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Chapter 10: Problem 68

Two chemical companies can supply a raw material. The concentration of a particular element in this material is important. The mean concentration for both suppliers is the same, but you suspect that the variability in concentration may differ for the two companies. The standard deviation of concentration in a random sample of \(n_{1}=10\) batches produced by company 1 is \(s_{1}=4.7\) grams per liter, and for company \(2,\) a random sample of \(n_{2}=16\) batches yields \(s_{2}=5.8\) grams per liter. Is there sufficient evidence to conclude that the two population variances differ? Use \(\alpha=0.05\).

Short Answer

Expert verified
No, there is not sufficient evidence to conclude that the variances differ.

Step by step solution

01

Identify the Hypotheses

We are testing if there is a difference in variances. Therefore, our hypotheses are: **Null Hypothesis (H_0):** \( \sigma_1^2 = \sigma_2^2 \) and **Alternative Hypothesis (H_a):** \( \sigma_1^2 eq \sigma_2^2 \).
02

Determine the Test Statistic

For comparing variances, we use the F-test statistic: \( F = \frac{s_1^2}{s_2^2} \), where \( s_1 = 4.7 \) and \( s_2 = 5.8 \).
03

Calculate the F-statistic

Compute \( s_1^2 = (4.7)^2 = 22.09 \) and \( s_2^2 = (5.8)^2 = 33.64 \). Then, calculate: \[ F = \frac{22.09}{33.64} \approx 0.656. \]
04

Determine the Critical Value

Look up critical values in the F-distribution table with \( df_1 = n_1 - 1 = 9 \) (numerator degrees of freedom) and \( df_2 = n_2 - 1 = 15 \) (denominator degrees of freedom) for \( \alpha = 0.05 \).
05

Decision Rule

The F-distribution is not symmetric, thus we have two critical values: \( F_{low} \) and \( F_{high} \). If \( F < F_{low} \) or \( F > F_{high} \), reject \( H_0 \). For \( \alpha = 0.05 \), the critical values are approximately \( F_{low} = 0.326 \) and \( F_{high} = 3.178 \).
06

Make the Decision

Since \( 0.656 \) is not less than \( 0.326 \) or greater than \( 3.178 \), we fail to reject the null hypothesis. There is not sufficient evidence to conclude that the variances differ.

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Key Concepts

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

Hypothesis Testing
Hypothesis testing is a crucial method in statistics that helps us decide whether there is enough evidence to support a certain belief about a population parameter.
It aims to test a hypothesis by using sample data.
In the context of the F-test for variance, hypothesis testing involves comparing two population variances to see if they differ significantly.

A typical hypothesis test involves:
  • Formulating two contrasting hypotheses:
    • The Null Hypothesis (H_0): Assumes no change or effect; in our case, it suggests that the two population variances are equal (\( \sigma_1^2 = \sigma_2^2 \)).
    • The Alternative Hypothesis (H_a): Assumes a change or effect; here, it implies that the variances differ (\( \sigma_1^2 eq \sigma_2^2 \)).
  • Gathering and analyzing sample data
  • Using statistical methods to assess the hypothesis - the F-test is particularly used for comparing variances.
A decision is made whether there's sufficient evidence to reject the null in favor of the alternative.
Critical Value
The critical value is a point on the test distribution that is compared to the test statistic to decide whether to reject the null hypothesis.
In hypothesis testing, the critical value helps set the boundaries for the decision-making process.

When performing an F-test:
  • You determine critical values using the F-distribution table that corresponds to your chosen significance level, \( \alpha \), often set at 0.05 for a 5% risk of error.
  • The degrees of freedom (\( df_1 \) for numerator and \( df_2 \) for denominator) are also important since they influence the shape of the distribution.
  • In our example, with \( df_1 = 9 \) and \( df_2 = 15 \), and \( \alpha = 0.05 \), we check an F-table to find the lower (F_{low}) and upper (F_{high}) critical values, approximately 0.326 and 3.178 respectively.
The test statistic is then compared against these boundaries to make a decision about the null hypothesis.
Statistical Decision Making
Statistical decision making is all about using the results from hypothesis testing to make informed conclusions about a population.
It guides us on whether to accept or reject the null hypothesis based on statistical evidence.

When employing this decision-making approach using the F-test:
  • Compute the F-statistic, which in our scenario was approximately 0.656.
  • Compare this F-statistic to the calculated critical values of the F-distribution, \( F_{low} = 0.326 \) and \( F_{high} = 3.178 \).
  • Make a decision:
    • If F falls outside the range (i.e., less than F_{low} or greater than F_{high}), reject the null hypothesis.
    • If F is within the range, as it was at 0.656, we fail to reject the null hypothesis.
Since our F-statistic was within the critical value range, we concluded insufficient evidence to show the variances differ.
Degrees of Freedom
Degrees of freedom are essential in statistical tests as they influence the shape and spread of the sampling distribution.
They can be thought of as the number of independent values available to estimate a parameter.

