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Headability is the ability of a cylindrical piece of material to be shaped into the head of a bolt, screw, or other cold-formed part without cracking. The article "New Methods for Assessing Cold Heading Quality" (Wire \(J\). Intl., Oct. 1996: 66-72) described the result of a headability impact test applied to 30 specimens of aluminum killed steel and 30 specimens of silicon killed steel. The sample mean headability rating number for the steel specimens was \(6.43\), and the sample mean for aluminum specimens was \(7.09\). Suppose that the sample standard deviations were \(1.08\) and \(1.19\), respectively. Do you agree with the article's authors that the difference in headability ratings is significant at the \(5 \%\) level (assuming that the two headability distributions are normal)?

Step by step solution

01

Define Hypotheses

To determine if there is a significant difference between the two sample means, we perform a hypothesis test. Let \( \mu_1 \) be the mean headability rating for aluminum killed steel and \( \mu_2 \) be the mean for silicon killed steel. The null hypothesis \( H_0 \) is that there is no difference in means: \( H_0: \mu_1 = \mu_2 \). The alternative hypothesis \( H_a \) is that there is a difference: \( H_a: \mu_1 eq \mu_2 \). This is a two-tailed test.

02

Choose Significance Level

The significance level \( \alpha \) is given as 0.05. This is the threshold we use to determine whether or not the observed difference is statistically significant.

03

Calculate the Test Statistic

We use a two-sample t-test for comparing means since the standard deviations are known. The test statistic is calculated using the formula:\[t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}\]where \( \bar{x}_1 = 7.09 \), \( \bar{x}_2 = 6.43 \), \( s_1 = 1.08 \), \( s_2 = 1.19 \), \( n_1 = 30 \), and \( n_2 = 30 \). Substituting these values gives us:\[t = \frac{7.09 - 6.43}{\sqrt{\frac{1.08^2}{30} + \frac{1.19^2}{30}}} \approx 2.52\]

04

Determine the Critical Value

For a two-tailed test at the 5% significance level with degrees of freedom calculated using a conservative approach (\( n_1 - 1 + n_2 - 1 = 58 \)), we find the critical value from the t-distribution. Using a t-table or calculator, this critical value is approximately \( \pm2.002 \).

05

Make a Decision

Since the calculated test statistic \( t = 2.52 \) is greater than the critical value \( 2.002 \) (in magnitude), we reject the null hypothesis. This suggests that the difference in headability ratings is statistically significant at the 5% level.

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Key Concepts

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

Hypothesis Testing

When conducting hypothesis testing, we aim to make informed decisions about population parameters based on sample data. This technique helps us determine if observed differences between groups are significant or merely due to random chance. In our context, we are comparing headability between aluminum killed steel and silicon killed steel. By setting up our hypotheses, we differentiate between the assumption of no effect (null hypothesis) and the presence of an effect (alternative hypothesis). This serves as the foundation for our statistical evaluation.

Significance Level

The significance level, represented by \( \alpha \), is a crucial component in hypothesis testing. It indicates the probability threshold for rejecting the null hypothesis. In simple terms, \( \alpha \) helps us control the chances of making a Type I error, which is accepting an effect that doesn't exist. For this exercise, a common choice of \( \alpha = 0.05 \) implies that we allow a 5% chance of incorrectly rejecting the null hypothesis. This balance ensures that our findings are carry trustworthy implications and minimize erroneous conclusions.

Critical Value

Critical values play an essential role in hypothesis testing. At a 5% significance level, critical values define boundary points where we reject the null hypothesis if the test statistic falls outside this range. Specifically, for a two-tailed test like ours, critical values indicate where significant differences might lie on both ends of the distribution. By referring to statistical tables or software, we found our critical values to be approximately \( \pm2.002 \), determining if the observed effects surpass random variation.

Test Statistic

The test statistic quantifies the degree to which sample data deviate from the null hypothesis. It's calculated using a specific formula that considers sample means, standard deviations, and sizes. For our two-sample t-test, we used:

• \( \bar{x}_1 = 7.09 \) (mean of aluminum killed steel)
• \( \bar{x}_2 = 6.43 \) (mean of silicon killed steel)
• \( s_1 = 1.08 \) (standard deviation for aluminum)
• \( s_2 = 1.19 \) (standard deviation for silicon)
• \( n_1 = n_2 = 30 \) (sample sizes)

Substituting into the t-test formula, \[ t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \approx 2.52 \] This computed value is crucial as it assists in determining the significance of the difference between our groups.

Null Hypothesis

The null hypothesis, symbolized by \( H_0 \), proposes that there is no meaningful difference between the groups being compared. Contrary to the alternative hypothesis, it serves as the status quo we seek to disprove. For this exercise, the null hypothesis is that the mean headability ratings for aluminum and silicon killed steel are equal: \( \mu_1 = \mu_2 \). This assumption acts as our baseline, directing our test toward identifying whether the observed variations are statistically significant.