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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: 9780137493470Non è quello che usi tu?Cambia libro di testo
Capitolo 8, Problema 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.

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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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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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Correlation Coefficient

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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Performing Hypothesis Tests: Proportions

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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Variance & Standard Deviation of Discrete Random Variables
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