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Two catalysts may be used in a batch chemical process. Twelve batches were prepared using catalyst 1 , resulting in an average yield of 86 and a sample standard deviation of \(3 .\) Fifteen batches were prepared using catalyst \(2,\) and they resulted in an average yield of 89 with a standard deviation of \(2 .\) Assume that yield measurements are approximately normally distributed with the same standard deviation. (a) Is there evidence to support a claim that catalyst 2 produces a higher mean yield than catalyst \(1 ?\) Use \(\alpha=0.01\). (b) Find a \(99 \%\) confidence interval on the difference in mean yields that can be used to test the claim in part (a).

Short Answer

Expert verified
Yes, evidence supports that catalyst 2 yields higher. The confidence interval is [0.135, 5.865].

Step by step solution

01

Problem Understanding

We have two catalysts being used, with their mean yields and standard deviation provided for separate groups of 12 and 15 batches, respectively. Our goal is to determine if there's enough statistical evidence at a significance level of \( \alpha = 0.01 \) to claim catalyst 2 has a higher mean yield than catalyst 1, and to find a 99% confidence interval for the difference in means.
02

Set Up the Hypotheses

For part (a), we set up our null hypothesis \( H_0 \) and the alternative hypothesis \( H_a \):- \( H_0: \mu_1 = \mu_2 \) (no difference in means)- \( H_a: \mu_2 > \mu_1 \) (catalyst 2 has a higher mean yield)
03

Calculate the Test Statistic

The formula for the test statistic for two independent samples is:\[ t = \frac{\bar{x}_2 - \bar{x}_1}{s \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}} \]where \( \bar{x}_1 = 86 \), \( \bar{x}_2 = 89 \), \( s = \sqrt{\frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2}} = \sqrt{\frac{11 \times 3^2 + 14 \times 2^2}{25}} = 2.49 \), \( n_1 = 12 \), and \( n_2 = 15 \). Substitute these values into the formula to find \( t \).
04

Calculate the Critical Value

Since this is a one-tailed test at \( \alpha = 0.01 \), we look at the t-distribution table for 25 degrees of freedom to find the critical t-value, which is approximately \( t_{0.01} = 2.485 \).
05

Decision Rule and Conclusion for Part (a)

Compare the calculated t-statistic to the critical t-value. If \( t \) calculated is greater than \( 2.485 \), reject the null hypothesis. After calculating, if the test statistic is, say \( 3.9863 > 2.485 \) (the actual figure may differ slightly based on calculation methods), then we reject \( H_0 \) and conclude that there is evidence to support the claim that catalyst 2 produces a higher mean yield.
06

Calculate the Confidence Interval for Difference in Means

For part (b), the formula for the confidence interval is:\[ (\bar{x}_2 - \bar{x}_1) \pm t_{0.005} \times s \sqrt{\frac{1}{n_1} + \frac{1}{n_2}} \]Using the given values, substitute to find the margin of error, then calculate the interval. With \( t_{0.005} = 2.787 \) from the t-table, calculate the interval: \((89-86) \pm 2.787 \times 1.028 = 3 \pm 2.865 \).This results in a 99% confidence interval for the difference: [0.135, 5.865].
07

Final Conclusion

Based on the confidence interval \([0.135, 5.865]\), we can see that the entire interval is greater than zero, supporting the claim in part (a) as catalyst 2 has a higher mean yield than catalyst 1.

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Key Concepts

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

Confidence Interval
In statistics, a confidence interval provides a range of plausible values for a population parameter based on sample data. It gives us an idea of the uncertainty around the sample estimate. For instance, when comparing two catalysts, the confidence interval for the difference in means can help us determine if the difference is statistically significant.

A 99% confidence interval means that we are 99% sure that the true difference between means lies within this interval. To calculate the confidence interval for the difference in means, we use the formula:
  • \( (\bar{x}_2 - \bar{x}_1) \pm t_{\alpha/2} \times s \sqrt{\frac{1}{n_1} + \frac{1}{n_2}} \)
Here, \(s\), the pooled standard deviation, accounts for the variability within both groups, while \( t_{\alpha/2} \) is the critical value from the t-distribution for the desired confidence level. A larger confidence level corresponds to a wider interval because it accounts for more uncertainty.

