/*! 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 13 A manufacturer can tolerate a sm... [FREE SOLUTION] | 91Ó°ÊÓ

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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 ten 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 ten test readings is \(.058 \mathrm{mg} / \mathrm{l}\), with a standard deviation of \(.012 \mathrm{mg} / \mathrm{l}\). Do the data provide sufficient evidence to indicate that the amount of impurities in the batch exceeds \(.05 \mathrm{mg} / 1 ?\) Find the \(p\) -value for the test and interpret its value.

Short Answer

Expert verified
Answer: Yes, there is sufficient evidence to indicate that the amount of impurities in the batch exceeds 0.05 mg/l at a 5% significance level.

Step by step solution

01

State the null and alternative hypotheses

Null hypothesis \((H_0)\): \(\mu \leq 0.05\) mg/l. Alternative hypothesis \((H_1)\): \(\mu > 0.05\) mg/l.
02

Calculate the test statistic

To calculate the test statistic, we will use the formula: $$t = \frac{\bar{x} - \mu_0}{\frac{s}{\sqrt{n}}},$$ where \(\bar{x}\) is the sample mean, \(\mu_0\) is the population mean under \(H_0\), \(s\) is the sample standard deviation, and \(n\) is the sample size. Given the information in the problem, we have \(\bar{x} = 0.058\) mg/l, \(s = 0.012\) mg/l, and \(n = 10\). The null hypothesis states that the population mean is \(\mu_0 = 0.05\) mg/l. Inserting the values: $$t = \frac{0.058 - 0.05}{\frac{0.012}{\sqrt{10}}} = \frac{0.008}{0.0038} \approx 2.10.$$ The test statistic is \(t \approx 2.10\).
03

Find the p-value

As the alternative hypothesis states that the true mean impurity level is greater than \(0.05\) mg/l, this is a one-tailed test, and we want to find the probability of getting a test statistic greater than or equal to \(2.10\). To find the p-value, we use the t-distribution with \(9\) degrees of freedom (since \(n-1 = 10-1 = 9\)). Using a t-distribution table or calculator, we find the p-value to be approximately \(0.03\).
04

Interpret the p-value

The p-value of \(0.03\) indicates that there is a \(3\%\) chance of observing a test statistic as extreme as \(t = 2.10\) or even more extreme if the null hypothesis was true. If we were to use a significance level of \(5\%\) (commonly used in practice), the p-value of \(0.03\) would be less than the significance level, and we would reject the null hypothesis. In conclusion, the data provide sufficient evidence (at \(5\%\) significance level) to indicate that the amount of impurities in the batch exceeds \(0.05\) mg/l.

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Key Concepts

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

t-distribution
When we perform hypothesis testing, particularly when dealing with small sample sizes, it's crucial to understand the role of the t-distribution. Unlike the normal distribution, which we use when the sample size is large, the t-distribution is more suitable for smaller samples because it accounts for additional variability.
The t-distribution is similar in shape to the normal distribution but has heavier tails. This means that it is more likely to produce values further from the mean. This characteristic is crucial because it provides a more conservative estimate of the population parameter, reflecting the increased uncertainty inherent in small sample sizes.
In our context, the sample consisted of 10 test readings, which qualifies as a small sample. Thus, using the t-distribution provides a better estimate of the true mean impurity level of the batch. The degrees of freedom, which in this case is 9 (calculated as the sample size minus one), affects the shape of the t-distribution. As the degrees of freedom increase, the t-distribution becomes more like the normal distribution.
p-value interpretation
The p-value is a key concept in hypothesis testing, acting as a measure of the evidence against the null hypothesis. A p-value helps us understand the likelihood of observing the test results, or something more extreme, assuming the null hypothesis is true.
In the given problem, the calculated p-value is approximately 0.03. This p-value is compared against a significance level, traditionally set at 0.05. When the p-value is less than the significance level, it suggests that the observed data is unlikely under the null hypothesis. Consequently, we might consider rejecting the null hypothesis in favor of the alternative.
For our batch impurities test, the p-value of 0.03 implies that there's a 3% probability of observing a test statistic as extreme as we did, or even more extreme, if the true impurity level were \(.05 \, \text{mg/l}\). Since 0.03 is less than the typical significance threshold of 0.05, we would reject the null hypothesis, supporting the claim that the impurities exceed the permissible limit.
experimental error
Experimental error refers to the possible inaccuracies that may occur during experimentation or measurement in the practical study. These errors can stem from various sources, such as instrument limitations or human mistakes during data collection.
In the exercise at hand, the laboratory test for impurities inherently involves experimental error. This means that the test readings may not perfectly reflect the true impurity levels due to various factors, such as fluctuations in equipment calibration or differing conditions across trials. To minimize the impact of such errors, multiple tests (10 in this case) are conducted, and their mean provides a more reliable estimate.
The assumption made in this problem is that the mean experimental error is zero, suggesting that while individual tests may deviate from the true value, over numerous trials, these deviations will average out. Hence, the mean of the test results serves as an unbiased estimate of the true impurity level.
  • Conducting repeated measurements helps to average out errors, leading to a more accurate mean estimate.
  • An unbiased estimate means that systematic error is minimized, and the average of measurements is expected to approach the true value.
Understanding and accounting for experimental error is crucial for making informed decisions based on experimental data.

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Most popular questions from this chapter

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