Question

A manufacturer can tolerate a small amount (.05 milligrams per liter (mg/I)) of impurities in a raw material needed for manufacturing its product. Because the laboratory test for the impurities is subject to experimental error, the manufacturer tests each batch 10 times. Assume that the mean value of the experimental error is 0 and hence that the mean value of the 10 test readings is an unbiased estimate of the true amount of the impurities in the batch. For a particular batch of the raw material, the mean of the 10 test readings is \(.058 \mathrm{mg} / \mathrm{l}\), with a standard deviation of \(.012 \mathrm{mg} /\) l. Do the data provide sufficient evidence to indicate that the amount of impurities in the batch exceeds \(.05 \mathrm{mg} / \mathrm{l} ?\) Find the \(p\) -value for the test and interpret its value.

Short answer

Based on our one-sample t-test, we find that the p-value is approximately 0.031. This indicates that there is a 3.1% chance of observing a sample mean concentration as extreme as 0.058 mg/l or more if the true mean concentration is indeed 0.05 mg/l. Since this p-value (0.031) is less than the commonly used significance level of 0.05, we reject the null hypothesis, concluding that the amount of impurities in the batch exceeds the allowed limit of 0.05 mg/l.

Step-by-step solution

Identify null and alternative hypothesis

Our null hypothesis (\(H_0\)) will be that the concentration of impurities in the batch is within the acceptable limit, and the alternative hypothesis (\(H_a\)) will be that the concentration of impurities exceeds the allowed limit. $$ H_0: \mu = 0.05 \, mg/l \\ H_a: \mu > 0.05 \, mg/l $$ We will test these hypotheses using a one-sample t-test.

Calculate the test statistic

In order to calculate the test statistic, we can use the formula: $$ t = \frac{\bar{x} - \mu_0}{\frac{s}{\sqrt{n}}} $$ Where \(\bar{x}\) is the sample mean, \(\mu_0\) is the value given in the null hypothesis, \(s\) is the sample standard deviation, and \(n\) is the sample size. Plugging in the given values, we have: $$ t = \frac{0.058 - 0.05}{\frac{0.012}{\sqrt{10}}} \approx 2.116 $$

Find the p-value

Since the alternative hypothesis states that the concentration of impurities is greater than the allowed limit, we will be looking at a one-tailed test. Using a t-distribution table or calculator, we can find the p-value associated with our test statistic and degrees of freedom \(\left(n - 1 = 9\right)\): $$ p \approx 0.031 $$

Interpret the p-value

A p-value of 0.031 means that there is a 3.1% chance of observing a sample mean concentration as extreme as 0.058 mg/l or more in favor of the alternative hypothesis, assuming that the null hypothesis is true (i.e., the true mean concentration is 0.05 mg/l). Since the p-value is less than a commonly used significance level of 0.05, we have sufficient evidence to reject the null hypothesis and conclude that the amount of impurities in the batch exceeds the allowed limit of 0.05 mg/l.

Key concepts

One-Sample T-Test

Imagine you're cooking a dish and want to ensure it's just the right sweetness. Just as you might taste and adjust seasonings, a one-sample t-test helps us compare the average of a sample to a known value. This test is like checking if the average sweetness matches your ideal sweetness. The one-sample t-test is particularly useful in determining whether a sample mean is significantly different from a hypothesized population mean. In our example, it helps the manufacturer assess if the impurity level averages more than the accepted limit of 0.05 mg/l. By calculating a 't' value, we assess how far off our sample mean is from the population mean when accounting for sample variability. The use of a t-test is appropriate when the sample size is small – typically less than 30 – and the population standard deviation is unknown, as in this case. Results from these tests help businesses to make data-driven decisions, like ensuring the quality of their products.

Significance Level

Imagine you’re making guesses in a game, and you want to be really sure your guess is correct before making a move. This caution is similar to what the significance level ( α ) represents in hypothesis testing. The significance level is a threshold set by researchers to determine how extreme data must be before we reject the null hypothesis. A common choice for the significance level is 0.05, which corresponds to a 5% risk of concluding that a difference exists when there is no actual difference. In simpler terms, it's like saying, "I will only make a big decision like this if I'm at least 95% sure." In our impurity test scenario, the significance level of 0.05 means we're willing to accept a 5% chance of incorrectly concluding that the impurities exceed the allowable limit. Setting this level guides us in interpreting the p-value and making informed decisions.

P-Value

Think of the p-value as a way to measure surprise. How surprised are we by our test results under the assumption that the null hypothesis is true? In hypothesis testing, the p-value tells us the probability of obtaining test results at least as extreme as the results actually observed, assuming that the null hypothesis is true. A small p-value indicates that the observed data is unusual under the null hypothesis, leading to its rejection. For our impurity test, a p-value of 0.031 tells us there is only a 3.1% chance that the impurity levels we observed would occur if the actual mean were indeed 0.05 mg/l. Because this is lower than our significance level of 0.05, it suggests the observed mean is unlikely to have happened by random chance alone.

Null and Alternative Hypothesis

In any investigation, we begin with questions or assumptions to verify. The null hypothesis ( H_0 ) and alternative hypothesis ( H_a ) are fundamental parts of hypothesis testing, framing the predictions we test. The null hypothesis typically states that there is no effect or difference, and in our example, it states the impurity level is exactly 0.05 mg/l. It's like saying, "Nothing unordinary is happening." On the other hand, the alternative hypothesis claims that some effect exists; here, it suggests that the impurity level exceeds 0.05 mg/l, signaling something out of the ordinary might indeed be occurring. Setting up these hypotheses is crucial as they form the basis of what we're testing. It helps guide data analysis, determining whether to maintain or reject our initial assumptions based on the observed data.