/*! 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 6 The slant shear test is widely a... [FREE SOLUTION] | 91Ó°ÊÓ

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The slant shear test is widely accepted for evaluating the bond of resinous repair materials to concrete; it utilizes cylinder specimens made of two identical halves bonded at \(30^{\circ}\). The article "Testing the Bond Between Repair Materials and Concrete Substrate" (ACI Materials J., 1996: 553-558) reported that for 12 specimens prepared using wire-brushing, the sample mean shear strength \(\left(\mathrm{N} / \mathrm{mm}^{2}\right)\) and sample standard deviation were \(19.20\) and \(1.58\), respectively, whereas for 12 hand-chiseled specimens, the corresponding values were \(23.13\) and \(4.01\). Does the true average strength appear to be different for the two methods of surface preparation? State and test the relevant hypotheses using a significance level of 05 . What are you assuming about the shear strength distributions?

Short Answer

Expert verified
Perform a two-sample t-test assuming normal distribution and equal variance; reject the null hypothesis if t exceeds critical value.

Step by step solution

01

State Hypotheses

We need to determine if the true average shear strength for wire-brushing differs from hand-chiseled preparation. Let's denote \( \mu_1 \) as the mean shear strength of wire-brushed specimens and \( \mu_2 \) as the mean shear strength of hand-chiseled specimens. - Null Hypothesis \( H_0: \mu_1 = \mu_2 \) - Alternative Hypothesis \( H_a: \mu_1 eq \mu_2 \) The null hypothesis assumes there is no difference in means, while the alternative suggests a difference exists.
02

Identify Test and Assumptions

We will use a two-sample t-test to compare the means. Assumptions include: 1. Both samples are randomly selected and independent. 2. Shear strengths are normally distributed. 3. The variances of the two populations are equal. This assumption will allow us to use a pooled variance for the test.
03

Calculate Test Statistic

First, compute the pooled standard deviation \( s_p \):\[ s_p = \sqrt{\frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2}} \]Given: \( n_1 = 12, \) \( s_1 = 1.58, \) \( n_2 = 12, \) \( s_2 = 4.01 \), calculate \( s_p \). Then calculate the t-statistic:\[ t = \frac{\bar{x}_1 - \bar{x}_2}{s_p \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}} \]Substitute \(\bar{x}_1 = 19.20\) and \(\bar{x}_2 = 23.13\) to find \( t \).
04

Determine Critical Value and Compare

With \( df = n_1 + n_2 - 2 = 22 \) and a significance level of 0.05, find the critical t-value for a two-tailed test using a t-distribution table. Compare the calculated t-statistic to these critical values to decide whether to reject or fail to reject the null hypothesis.
05

Conclusion

If the absolute t-statistic value exceeds the critical value, reject the null hypothesis, indicating a significant difference in mean shear strengths. If not, fail to reject the null hypothesis, implying no significant difference regarding the significance level specified.
06

Assumption Examination

For the validity of the t-test results, ensure: 1. Both samples appear from normal distributions, or sample sizes are large enough for the Central Limit Theorem to apply. 2. Equality of variances holds, which can be checked using an F-test or by inspecting standard deviation similarity.

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Key Concepts

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

Hypothesis Testing
To investigate whether the average shear strengths are different for two surface preparation methods, we apply hypothesis testing.
Hypothesis testing is a statistical method of testing a hypothesis about the parameters of a population, based on sample data.
This involves setting up two opposing hypotheses: the null hypothesis (H_0) and the alternative hypothesis (H_a). - **Null Hypothesis (H_0):** Suggests there is no effect or difference. In this case, the average shear strengths for both methods are the same, i.e., \( \mu_1 = \mu_2 \).- **Alternative Hypothesis (H_a):** Indicates a difference exists. Here, it means that the shear strengths are different, i.e., \( \mu_1 eq \mu_2 \).We then determine whether to reject the null hypothesis in favor of the alternative using statistical evidence from our sample.
This decision is based on the computation of a test statistic and comparing it with a critical value that is derived from a specified significance level.
Assumptions of Normality
Ensuring assumptions are met is crucial in hypothesis testing. One key assumption for the t-test is normality.
Normality means the data should follow a normal distribution, or at least approximately, when sample sizes are small. This allows for the application of the Central Limit Theorem, which ensures that the sampling distribution of the sample mean is close to normal as the sample size increases.
Here’s what you need to check:
  • Each group should be independently sampled from the population.
  • The data in each group should come from a normally distributed population, especially important if the sample size is small (typically, n < 30).
If data is not normally distributed and the sample size is small, the t-test might not be appropriate.
However, when samples are large enough or nearly normal, the results are reliable due to the Central Limit Theorem.
Pooled Variance
In a two-sample t-test, the concept of pooled variance is often used to find a common measure of variance across samples.
Pooled variance assumes equality of variances between the populations from which samples are drawn. This helps in achieving a more accurate estimate of the overall variance when we suspect or are told that both variances should be similar.To calculate pooled variance:\[s_p = \sqrt{\frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2}}\]where:
  • \(n_1\) and \(n_2\) are the sample sizes,
  • \(s_1^2\) and \(s_2^2\) are the sample variances.
The pooled variance gives a weighted average of the variances from two groups, taking into account their respective sample sizes. Once pooled variance is calculated, it’s used to find the t-statistic, which helps in determining statistical significance.
Significance Level
The significance level is a probability threshold used in hypothesis testing to determine whether an observed difference is statistically significant.
It represents the risk of rejecting the null hypothesis when it is actually true, often denoted by \( \alpha \).For this exercise, we use a significance level of 0.05, which means:- There is a 5% risk of concluding that there is a difference between mean shear strengths when there actually isn’t.- The critical region for rejecting the null hypothesis is determined by this level. To decide whether to reject \( H_0 \), compare the t-statistic to critical values from the t-distribution based on \( \alpha = 0.05 \).If the t-statistic falls into the critical region, \( H_0 \) is rejected, indicating significant evidence that the true mean shear strengths are different across preparation methods.
If not, \( H_0 \) is not rejected, suggesting any observed difference is likely due to chance.

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Most popular questions from this chapter

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