Question

To properly treat patients, drugs prescribed by physicians must not only have a mean potency value as specified on the drug's container, but also the variation in potency values must be small. Otherwise, pharmacists would be distributing drug prescriptions that could be harmfully potent or have a low potency and be ineffective. A drug manufacturer claims that his drug has a potency of \(5 \pm .1\) milligram per cubic centimeter \((\mathrm{mg} / \mathrm{cc})\). A random sample of four containers gave potency readings equal to 4.94,5.09 , 5.03, and \(4.90 \mathrm{mg} / \mathrm{cc} .\) a. Do the data present sufficient evidence to indicate that the mean potency differs from \(5 \mathrm{mg} / \mathrm{cc} ?\) b. Do the data present sufficient evidence to indicate that the variation in potency differs from the error limits specified by the manufacturer? (HINT: It is sometimes difficult to determine exactly what is meant by limits on potency as specified by a manufacturer. Since he implies that the potency values will fall into the interval \(5 \pm .1 \mathrm{mg} / \mathrm{cc}\) with very high probability - the implication is almost always-let us assume that the range \(.2 ;\) or 4.9 to \(5.1,\) represents \(6 \sigma\) as suggested by the Empirical Rule).

Short answer

Is there evidence that the variation in potency differs from the error limits specified by the manufacturer? Answer: No, there is not enough evidence to suggest that the mean potency is different from 5 mg/cc. However, there is sufficient evidence to indicate that the variation in potency differs from the error limits specified by the manufacturer.

Step-by-step solution

Hypothesis Test on the Mean Potency

First, we need to perform a hypothesis test on whether the mean potency is different from 5 mg/cc. To do this, we will use the t-distribution as the sample size is small (n=4). The null and alternative hypotheses are as follows: $$H_0: \mu = 5 \, mg/cc \quad \text{vs.} \quad H_1: \mu \neq 5 \, mg/cc$$ where \(\mu\) is the true mean potency of the drug.

Calculate the Sample Mean and Standard Deviation

Now, we calculate the sample mean (\(\overline{x}\)) and the standard deviation (\(s\)) from the given data: \((4.94, 5.09, 5.03, 4.90)\) Sample mean \(\overline{x} = \frac{(4.94 + 5.09 + 5.03 + 4.90)}{4} = 5.0 \, mg/cc\) Sample standard deviation \(s = \sqrt{ \frac{(4.94-5)^2 + (5.09-5)^2 + (5.03-5)^2 + (4.90-5)^2}{4-1} } = 0.0955 \, mg/cc\)

Calculate t-statistic and Perform the t-test

With a sample size of 4, our degrees of freedom (df) is (n-1), or 3. The t-statistic for the hypothesis test is calculated as follows: $$t = \frac{\overline{x} - \mu}{s / \sqrt{n}} = \frac{5.0 - 5}{0.0955 / 2} = 0$$ Given the two-tailed nature of our test, look up in the t-table the critical value for a 3 df with a level of significance of 0.05 (alpha=0.05). We find this to be \(\pm 3.182\). Since the calculated t-statistic is 0, which falls within the range of \([-3.182,3.182]\), we fail to reject the null hypothesis. Thus, there is not enough evidence to suggest that the mean potency is different from 5 mg/cc.

Hypothesis Test on Variation in Potency

For part (b), the goal is to test whether the variation in potency conforms to the claim made by the manufacturer. Based on the hint, we can assume that 6 standard deviations correspond to a range of 0.2 mg/cc. So, we have: $$\sigma = \frac{0.2}{6}=0.0333 \, mg/cc$$ The null and alternative hypotheses for variation in potency are: $$H_0: \sigma^2 = (0.0333)^2 \quad \text{vs.} \quad H_1: \sigma^2 \neq (0.0333)^2$$ where \(\sigma^2\) is the variance of the true population.

