknown. It is similar to the standard normal distribution but has thicker tails, which means it is more prone to producing values that fall far from its mean. This is especially important when the sample size is small, as the t-distribution accounts for the additional uncertainty. When performing a t-test, the calculated t-value is compared with values from the t-distribution to determine whether to reject the null hypothesis. In the evaluation of the antibiotic's potency, the t-distribution was used in conjunction with the predetermined significance level and degrees of freedom to reach a conclusion.

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