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Acetaminophen is an active ingredient found in more than 600 over-the-counter and prescription medicines, such as pain relievers, cough suppressants, and cold medications. It is safe and effective when used correctly, but taking too much can lead to liver damage. A researcher believes the mean amount of acetaminophen per tablet in a particular brand of cold tablets is different from the 600 mg claimed by the manufacturer. A random sample of 30 tablets had a mean acetaminophen content of \(596.3 \mathrm{mg}\) with a standard deviation of \(4.7 \mathrm{mg}\). a. Is the assumption of normality reasonable? Explain. b. Construct a \(99 \%\) confidence interval for the estimate of the mean acetaminophen content. c. What does the confidence interval found in part b suggest about the mean acetaminophen content of one pill? Do you believe there is 600 mg per tablet? Explain.

Short Answer

Expert verified
The short answer depends on the results of the calculations. If the computed confidence interval includes 600 mg, then yes, the claim of 600 mg per tablet is plausible. If not, the claim is unlikely to be true.

Step by step solution

01

Assess the Assumption of Normality

Given a sample size of 30, which is more than the threshold of about 30 for the Central Limit Theorem, it is reasonable to assume normality for this sample.
02

Compute the Standard Error

Standard Error (SE) can be computed using the formula \( SE = \frac{s}{\sqrt{n}} \), where 's' is the sample standard deviation, and 'n' is the sample size. Plug in the given 's' value of 4.7 mg and 'n' of 30 into the formula to compute the standard error.
03

Find the z-score

We seek to construct a 99% confidence interval, leaving 1% in the tails of the normal distribution. Given symmetry, half of this, or 0.5%, will be left in the tail on the right. A z-table or calculator can be used to find the z-score that leaves 99.5% to the left under the standard normal curve, which is approximately 2.576.
04

Calculate the Confidence Interval

Substitute the computed standard error and the detected z-score into the confidence interval formula: \[ \overline{x} \pm z \times SE \], where \( \overline{x} \) is the sample mean, to compute your confidence interval.
05

Interpret the Confidence Interval

The computed confidence interval shows the range of values for the true population mean with our level of confidence of 99%. If this interval includes the claimed 600 mg, it is plausible that the manufacturer's claim is true. If the confidence interval does not contain 600 mg, then there is strong evidence to suggest that the manufacturer's claim is false.

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Key Concepts

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

Central Limit Theorem
The Central Limit Theorem (CLT) is a fundamental principle in statistics that offers an explanation as to why the normal distribution arises so commonly in various datasets, especially when dealing with sample averages. The theorem states that when an independent and identically distributed (i.i.d.) sample with a sufficiently large size is taken from a population (usually n > 30 is considered large enough), the sampling distribution of the sample mean will approximate a normal distribution, regardless of the shape of the population distribution.

This principle is crucial when addressing problems like the one presented in the exercise where the researcher is trying to establish the average content of acetaminophen per tablet. With the sample size of 30 tablets, we utilize the CLT to justify the assumption of normality, which allows us to use methods that rely on normal distributions, such as constructing confidence intervals or conducting hypothesis tests.
Normal Distribution
A normal distribution, also known as a Gaussian distribution, is a continuous probability distribution that is symmetric around the mean, showing that data near the mean are more frequent in occurrence than data far from the mean. In the context of the exercise, assessing the normality of the sample distribution is essential because many statistical tests and confidence intervals rely on the assumption of normality.

The properties of a normal distribution include the mean, median, and mode being equal and the distribution being defined entirely by its mean and standard deviation. The center of the normal distribution is at the mean, and the spread or dispersion is determined by the standard deviation. Normal distributions are denoted as N(mean, standard deviation) and are the basis for z-scores, which are used to calculate confidence intervals.
Sample Size
Sample size refers to the number of observations or measurements taken from a population to form a sample. It plays a critical role in statistical analyses because it affects the reliability of the results. A larger sample size reduces the margin of error and yields more precise estimates of population parameters.

In our exercise scenario, 30 cold tablets make up the sample size. This number, being greater than 30, is typically seen as a sufficient sample size for the CLT to hold, ensuring that the distribution of the sample means approaches a normal distribution. The sample size is also used in calculating the standard error, which is necessary for creating confidence intervals and conducting hypothesis tests.
Standard Error
The standard error (SE) of a statistic (most commonly the mean) is essentially an estimate of that statistic's standard deviation if the same experiment or study were to be repeated multiple times. It is calculated as the standard deviation of the sample divided by the square root of the sample size. The SE gives us a measure of the variability or precision of the sample mean as an estimate of the population mean.

In our exercise, the standard deviation is provided, and we have the sample size; thus, we can compute the SE. The smaller the SE, the more accurate our estimate of the population mean is likely to be. Since the sample size exists in the denominator of the SE formula, the larger the sample, the smaller the SE, and hence, the more precise our estimate will be.
Z-score
A z-score, or standard score, represents the number of standard deviations a data point is from the mean. In the context of constructing confidence intervals, the z-score indicates how many standard deviations we need to go from the mean of a normal distribution to capture the desired confidence level. For example, a z-score of 2 would mean a data point is 2 standard deviations away from the mean.

