/*! 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 52 An aspirin manufacturer fills bo... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 52

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
The P-value is 0.0002; reject the null hypothesis.

Step by step solution

01

Identify the Hypotheses

First, define the null and alternative hypotheses. The null hypothesis \( H_0 \) states that the true average weight of a tablet is 5 grains, so \( H_0: \mu = 5 \). The alternative hypothesis \( H_a \) is that the average weight is not 5 grains, so \( H_a: \mu eq 5 \). This will be a two-tailed test.
02

Calculate the Test Statistic

Compute the test statistic using the formula for the sample mean: \( z = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}} \), where \( \bar{x} = 4.87 \), \( \mu = 5 \), \( \sigma = 0.35 \), and \( n = 100 \). Substitute the values to get \( z = \frac{4.87 - 5}{0.35/\sqrt{100}} = \frac{-0.13}{0.035} \approx -3.71 \).
03

Determine the P-Value

The calculated \( z \)-value is about \(-3.71\). Using the standard normal distribution, find the corresponding \( P \)-value for a two-tailed test. This \( P \)-value, based on \( z \approx -3.71 \), is approximately \( 0.0002 \).
04

Compare P-Value to Significance Level

Compare the \( P \)-value of \( 0.0002 \) with the significance level \( \alpha = 0.01 \). Since \( 0.0002 < 0.01 \), we reject the null hypothesis \( H_0 \).
05

Conclusion

The \( P \)-value is significantly lower than \( \alpha \), providing strong evidence against the null hypothesis. Thus, we conclude that there is strong evidence to suggest that the company is not filling its bottles as advertised.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

P-Value
The P-value is a crucial component of hypothesis testing. It's a probability measurement that helps us determine the significance of our test results. In simple terms, the P-value indicates the chance of obtaining the observed data, or more extreme, assuming that the null hypothesis is true.

For example, in our aspirin bottle test, we got a P-value of 0.0002. This tiny value suggests that there is only a 0.02% chance of getting a sample average weight as extreme as 4.87 grains if the true average is indeed 5 grains.
  • The smaller the P-value, the stronger the evidence against the null hypothesis.
  • In many cases, a P-value less than the significance level suggests rejecting the null hypothesis.
This makes the P-value a handy tool to decide whether a hypothesis holds up under scrutiny. It quantifies the probability and helps us make informed decisions.
Null Hypothesis
The null hypothesis, often denoted as \( H_0 \), is the starting assumption for hypothesis testing. It represents a baseline statement that there is no effect or difference. In our example, the null hypothesis was that the average weight of the tablets is exactly 5 grains (\( \mu = 5 \)).

Establishing a null hypothesis allows us to test if there's sufficient evidence to reject it.
  • It serves as the default position that the test seeks to challenge.
  • We either reject or fail to reject the null hypothesis based on data.
The interesting part is that failing to reject the null does not prove it to be true, it simply shows that there wasn't enough evidence to dispute it. Thus, it's fundamental to define it clearly for any hypothesis test.
Two-tailed Test
A two-tailed test is a type of hypothesis test that determines if there is a significant effect or difference. It checks both directions for deviation from the null hypothesis. For an aspirin test, the null considered average weight as 5 grains. But, the alternative hypothesis \((H_a)\) considered two possibilities: the weight could be less than or greater than 5 grains.

This makes the test two-tailed since we are interested in deviations in either direction.
  • Two-tailed tests require enough evidence to reject the null in either direction.
  • P-value calculations for two-tailed tests involve both tails of the distribution curve.
This type of test offers a more comprehensive examination of the data, providing more insight into variations from the expected outcomes.
Significance Level
The significance level, symbolized as \( \alpha \), is the threshold at which you decide whether to reject the null hypothesis. It represents the probability of committing a type I error, which is incorrectly rejecting a true null hypothesis. In our scenario, we set \( \alpha \) at 0.01 or 1%.

