91Ó°ÊÓ

An experiment to compare the tension bond strength of polymer latex modified mortar (Portland cement mortar to which polymer latex emulsions have been added during mixing) to that of unmodified mortar resulted in \(\bar{x}=18.12 \mathrm{kgf} / \mathrm{cm}^{2}\) for the modified mortar \((m=40)\) and \(\bar{y}=16.87 \mathrm{kgf} / \mathrm{cm}^{2}\) for the unmodified mortar ( \(n=32\) ). Let \(\mu_{1}\) and \(\mu_{2}\) be the true average tension bond strengths for the modified and unmodified mortars, respectively. Assume that the bond strength distributions are both normal. a. Assuming that \(\sigma_{1}=1.6\) and \(\sigma_{2}=1.4\), test \(H_{0}\) : \(\mu_{1}-\mu_{2}=0\) versus \(H_{\mathrm{a}}: \mu_{1}-\mu_{2}>0\) at level \(.01\). b. Compute the probability of a type II error for the test of part (a) when \(\mu_{1}-\mu_{2}=1\). c. Suppose the investigator decided to use a level \(.05\) test and wished \(\beta=.10\) when \(\mu_{1}-\mu_{2}=1\). If \(m=40\), what value of \(n\) is necessary? d. How would the analysis and conclusion of part (a) change if \(\sigma_{1}\) and \(\sigma_{2}\) were unknown but \(s_{1}=1.6\) and \(s_{2}=1.4\) ?

Step by step solution

01

Define Hypotheses

The null hypothesis is \( H_0: \mu_1 - \mu_2 = 0 \), which suggests that there is no difference in the true average tension bond strength between modified and unmodified mortar. The alternative hypothesis is \( H_a: \mu_1 - \mu_2 > 0 \), suggesting that the modified mortar has a greater true average tension bond strength than the unmodified.

02

Calculate Test Statistic

Using the given means \( \bar{x} = 18.12 \) and \( \bar{y} = 16.87 \), and the population standard deviations \( \sigma_1 = 1.6 \) and \( \sigma_2 = 1.4 \), we can calculate the test statistic with:\[z = \frac{(\bar{x} - \bar{y}) - 0}{\sqrt{\frac{\sigma_1^2}{m} + \frac{\sigma_2^2}{n}}}\]Substituting the values:\[z = \frac{(18.12 - 16.87)}{\sqrt{\frac{1.6^2}{40} + \frac{1.4^2}{32}}} \approx 3.97\]

03

Determine Critical Value and Make Decision

For a one-tailed test at \( \alpha = 0.01 \), the critical z-value is approximately 2.33 (from z-table). Since the calculated z-value (3.97) is greater than 2.33, we reject the null hypothesis at the 0.01 level, concluding there is sufficient evidence that \( \mu_1 > \mu_2 \).

04

Calculate Probability of Type II Error (part b)

To find the probability of Type II error (\( \beta \)), we need to determine the non-rejection region and calculate: \[z = \frac{(\bar{x} - \bar{y}) - d}{\sqrt{\frac{\sigma_1^2}{m} + \frac{\sigma_2^2}{n}}}\]If \( \mu_1 - \mu_2 = 1 \), calculate:\[z = \frac{18.12 - 16.87 - 1}{\sqrt{\frac{1.6^2}{40} + \frac{1.4^2}{32}}} \approx -0.137\]Find the probability that \( Z \leq 2.33 \) when this \( z \) is \(-0.137\), which involves finding \( \beta \) from the normal distribution.

05

Calculate Sample Size for given \( \beta = 0.10 \) (part c)

Given the power (1-\( \beta \)) = 0.90, calculate the required sample size \( n \). Set:\[1.645 + 2.33 = \frac{1}{\sqrt{\frac{1.6^2}{40} + \frac{1.4^2}{n}}}\]Rearrange and solve for \( n \). This will involve algebraic manipulations to approximate the required sample size \( n \) to achieve the desired power.

06

Analysis with Unknown Standard Deviations (part d)

If \( \sigma_1 \) and \( \sigma_2 \) are unknown but sample standard deviations \( s_1 = 1.6 \) and \( s_2 = 1.4 \) are given, use a t-test instead of a z-test. Calculate:\[t = \frac{\bar{x} - \bar{y}}{\sqrt{\frac{s_1^2}{m} + \frac{s_2^2}{n}}}\]Given \( m-1 + n-1 = 70 \) degrees of freedom, find the critical t-value for \( \alpha = 0.01 \). If \( t \) is greater, reject \( H_0 \).

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Key Concepts

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

Tension Bond Strength

Tension bond strength refers to the capacity of mortar to withstand tension forces without breaking apart. It is a crucial property when evaluating the effectiveness of construction materials, particularly mortars used in masonry. Tension is the pulling force that can act on the mortar bond between the cement and other surfaces. A higher tension bond strength indicates a more durable and resistant material.

In the context of the exercise, we compared the tension bond strengths of polymer latex modified mortar versus unmodified mortar. Adding polymer latex emulsions to mortar potentially increases its tension bond strength, making it more resilient. This is evident through hypothesis testing, which provides statistical evidence on whether the modification improves bond strength.

Normal Distribution

Normal distribution is a fundamental concept in statistics, representing how the values of a variable are dispersed across a range. It is often depicted as a symmetric, bell-shaped curve. Many natural phenomena and measurement errors naturally follow this pattern, making it an essential tool for statistical analysis.

For hypothesis testing, assuming that the data follows a normal distribution allows us to use statistical tests, like the z-test, to determine the likelihood of different outcomes. In this problem, the tension bond strengths of both modified and unmodified mortars are assumed to follow normal distributions. This assumption simplifies calculations and supports the validity of using z-scores and critical values in decision-making processes.

Type II Error

A Type II error occurs when we fail to reject a false null hypothesis. In simpler terms, it means incorrectly accepting that no effect or difference exists when, in fact, it does. The probability of making a Type II error is denoted by \( \beta \).

In the exercise, to find \( \beta \), we calculated the z-value for an assumed difference in tension bond strength between mortars. Knowing \( \beta \) is useful for understanding the test's reliability and ensuring that significant differences are not overlooked. A small \( \beta \) indicates a low probability of failing to detect true differences, increasing the test's effectiveness.

Sample Size Calculation

Sample size calculation is critical in designing experiments. It refers to determining the number of observations needed to confidently detect an effect if there is one. A sufficiently large sample size ensures the study has enough power to avoid Type II errors.

In our exercise, we needed to calculate the sample size to achieve a specific power magnitude, ensuring that the probability of committing a Type II error is kept to a minimum. Calculating the sample size involves using both the desired power level and significance level to solve for the necessary number of samples. Appropriate sample size not only improves the test's reliability but also optimizes resources by using only as many samples as necessary.