/*! 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 53 An aspirin manufacturer fills bo... [FREE SOLUTION] | 91Ó°ÊÓ

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An aspirin manufacturer fills bottles by weight rather than by count. Since each bottle should contain 100 tablets, the average weight per tablet should be 5 grains. Each of 100 tablets taken from a very large lot is weighed, resulting in a sample average weight per tablet of \(4.87\) grains and a sample standard deviation of \(.35\) grain. Does this information provide strong evidence for concluding that the company is not filling its bottles as advertised? Test the appropriate hypotheses using \(\alpha=.01\) by first computing the \(P\)-value and then comparing it to the specified significance level.

Short Answer

Expert verified
There is strong evidence that the bottling process deviates from the advertised mean of 5 grains per tablet.

Step by step solution

01

State the Hypotheses

Start by stating the null hypothesis \[ H_0: \mu = 5 \] and the alternative hypothesis \[ H_a: \mu eq 5 \], where \( \mu \) is the true mean weight of a tablet. We are testing whether the mean weight deviates from 5 grains, so we opt for a two-tailed test.
02

Determine the Test Statistic

Calculate the test statistic using the formula for the test of a single mean:\[ t = \frac{\bar{x} - \mu}{s/\sqrt{n}} \]Where:- \( \bar{x} = 4.87 \) is the sample mean- \( \mu = 5 \) is the population mean (as per the claim)- \( s = 0.35 \) is the sample standard deviation- \( n = 100 \) is the sample size.Substitute the values to get:\[ t = \frac{4.87 - 5}{0.35/\sqrt{100}} = \frac{-0.13}{0.035} \approx -3.71 \]
03

Find the P-value

Using a t-distribution table or calculator with 99 degrees of freedom (\(n-1\)), find the P-value for \(t = -3.71\). Since this is a two-tailed test, compute the probability of getting a value as extreme as or more extreme than the observed value on either side of the distribution.The P-value is approximately 0.0003.
04

Make a Decision

Compare the P-value to the significance level \(\alpha = 0.01\). Since 0.0003 is less than 0.01, we reject the null hypothesis.
05

Conclusion

Conclude that there is strong evidence to suggest that the mean weight per aspirin tablet is different from 5 grains. The manufacturer is likely not filling the bottles correctly as advertised under the given significance level \(\alpha = 0.01\).

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Key Concepts

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

P-value calculation
The P-value is an essential tool in hypothesis testing as it helps us decide whether to reject the null hypothesis. It measures the probability of obtaining test results at least as extreme as the ones observed, assuming the null hypothesis is true. In our example, the P-value is a probability related to the t-distribution. When we calculated the t-statistic to be approximately -3.71, we wanted to determine how extreme this result was under the assumption that the null hypothesis (\(\mu = 5\)) is correct. We did this by looking it up in a t-distribution table or using a calculator.Since the test is two-tailed, we must consider both tails of the distribution. By doing this, we found a P-value of approximately 0.0003. This small P-value indicates that observing such extreme values would be very improbable if the null hypothesis were true. That's why we compared the P-value to our significance level, \(\alpha = 0.01\), and since 0.0003 < 0.01, we rejected the null hypothesis.
t-distribution
The t-distribution is a key component of hypothesis testing, especially when dealing with small sample sizes. It's similar to a normal distribution but has thicker tails, making it more prone to yield values far from its mean. This characteristic is helpful when you do not know the population standard deviation and need to work with the sample standard deviation instead. For this exercise, we used the t-distribution to calculate the test statistic because we aimed to determine how far our sample mean (\(\bar{x}=4.87\)) was from the claimed population mean (\(\mu=5\)), given the sample standard deviation and sample size. By substituting the given values into the t-test statistic formula, we found a value of -3.71. This indicates the number of standard errors that the sample mean is away from the hypothesized population mean. When using 99 degrees of freedom (\(n-1=100-1\)), the t-distribution allows us to find the P-value associated with our test statistic.
null hypothesis
The null hypothesis is a fundamental concept in hypothesis testing. It represents the default or "no effect" assumption, stating that there is no difference or change. In this case, the null hypothesis (\(H_0: \mu = 5\)) suggests the mean weight per tablet is exactly 5 grains, as advertised.The purpose of setting up a null hypothesis is to provide a baseline against which we can compare our sample data. Under the null hypothesis assumption, we calculate probabilities to decide whether the data we have is consistent with this hypothesis. If our test results show that such data is highly unlikely, given the null hypothesis is true, we reject it. Otherwise, we fail to reject it and generally accept the null hypothesis. Rejecting the null hypothesis in our aspirin weight example suggests that the actual mean weight differs from the claimed 5 grains.
alternative hypothesis
The alternative hypothesis acts as a counterpoint to the null hypothesis in hypothesis testing. It proposes what we suspect might be true instead of the null hypothesis. In the aspirin exercise, the alternative hypothesis (\(H_a: \mu eq 5\)) suggests that the mean weight per tablet is not equal to 5 grains.Selecting the correct form of the alternative hypothesis is dependent on the research question. Here, we opt for a two-tailed test because we want to know if the weight is either more or less than the advertised 5 grains, not just one over the other. By rejecting the null hypothesis in favor of the alternative hypothesis, our analysis provides strong evidence suggesting that the tablets' mean weight does indeed differ from the claimed 5 grains, hinting that the manufacturer might need to revisit their filling process.

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Most popular questions from this chapter

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