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Use the following information to answer the next five exercises. Two metal alloys are being considered as material for ball bearings. The mean melting point of the two alloys is to be compared. 15 pieces of each metal are being tested. Both populations have normal distributions. The following table is the result. It is believed that Alloy Zeta has a different melting point. $$\begin{array}{|l|l|l|}\hline & {\text { Sample Mean Melting Temperatures }\left(^{\circ} F\right)} & {\text { Population Standard Deviation }} \\\ \hline \text { Alloy Gamma } & {800}&{95} \\ \hline \text { Alloy zeta } & {900} &{105} \\ \hline\end{array}$$ At the 1% significance level, what is your conclusion?

Step by step solution

01

State the Hypotheses

Let's set up the null and alternative hypotheses. The null hypothesis \( H_0 \) states that the mean melting point of Alloy Gamma and Alloy Zeta is the same, i.e., \( \mu_1 = \mu_2 \). The alternative hypothesis \( H_a \) states that the mean melting point of Alloy Gamma and Alloy Zeta are different, i.e., \( \mu_1 eq \mu_2 \).

02

Determine the Significance Level

The problem states that the test should be conducted at the 1% significance level, \( \alpha = 0.01 \).

03

Calculate the Test Statistic

The formula for the test statistic \( z \) in this scenario (comparison of two means with known standard deviations) is:\[ z = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}} \]where \( \bar{x}_1 = 800 \), \( \bar{x}_2 = 900 \), \( \sigma_1 = 95 \), \( \sigma_2 = 105 \), and both sample sizes \( n_1 = n_2 = 15 \). Plugging in the values gives:\[ z = \frac{800 - 900}{\sqrt{\frac{95^2}{15} + \frac{105^2}{15}}} \]

04

Solve for the Test Statistic

Continuing the calculation:\[ z = \frac{-100}{\sqrt{\frac{9025}{15} + \frac{11025}{15}}} = \frac{-100}{\sqrt{601.67 + 735}} = \frac{-100}{33.48} \approx -2.99 \]

05

Determine the Critical Value

Since this is a two-tailed test at \( \alpha = 0.01 \), the critical z-values are \( -2.576 \) and \( 2.576 \). This is often found using a z-table or standard normal distribution calculator.

06

Make the Decision

Compare the calculated \( z \)-value of \(-2.99\) with the critical values of \(-2.576\) and \(2.576\). Since \(-2.99\) is less than \(-2.576\), we reject the null hypothesis.

07

Conclusion

At the 1% significance level, we have enough evidence to conclude that there is a significant difference between the mean melting points of Alloy Gamma and Alloy Zeta.

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Key Concepts

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

Null Hypothesis

The concept of a null hypothesis is the foundation of hypothesis testing. In our scenario, the null hypothesis is denoted by \( H_0 \). It claims there is no difference in the mean melting points of the alloys being compared. The null hypothesis thus asserts that Alloy Gamma and Alloy Zeta have equal melting points, mathematically expressed as \( \mu_1 = \mu_2 \).
- Represents no effect or no difference.- Assumed true until evidence suggests otherwise.
Rejecting the null hypothesis implies that there is sufficient statistical evidence to support an alternative claim.

Alternative Hypothesis

While the null hypothesis implies no difference, the alternative hypothesis challenges it. In this case, the alternative hypothesis \( H_a \) suggests that the mean melting points of Alloy Gamma and Alloy Zeta are not the same. This means that there is a discernible difference, which we express mathematically as \( \mu_1 eq \mu_2 \).
- Contradicts the null hypothesis.- Suggests there is an effect or a difference present.
When hypothesis testing results in rejecting \( H_0 \), it indicates that the alternative hypothesis (\( H_a \)) is more consistent with the data.

Significance Level

The significance level, denoted as \( \alpha \), is a critical threshold in hypothesis testing. For our exercise, the significance level is set at 1%, expressed as \( \alpha = 0.01 \). This level defines the probability of rejecting the null hypothesis when it is true, also known as a Type I error.
- Acts as a benchmark for decision-making.- Smaller \( \alpha \) values demand more substantial evidence to reject \( H_0 \).
In essence, the significance level gives us control over how stringently we test the null hypothesis.

Test Statistic

The test statistic is a calculated value we use to test our hypotheses. In this context, the test statistic is represented by \( z \), derived from the difference in sample means. For the alloys, it's calculated using this formula:\[ z = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}} \].
- Measures how far the sample data is from the null hypothesis.- Helps determine the likelihood of observing the data assuming \( H_0 \) is true.
Our calculation gave us \( z \approx -2.99 \), informing us whether the observed difference is significant.

Critical Value

The critical value is a point on the scale of the test statistic beyond which we reject the null hypothesis. For this two-tailed test at a 1% significance level, our critical z-values are \(-2.576\) and \(2.576\). These values mark the boundaries of the critical region:
- Determined by the chosen \( \alpha \) value.- If the test statistic exceeds the critical values, \( H_0 \) is rejected.
Since our test statistic \(-2.99\) is outside the critical value range, we reject the null hypothesis, confirming a significant difference in melting points.