This means that we accept a 1% risk of incorrectly rejecting the null hypothesis.
  • Common significance levels are 0.05 (5%) and 0.01 (1%), each choice depending on the context and acceptable risk.
  • A smaller significance level demands stronger evidence to reject the null, thus reducing the chance of errors.
By comparing the calculated P-value with the significance level, we determine whether our findings are statistically significant. If our P-value is less than 0.01, we conclude that the evidence is strong enough to reject the null with high confidence.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Chapter 8 presented a CI for the variance \(\sigma^{2}\) of a normal population distribution. The key result there was that the rv \(\chi^{2}=(n-1) S^{2} / \sigma^{2}\) has a chi-squared distribution with \(n-1\) df. Consider the null hypothesis \(H_{0}: \sigma^{2}=\sigma_{0}^{2}\) (equivalently, \(\sigma=\sigma_{0}\) ). Then when \(H_{0}\) is true, the test statistic \(\chi^{2}=(n-1) S^{2} / \sigma_{0}^{2}\) has a chi-squared distribution with \(n-1\) df. If the relevant alternative is \(H_{\mathrm{a}}: \sigma^{2}>\sigma_{0}^{2}\), rejecting \(H_{0}\) if \((n-1) S^{2} / \sigma_{0}^{2} \geq\) \(\chi_{\alpha, n-1}^{2}\) gives a test with significance level \(\alpha\). To ensure reasonably uniform characteristics for a particular application, it is desired that the true standard deviation of the softening point of a certain type of petroleum pitch be at most \(.50^{\circ} \mathrm{C}\). The softening points of ten different specimens were determined, yielding a sample standard deviation of \(.58^{\circ} \mathrm{C}\). Does this strongly contradict the uniformity specification? Test the appropriate hypotheses using \(\alpha=.01\).

The calibration of a scale is to be checked by weighing a 10-kg test specimen 25 times. Suppose that the results of different weighings are independent of one another and that the weight on each trial is normally distributed with \(\sigma=.200 \mathrm{~kg}\). Let \(\mu\) denote the true average weight reading on the scale. a. What hypotheses should be tested? b. Suppose the scale is to be recalibrated if either \(\bar{x} \geq 10.1032\) or \(\bar{x} \leq 9.8968\). What is the probability that recalibration is carried out when it is actually unnecessary? c. What is the probability that recalibration is judged unnecessary when in fact \(\mu=10.1\) ? When \(\mu=9.8\) ? d. Let \(z=(\bar{x}-10) /(\sigma / \sqrt{n})\). For what value \(c\) is the rejection region of part (b) equivalent to the "two-tailed" region either \(z \geq\) cor \(z \leq-c\) ? e. If the sample size were only 10 rather than 25 , how should the procedure of part (d) be altered so that \(\alpha=.05\) ? f. Using the test of part (e), what would you conclude from the following sample data? \(\begin{array}{lllll}9.981 & 10.006 & 9.857 & 10.107 & 9.888 \\ 9.728 & 10.439 & 10.214 & 10.190 & 9.793\end{array}\) g. Re-express the test procedure of part (b) in terms of the standardized test statistic \(Z=(\bar{X}-10) /(\sigma / \sqrt{n)} .\)

Let \(\mu\) denote the mean reaction time to a certain stimulus. For a large- sample \(z\) test of \(H_{0}: \mu=5\) versus \(H_{\mathrm{a}}: \mu>5\), find the \(P\)-value associated with each of the given values of the \(z\) test statistic. a. \(1.42\) b. \(.90\) c. \(1.96\) d. \(2.48\) e. \(-.11\)

A sample of 12 radon detectors of a certain type was selected, and each was exposed to \(100 \mathrm{pCi} / \mathrm{L}\) of radon. The resulting readings were as follows: \(\begin{array}{rrrrrr}105.6 & 90.9 & 91.2 & 96.9 & 96.5 & 91.3 \\ 100.1 & 105.0 & 99.6 & 107.7 & 103.3 & 92.4\end{array}\) a. Does this data suggest that the population mean reading under these conditions differs from 100 ? State and test the appropriate hypotheses using \(\alpha=.05\). b. Suppose that prior to the experiment, a value of \(\sigma=7.5\) had been assumed. How many determinations would then have been appropriate to obtain \(\beta=.10\) for the alternative \(\mu=95\) ?

State DMV records indicate that of all vehicles undergoing emissions testing during the previous year, \(70 \%\) passed on the first try. A random sample of 200 cars tested in a particular county during the current year yields 124 that passed on the initial test. Does this suggest that the true proportion for this county during the current year differs from the previous statewide proportion? Test the relevant hypotheses using \(\alpha=.05\).
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Calculus

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Discrete Mathematics

Read Explanation

Pure Maths

Read Explanation

Geometry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.