91Ó°ÊÓ

The melting points of two alloys used in formulating solder were investigated by melting 21 samples of each material. The sample mean and standard deviation for alloy 1 was \(\bar{x}_{1}=420^{\circ} \mathrm{F}\) and \(s_{1}=4^{\circ} \mathrm{F},\) and for alloy \(2,\) they were \(\bar{x}_{2}=426^{\circ} \mathrm{F}\) and \(s_{2}=3^{\circ} \mathrm{F}\) (a) Do the sample data support the claim that both alloys have the same melting point? Use \(\alpha=0.05\) and assume that both populations are normally distributed and have the same standard deviation. Find the \(P\) -value for the test. (b) Suppose that the true mean difference in melting points is \(3^{\circ} \mathrm{F}\). How large a sample would be required to detect this difference using an \(\alpha=0.05\) level test with probability at least \(0.9 ?\) Use \(\sigma_{1}=\sigma_{2}=4\) as an initial estimate of the common standard deviation.

Step by step solution

01

Understand the Hypotheses for Part (a)

For part (a), we want to test the null hypothesis \( H_0: \mu_1 = \mu_2 \) against the alternative hypothesis \( H_1: \mu_1 eq \mu_2 \). This is a two-tailed test where we want to see if the means of the melting points of the two alloys are equal.

02

Determine the Test Statistic for Part (a)

Since the problem assumes equal variances, we will use a pooled variance t-test. The test statistic \( t \) is given by:\[t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{s_p^2 \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}}\]where \[ s_p^2 = \frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2} \]Substitute \( \bar{x}_1 = 420, \bar{x}_2 = 426, s_1 = 4, s_2 = 3, n_1 = 21, n_2 = 21 \) to calculate \( t \).

03

Calculate the Pooled Variance and t-statistic for Part (a)

First, calculate the pooled variance:\[s_p^2 = \frac{(21-1)(4^2) + (21-1)(3^2)}{21+21-2} = \frac{20 \times 16 + 20 \times 9}{40} = \frac{320 + 180}{40} = 12.5\]Now calculate the t-statistic:\[t = \frac{420 - 426}{\sqrt{12.5 \left(\frac{1}{21} + \frac{1}{21}\right)}} = \frac{-6}{\sqrt{12.5 \times \frac{2}{21}}} = \frac{-6}{\sqrt{1.1905}} = \frac{-6}{1.0911} \approx -5.50\]

04

Find the P-value for Part (a)

The degrees of freedom \( df = 21 + 21 - 2 = 40 \). Using a t-table or calculator, find the P-value for \( t = -5.50 \) with \( 40 \) df. The P-value will be less than the critical value for a two-tailed test at \( \alpha = 0.05\) (which approximately corresponds to \( P < 0.001 \)).

05

Conclusion for Part (a)

Since the P-value is less than \( \alpha = 0.05 \), we reject the null hypothesis. The data supports the claim that there is a difference in the melting points of the two alloys.

06

Determine Sample Size for Part (b)

For part (b), use the formula for sample size required in a two-sample t-test to detect a specified difference \( \Delta = 3 \) with given \( \alpha \) and power \( 1-\beta \). The formula is:\[n = \left(\frac{2(z_{\alpha/2} + z_{\beta})^2 \sigma^2}{\Delta^2} \right)\]where \( z_{\alpha/2} \) is the critical value from the Z-distribution for \( \alpha = 0.05 \), and \( z_{\beta} \) is for \( 1-\beta = 0.9 \). Assume \( \sigma = 4 \).

07

Calculate the Required Sample Size for Part (b)

Given \( \alpha = 0.05 \) and \( 1-\beta = 0.9 \), find \( z_{0.025} \approx 1.96 \) and \( z_{0.1} \approx 1.28 \). Substitute into the sample size formula:\[n = \left( \frac{2(1.96 + 1.28)^2 \times 4^2}{3^2} \right) = \frac{2 \times 10.4976 \times 16}{9} = \frac{335.9232}{9} \approx 37.33\]Thus, the required sample size is \( n \approx 38 \) per group.

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Key Concepts

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

t-test

A t-test is a statistical method used to determine if there is a significant difference between the means of two groups. It is particularly useful when dealing with small sample sizes or unknown population variances. In this context, we use a t-test to assess whether the melting points of two alloys are different. We compare the sample means of the two groups to test this hypothesis.

The t-test helps us decide whether any observed difference is due to random chance or if it indicates a true difference in population means. A common variant is the two-sample t-test, which we apply when comparing two independent groups, as is the case in our alloys example.

sample size calculation

Calculating the sample size is crucial in hypothesis testing to ensure that the study has enough power to detect a true effect. Power refers to the probability of correctly rejecting the null hypothesis when it is false. A higher sample size increases the power of the test, reducing the risk of type II errors (failing to detect a true effect).

In our exercise, we calculated the sample size needed to detect a specified mean difference in melting points, with a known standard deviation and desired power level. The formula used factors in the critical values from the Z-distribution, which correspond to our predetermined significance level (alpha) and power (1-beta).

pooled variance

Pooled variance is an estimate of the common variance shared by two populations. It's useful in t-tests when assuming equal population variances, allowing us to combine data from both samples to gain a better estimate of this variance.

The calculation involves summing the weighted variances of each group, with weights corresponding to each group's degrees of freedom. For example, in our alloy test, pooled variance equals the weighted average of the variances from the two samples, providing a more stable estimate for the t-test.

two-sample test

The two-sample test is a statistical technique used to compare the means of two independent groups, allowing us to determine if they differ significantly. It's often implemented as a two-sample t-test when dealing with small sample sizes and unknown variances.

In our scenario with alloys, we conduct a two-sample t-test to assess whether the melting points from the two samples significantly differ. We start by setting up null and alternative hypotheses, calculate the test statistic using the sample data, and finally, derive a P-value to draw conclusions about the hypotheses.