For our exercise, the 99% confidence level leads us to seek a z-score that captures 99% of the data within a normal distribution. The respective z-score is used in conjunction with the standard error to construct the confidence interval, indicating the range within which we expect the population mean to lie with a specified level of confidence.
Statistical Significance
Statistical significance is a determination about the likelihood of the observed result being due to chance. It's typically assessed through p-values, where a low p-value indicates that the obtained result would be very unlikely if there were no true effect (for example, no difference between groups or no association between variables).

In the context of our exercise, determining whether the mean acetaminophen content per tablet is statistically significantly different from the claimed 600 mg would involve hypothesis testing. For example, if the calculated confidence interval does not contain the manufacturer's claimed amount, we might conclude there is significant evidence to doubt the claim, albeit with a certain risk of being incorrect, known as the significance level.
Hypothesis Testing
Hypothesis testing is a structured process used to determine if there is enough statistical evidence in a sample of data to infer that a certain condition is true for the entire population. It begins with the formulation of two opposing statements: the null hypothesis, which usually represents the status quo or a baseline value, and the alternative hypothesis, which represents what the researcher aims to support.

In this exercise, a researcher might set up the null hypothesis to assert that the true mean content is 600 mg per tablet, whereas the alternative hypothesis might claim it is different from 600 mg. Using the distribution of the sample mean, standard error, and an appropriate test statistic, we would then make a decision to either reject or fail to reject the null hypothesis based on the level of statistical significance and the data at hand.

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Most popular questions from this chapter

Although most people are aware of minor dehydration symptoms such as dry skin and headaches, many are less knowledgeable about the causes of dehydration. According to a poll done for the Nutrition Information Center, the results of a random sample of 3003 American adults showed that \(20 \%\) did not know that caffeine dehydrates. The survey listed a margin of error of plus or minus \(1.8 \%\). a. Describe how this survey of 3003 American adults fits the properties of a binomial experiment. Specifically identify \(n,\) a trial, success, \(p,\) and \(x .\) b. What is the point estimate for the proportion of all Americans who did not know that caffeine dehydrates? Is it a parameter or a statistic? c. Calculate the \(95 \%\) confidence maximum error of estimate for a binomial experiment of 3003 trials that result in an observed proportion of \(0.20 .\) d. How is the maximum error, found in part c, related to the \(1.8 \%\) margin of error quoted in the survey report? e. Find the \(95 \%\) confidence interval for the true proportion \(p\) based on a binomial experiment of 3003 trials that results in an observed proportion of 0.20.

The uniform length of nails is very important to a carpenter-the length of the nails being used are matched to the materials being fastened together, thereby making a small standard deviation an important property of the nails. A sample of 35 randomly selected 2-inch nails is taken from a large quantity of Nails, Inc.'s, recent production run. The resulting length measurements have a mean length of 2.025 inches and a standard deviation of 0.048 inch. a. Determine whether an assumption of normality is reasonable. Explain. b. Is the sample evidence sufficient to reject the idea that the nails have a mean length of 2 inches? Use \(\alpha=0.05\). c. Is there sufficient evidence, at the 0.05 level, to show that the length of nails from this production run has a standard deviation greater than the advertised 0.040 inch? d. Write a short report outlining the findings and recommendations as to whether or not the carpenter should use these nails for an application that requires 2 -inch nails.

Length is not very important in evaluating the quality of corks because it has little to do with the effectiveness of a cork in preserving wine. Winemakers have several lengths to choose from and order the length of cork they prefer (long corks tend to make a louder pop when the bottle is uncorked). Length is monitored very closely, though, because it is a specified quality of the cork. The lengths of no. 9 natural corks \((24 \mathrm{mm}\) diameter by \(45 \mathrm{mm}\) length) have a normal distribution. Twelve randomly selected corks were measured to the nearest hundredth of a millimeter. $$\begin{array}{llllll}\hline 44.95 & 44.95 & 44.80 & 44.93 & 45.22 & 44.82 \\\45.12 & 44.62 & 45.17 & 44.60 &44.60 & 44.75 \\\\\hline\end{array}$$ a. Does the preceding sample give sufficient reason to show that the mean length is different from \(45.0 \mathrm{mm}\) at the 0.02 level of significance? A different random sample of 18 corks is taken from the same batch. $$\begin{array}{lllllllll}\hline 45.17 & 45.02 & 45.30 & 45.14 & 45.35 & 45.50 & 45.26 & 44.88 & 44.71 \\\44.07 & 45.10 & 45.01 & 44.83 & 45.13 & 44.69 & 44.89 & 45.15 & 45.13 \\\\\hline\end{array}$$ b. Does the preceding sample give sufficient reason to show that the mean length is different from \(45.0 \mathrm{mm}\) at the 0.02 level of significance? c. What effect did the two different sample means have on the calculated test statistic in parts a and b? Explain. d. What effect did the two different sample sizes have on the calculated test statistic in parts a and b? Explain. e. What effect did the two different sample standard deviations have on the calculated test statistic in parts a and b? Explain.

An insurance company states that \(90 \%\) of its claims are settled within 30 days. A consumer group selected a random sample of 75 of the company's claims to test this statement. If the consumer group found that 55 of the claims were settled within 30 days, does it have sufficient reason to support the contention that less than \(90 \%\) of the claims are settled within 30 days? Use \(\alpha=0.05\). a. Solve using the \(p\) -value approach. b. Solve using the classical approach.

A bank randomly selected 250 checking account customers and found that 110 of them also had savings accounts at the same bank. Construct a \(95 \%\) confidence interval for the true proportion of checking account customers who also have savings accounts.

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