In the context of an F-test for variance:
  • Degrees of freedom for the numerator (\( df_1 \)) are calculated as \( n_1 - 1 \); in our example, \( df_1 = 10 - 1 = 9 \).
  • Degrees of freedom for the denominator (\( df_2 \)) are given by \( n_2 - 1 \); here, \( df_2 = 16 - 1 = 15 \).
  • These values are necessary to accurately determine the critical values from F-distribution tables, which, in turn, help make an informed statistical decision.
Understanding degrees of freedom helps in grasping why we use specific critical values for a given test.

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Most popular questions from this chapter

Consider the hypothesis test \(H_{0}: \mu_{1}=\mu_{2}\) against \(H_{1}: \mu_{1} \neq \mu_{2}\) with known variances \(\sigma_{1}=10\) and \(\sigma_{2}=5 .\) Suppose that sample sizes \(n_{1}=10\) and \(n_{2}=15\) and that \(\bar{x}_{1}=4.7\) and \(\bar{x}_{2}=7.8 .\) Use \(\alpha=0.05\) (a) Test the hypothesis and find the \(P\) -value. (b) Explain how the test could be conducted with a confidence interval (c) What is the power of the test in part (a) for a true difference in means of \(3 ?\) (d) Assume that sample sizes are equal. What sample size should be used to obtain \(\beta=0.05\) if the true difference in means is 3 ? Assume that \(\alpha=0.05\).

The brcaking strength of yarn supplicd by two manufacturers is being investigated. You know from experience with the manufacturers' processes that \(\sigma_{1}=5\) psi and \(\sigma_{2}=4\) psi. A random sample of 20 test specimens from each manufacturer results in \(\bar{x}_{1}=88\) psi and \(\bar{x}_{2}=91\) psi, respectively. (a) Using a \(90 \%\) confidence interval on the difference in mean breaking strength, comment on whether or not there is cvidence to support the claim that manufacturer 2 produces yarn with higher mean breaking strength. (b) Using a \(98 \%\) confidence interval on the difference in mean breaking strength, comment on whether or not there is evidence to support the claim that manufacturer 2 produces yarn with higher mean breaking strength. (c) Comment on why the results from parts (a) and (b) are different or the same. Which would you choose to make your decision and why?

The manufacturer of a hot tub is interested in testing two different heating elements for its product. The element that produces the maximum heat gain after 15 minutes would be preferable. The manufacturer obtains 10 samples of each heating unit and tests each one. The heat gain after 15 minutes (in \({ }^{\circ} \mathrm{F}\) ) follows. $$\begin{array}{l|l}\text { Unit 1: } & 25,27,29,31,30,26,24,32,33,38 \\\\\hline \text { Unit 2: } & 31,33,32,35,34,29,38,35,37,30\end{array}$$ (a) Is there any reason to suspect that one unit is superior to the other? Use \(\alpha=0.05\) and the Wilcoxon rank sum test. (b) Use the normal approximation for the Wilcoxon rank sum test. Assume that \(\alpha=0.05 .\) What is the approximate \(P\) -value for this test statistic?

A polymer is manufactured in a batch chemical process. Viscosity measurements are normally made on each batch, and long experience with the process has indicated that the variability in the process is fairly stable with \(\sigma=20 .\) Fifteen batch viscosity measurements are given as follows: $$\begin{array}{l}724,718,776,760,745,759,795,756,742,740,761, \\\749,739,747,742\end{array}$$ A process change that involves switching the type of catalyst used in the process is made. Following the process change, eight batch viscosity measurements are taken: $$735,775,729,755,783,760,738,780$$ Assume that process variability is unaffected by the catalyst change. If the difference in mean batch viscosity is 10 or less, the manufacturer would like to detect it with a high probability. (a) Formulate and test an appropriate hypothesis using \(\alpha=0.10 .\) What are your conclusions? Find the \(P\) -value. (b) Find a \(90 \%\) confidence interval on the difference in mean batch viscosity resulting from the process change. (c) Compare the results of parts (a) and (b) and discuss your findings.

Consider the hypothesis test \(H_{0}: \mu_{1}=\mu_{2}\) against \(H_{1}: \mu_{1}<\mu_{2}\) with known variances \(\sigma_{1}=10\) and \(\sigma_{2}=5 .\) Suppose that sample sizes \(n_{1}=10\) and \(n_{2}=15\) and that \(\bar{x}_{1}=14.2\) and \(\bar{x}_{2}=19.7 .\) Use \(\alpha=0.05\) (a) Test the hypothesis and find the \(P\) -value. (b) Explain how the test could be conducted with a confidence interval. (c) What is the power of the test in part (a) if \(\mu_{1}\) is 4 units less than \(\mu_{2} ?\) (d) Assume that sample sizes are equal. What sample size should be used to obtain \(\beta=0.05\) if \(\mu_{1}\) is 4 units less than \(\mu_{2} ?\) Assume that \(\alpha=0.05 .\)
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