Calculate the Chi-square Statistic and Perform the Chi-square Test

We will use the chi-square distribution to test the hypothesis about variation, where the degrees of freedom is 3 (n-1). The chi-square statistic for the hypothesis test is calculated as follows: $$\chi^2 = (n-1) \frac{s^2}{\sigma^2} = 3\frac{(0.0955)^2}{(0.0333)^2} = 32.66$$ The critical value at a 0.05 level of significance (alpha=0.05) and 3 df from the chi-square table is 7.815 for the lower tail and 0.352 for the upper tail. Since the calculated chi-square statistic (32.66) falls outside the range of \([0.352, 7.815]\), we reject the null hypothesis. In conclusion, based on the data, there is sufficient evidence to indicate that the variation in potency differs from the error limits specified by the manufacturer, while there is not enough evidence to suggest that the mean potency is different from 5 mg/cc.

Key concepts

T-Distribution

When conducting hypothesis tests, especially with small sample sizes, the t-distribution becomes incredibly useful. It is a type of probability distribution that is symmetric and bell-shaped, like the normal distribution. However, it has heavier tails. This means it gives more probability to values far away from the mean than the normal distribution.

This is particularly helpful when dealing with small sample sizes, usually less than 30, and when the population standard deviation is unknown. In the provided exercise, the t-distribution is used to test whether the mean potency is different from 5 mg/cc. The small sample size of 4 makes the t-distribution more appropriate than the normal distribution.

• The degrees of freedom (df) is calculated as the sample size minus one (n-1). In this scenario, it is 3.
• The critical value from the t-table at a 0.05 significance level for 3 df was given as \( \pm 3.182 \).
• The computed t-statistic was 0, which fell within this range, indicating there is not enough evidence to reject the null hypothesis that the mean is 5 mg/cc.

Understanding the application of t-distribution is essential for handling hypotheses involving small datasets.

Chi-Square Statistic

The chi-square statistic is pivotal for testing hypotheses about variances in a dataset. This technique is especially useful when comparing the observed variance to a theoretically expected variance, like in our exercise where we're testing the drug potency variation.

When the sample size (n) is small, the degrees of freedom in the chi-square test are determined by subtracting one from the sample size (n-1). In this problem, it is again 3. The chi-square statistic uses the formula:

• \( \chi^2 = (n-1) \frac{s^2}{\sigma^2} \)

With the exercise data, testing the variance against the manufacturer's claim showed a chi-square statistic of 32.66. Using chi-square tables at a significance level of 0.05 and 3 df reveals critical values:

• Lower tail: 0.352
• Upper tail: 7.815

The value of 32.66 sat outside this range, prompting rejection of the null hypothesis that the variance conforms to the claimed potency range. Thus, the chi-square statistic is a potent tool for assessing variance claims against empirical data.

Sample Standard Deviation

Sample standard deviation (\( s \)) is a measure of data dispersion, signifying how much data values deviate from the sample mean. It offers insight into variability, which is often critical in statistical testing, and serves as an estimator for population standard deviation when unknown.

In our drug potency exercise, it was vital to calculate the sample standard deviation to understand the spread of potency values among the sampled containers. The computed \( s \) was 0.0955 mg/cc, which was slightly high given the supposed narrow range specified by the manufacturer.

• The calculation of \( s \) involves finding the square root of the variance, determined by summing the squared deviations from the mean, divided by degrees of freedom (n-1).
• It acts as a fundamental component in both t-tests and chi-square tests, as seen in the provided steps.

A solid grasp of sample standard deviation equips you with understanding how sample data can represent population spread, crucial for hypothesis testing and variance analysis.

Empirical Rule

The Empirical Rule, also known as the 68-95-99.7 rule, describes how data spread out in a normal distribution. It tells us that approximately 68% of data falls within one standard deviation from the mean, 95% within two, and 99.7% within three.

In the context of the exercise, the manufacturer implies a potency range of \(5 \pm 0.1\) mg/cc, suggesting a span that should encompass extremely high probability within the interval. Assuming a normal distribution, the range \(4.9\) to \(5.1\) mg/cc implies a span of 6 standard deviations according to the Empirical Rule.

• Hence, each standard deviation \( \sigma \) is approximately 0.0333 mg/cc (\(\frac{0.2}{6}\)).
• This framework embodies the manufacturer's expressed confidence in the potency being tightly controlled around the mean of 5 mg/cc.

Familiarity with the Empirical Rule gives valuable insight into expected data behavior in normally distributed populations and helps interpret variance ranges, as done while assessing manufacturer's claims.