91Ó°ÊÓ

To properly treat patients, drugs prescribed by physicians must have a potency that is accurately defined. Consequently, not only must the distribution of potency values for shipments of a drug have a mean value as specified on the drug's container, but also the variation in potency 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 is marketed with a potency of \(5 \pm .1\) milligram per cubic centimeter \((\mathrm{mg} / \mathrm{cc}), \mathrm{A}\) 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 always-let us assume that the range \(.2 ;\) or \((4.9\) to 5.1\(),\) represents \(6 \sigma,\) as suggested by the Empirical Rule. Note that letting the range equal \(6 \sigma\) rather than \(4 \sigma\) places a stringent interpretation on the manufacturer's claim. We want the potency to fall into the interval \(5 \pm .1\) with very high probability.

Step by step solution

01

Calculate the sample mean and variance

Given the drug potency readings 4.94, 5.09, 5.03, and 4.90 mg/cc, calculate the sample mean: \(x_{mean} = \frac{4.94+5.09+5.03+4.90}{4} = 4.99\) mg/cc Next, calculate the sample variance (s^2): \(s^2 = \frac{(4.94-4.99)^2+(5.09-4.99)^2+(5.03-4.99)^2+(4.90-4.99)^2}{4-1} = 0.0083\) (approximately)

02

Hypothesis test for the mean (a)

Let's perform a hypothesis test to check if the claiming mean potency differs from 5 mg/cc. We will use a one-sample t-test. Null hypothesis (H0): \(\mu = 5\) Alternative hypothesis (H1): \(\mu \neq 5\) Using the t-test statistic with a significance level of \(\alpha = 0.05\): \(t = \frac{x_{mean} - \mu}{s/\sqrt{n}} = \frac{4.99-5}{\sqrt{0.0083}/\sqrt{4}} = -0.7583\) (approximately) With 3 degrees of freedom (df = n-1), use a t-distribution table to find the critical value. In this case, the critical value is \(t_{critical} = 2.353\) (approximately). Our t-test statistic falls between the critical values (-2.353 and 2.353), meaning there is not enough evidence to reject the null hypothesis. We can't conclude that the mean potency differs from 5 mg/cc.

03

Hypothesis test for the variance (b)

Now let's conduct a hypothesis test to check if the claiming variation in potency differs from the error limits specified by the manufacturer. We will use a chi-square test. From the hint given, assume that the range (\(6\sigma\)) to be 0.2 (\(4.9\) to \(5.1\)); hence, the variance (\(\sigma^2\)) must be equal to 0.001111 (approximately). Null hypothesis (H0): \(\sigma^2 = 0.001111\) Alternative hypothesis (H1): \(\sigma^2 \neq 0.001111\) Using the chi-square test statistic: \(\chi^2 = \frac{(n-1)s^2}{\sigma^2} = \frac{(4-1)*0.0083}{0.001111} = 22.5\) (approximately) Refer to a chi-square table with a significance level of \(\alpha = 0.05\), and 3 degrees of freedom (df = n-1). The critical values are \(\chi^2_{lower} = 1.213\) and \(\chi^2_{upper} = 7.815\) (approximately). Our chi-square test statistic (22.5) does not fall between the critical values, meaning there is enough evidence to reject the null hypothesis. We can conclude that there is sufficient evidence to indicate that the variation in potency differs from the error limits specified by the manufacturer.

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Key Concepts

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

Sample Mean

In hypothesis testing, the sample mean is a vital component as it represents the average value of a sample dataset. The sample mean provides a single value that summarizes the central tendency of a dataset. For instance, when determining the potency of a drug, calculating the sample mean can give an insight into whether the drug's potency is as expected.
The formula for the sample mean \(x_{mean}\) is:

• \(x_{mean} = \frac{\sum x_i}{n}\)

where \(\sum x_i\) is the sum of the individual measurements and \(n\) is the number of measurements.
In the given problem, the drug potency readings are added together and then divided by the number of samples to find the average potency. This helps in comparing whether the drug meets the expected potency value prescribed by the manufacturer.

Variance

Variance is a measure of how spread out the values in a data set are. It is crucial in determining consistency or variability in data points within a sample. The formula for variance \(s^2\) is:

• \(s^2 = \frac{\sum (x_i - x_{mean})^2}{n-1}\)

where \(x_i\) is each individual value, \(x_{mean}\) is the sample mean, and \(n-1\) is the degrees of freedom.
Variance helps in understanding if the data points deviate a lot from the mean; a higher variance means more spread, and potentially, more inconsistent data.
In the context of drug potency, calculating variance helps evaluate whether these potencies are closely aligned or scattered significantly. The example shows that the variance is roughly 0.0083, indicating a measure of dispersion in the potency readings.

t-test

The t-test is a statistical method used to determine if there is a significant difference between the mean of a sample and a known value (like a manufacturer's claim). The one-sample t-test compares the sample mean to the known population mean.
The formula for the t-test statistic is:

• \(t = \frac{x_{mean} - \mu}{s / \sqrt{n}}\)

where \(x_{mean}\) is the sample mean, \(\mu\) is the population mean, \(s\) is the sample standard deviation, and \(n\) is the sample size.
The calculated t-statistic is compared against the critical value from the t-distribution table. If the absolute value of the t-stat is greater than the critical value, the null hypothesis is rejected.
In the given exercise, the quick assessment revealed that the mean potency's t-value of -0.7583 falls within the range, showing no strong evidence to reject the claim of 5 mg/cc potency.

Chi-square test

The chi-square test for variance is used to ascertain if a sample's variance significantly differs from a specified population variance. This test is suitable when dealing with data that is normally distributed.
The chi-square test statistic is calculated as:

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

In this formula, \(n\) is the sample size; \(s^2\) is the sample variance, and \(\sigma^2\) is the population variance.
After computation, the test statistic is compared against critical values from the chi-square distribution table to determine whether the variance is significantly different.
In our case, with a calculated \(\chi^2\) of 22.5, exceeding the upper critical value of 7.815 indicates that the variance does significantly differ from the manufacturer's stipulated limits. It implies potential issues in the drug's potency consistency, needed to address safety and efficacy concerns.