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Ch. 8 - Hypothesis Testing with Two Samples
Larson - Elementary Statistics: Picturing the World 8th Edition
Larson8th EditionElementary Statistics: Picturing the WorldISBN: 9780137493470Not the one you use?Change textbook
All textbooksLarson 8th EditionCh. 8 - Hypothesis Testing with Two SamplesProblem 8.2.19
Chapter 8, Problem 8.2.19

[APPLET] Tensile Strength
The tensile strength of a metal is a measure of its ability to resist tearing when it is pulled lengthwise. An experimental method of treatment produced steel bars with the tensile strengths (in newtons per square millimeter) listed below.
Experimental Method:
391 383 333 378 368 401 339 376 366 348
The conventional method produced steel bars with the tensile strengths (in newtons per square millimeter) listed below.
Conventional Method:
362 382 368 398 381 391 400410 396 411 385 385 395 371
At , α=0.01 can you support the claim that the experimental method of treatment makes a difference in the tensile strength of steel bars? Assume the population variances are equal.

Verified step by step guidance
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Step 1: Define the null hypothesis (H₀) and the alternative hypothesis (H₁). H₀: The mean tensile strength of steel bars produced by the experimental method is equal to the mean tensile strength of steel bars produced by the conventional method (μ₁ = μ₂). H₁: The mean tensile strength of steel bars produced by the experimental method is different from the mean tensile strength of steel bars produced by the conventional method (μ₁ ≠ μ₂).
Step 2: Choose the appropriate statistical test. Since the problem states that the population variances are equal and we are comparing the means of two independent samples, use a two-sample t-test for equal variances.
Step 3: Calculate the test statistic. Use the formula for the two-sample t-test: t = (x̄₁ - x̄₂) / sqrt((s₁²/n₁) + (s₂²/n₂)), where x̄₁ and x̄₂ are the sample means, s₁² and s₂² are the sample variances, and n₁ and n₂ are the sample sizes for the experimental and conventional methods, respectively.
Step 4: Determine the critical value or p-value. Since α = 0.01 and this is a two-tailed test, find the critical t-value from the t-distribution table with degrees of freedom calculated as df = n₁ + n₂ - 2. Alternatively, calculate the p-value using statistical software or a calculator.
Step 5: Make a decision. Compare the test statistic to the critical value or compare the p-value to α. If the test statistic falls outside the critical value range or if the p-value is less than α, reject the null hypothesis. Otherwise, fail to reject the null hypothesis. Interpret the result in the context of the problem.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Tensile Strength

Tensile strength is the maximum amount of tensile (pulling) stress that a material can withstand before failure. It is measured in units such as newtons per square millimeter (N/mm²) and is crucial for understanding how materials behave under tension. In this context, it helps compare the effectiveness of different treatment methods on steel bars.
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Hypothesis Testing

Hypothesis testing is a statistical method used to determine whether there is enough evidence to reject a null hypothesis in favor of an alternative hypothesis. In this scenario, the null hypothesis would state that there is no difference in tensile strength between the two methods, while the alternative would claim that the experimental method results in higher tensile strength. The significance level (α) indicates the probability of making a Type I error.
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Equal Variances

The assumption of equal variances, also known as homoscedasticity, is important in statistical tests like the t-test. It means that the variability in tensile strengths for both the experimental and conventional methods is similar. This assumption allows for more accurate comparisons between the two groups, as unequal variances can lead to incorrect conclusions about the differences in means.
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Related Practice
Textbook Question

Test the claim about the mean of the differences for a population of paired data at the level of significance α. Assume the samples are random and dependent, and the populations are normally distributed.

Claim: μd<0 , α=0.05 , Sample statistics: d̄ =1.5 , sd=3.2 , n=14

Textbook Question

In Exercises 11–14, test the claim about the difference between two population means and at the level of significance . Assume the samples are random and independent, and the populations are normally distributed.

Claim: μ1>μ2; α=0.10

Population statistics:σ1=40 and σ2=15

Sample Statistics: x̅1=500, n1=100, x̅2=495, n2=75

Textbook Question

Explain how to perform a two-sample t-test for the difference between two population means.

Textbook Question

In Exercises 11–14, test the claim about the difference between two population means and at the level of significance . Assume the samples are random and independent, and the populations are normally distributed.

Claim: μ1<μ2; α=0.05

Population statistics:σ1=75 and σ2=105

Sample Statistics: x̅1=2435, n1=35, x̅2=2432, n2=90

Textbook Question

Parks and Mental Health In Exercises 13–18, use the figure, which shows the percentages from a survey of two hundred 18- to 24-year-olds in the United States who say that various park and recreation activities have a positive impact on their mental health. (Adapted from National Recreation and Park Association)



Socializing and Taking Classes At α=0.05, can you support the claim that the proportion of 18- to 24-year-olds who benefit mentally from socializing in parks is different from the proportion who benefit mentally from taking classes in parks?

Textbook Question

In Exercises 11–14, test the claim about the difference between two population means and at the level of significance . Assume the samples are random and independent, and the populations are normally distributed.

Claim: μ1=μ2; α=0.1

Population statistics:σ1=3.4 and σ2=1.5

Sample Statistics: x̅1=16, n1=29, x̅2=14, n2=28