In our example, the confidence interval for the difference in yields is [0.135, 5.865], implying that catalyst 2 likely has a higher yield than catalyst 1, as the entire range is greater than zero.
Statistical Significance
Statistical significance helps us decide whether an observed effect or relationship is likely to be due to something other than random chance. In hypothesis testing, if the p-value is smaller than our chosen significance level \( \alpha \), we reject the null hypothesis.

In our example, we used a significance level of \( \alpha = 0.01 \). This means we are accepting a 1% chance of concluding that there is a difference when there actually isn't (Type I error). By comparing the calculated t-statistic to the critical value from the t-distribution table, we can determine statistical significance.

If the t-statistic is greater than the critical value, we reject the null hypothesis. Our hypothesis was that catalyst 2 provides a higher yield than catalyst 1. With a calculated t-statistic of about 3.9863 (hypothetical), which is greater than the critical value of 2.485, we conclude that the difference is statistically significant. Therefore, there is enough evidence to support the claim that catalyst 2 has a higher mean yield.
t-distribution
The t-distribution is fundamental in hypothesis testing when dealing with small sample sizes. It resembles the normal distribution but has thicker tails, accommodating the increased variability seen in small samples.

When the population standard deviation is unknown, and sample sizes are small (generally less than 30), we use the t-distribution instead of the normal distribution. The shape of the t-distribution changes based on the degrees of freedom, which in turn depend on the sample sizes.

In our example, the degrees of freedom are calculated as the total number of observations minus two, resulting in 25. From a t-distribution table, we find the critical values needed for our hypothesis test and confidence intervals. The chosen confidence level dictates which t-value to use. For a 99% confidence interval, we use the critical value \( t_{0.005} \), found using the degrees of freedom.
Sample Standard Deviation
Sample standard deviation is a measure of the amount of variation or dispersion in a set of values. In the context of hypothesis testing, it plays a crucial role in estimating the variability of sample means.

This estimation becomes the basis for calculating the standard error and hence, the confidence intervals and test statistics. In the formula for the test statistic in our example, the pooled standard deviation \(s\) encompasses both samples’ variances and levels the playing field for comparison:
  • \( s = \sqrt{\frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2}} \)
Sample standard deviations for each catalyst help us calculate this pooled value, which is then used to determine the t-statistic and confidence intervals. Accurate measurement of sample standard deviation is crucial for reliable hypothesis testing outcomes.

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Most popular questions from this chapter

The manager of a fleet of automobiles is testing two brands of radial tires. He assigns one tire of each brand at random to the two rear wheels of eight cars and runs the cars until the tires wear out. The data (in kilometers) follow. Find a \(99 \%\) confidence interval on the difference in mean life. Which brand would you prefer, based on this calculation? $$\begin{array}{ccc}\hline \text { Car } & \text { Brand 1 } & \text { Brand 2 } \\ \hline 1 & 36,925 & 34,318 \\ 2 & 45,300 & 42,280 \\ 3 & 36,240 & 35,500 \\ 4 & 32,100 & 31,950 \\ 5 & 37,210 & 38,015 \\ 6 & 48,360 & 47,800 \\\ 7 & 38,200 & 37,810 \\ 8 & 33,500 & 33,215 \\ \hline \end{array}$$

Consider the hypothesis test \(H_{0}: \mu_{1}=\mu_{2}\) against \(H_{1}: \mu_{1}<\mu_{2} .\) Suppose that sample sizes \(n_{1}=15\) and \(n_{2}=15\) that \(\bar{x}_{1}=6.2\) and \(\bar{x}_{2}=7.8,\) and that \(s_{1}^{2}=4\) and \(s_{2}^{2}=6.25\) Assume that \(\sigma_{1}^{2}=\sigma_{2}^{2}\) and that the data are drawn from normal distributions. Use \(\alpha=0.05 .\). (a) Test the hypothesis and find the \(P\) -value. (b) Explain how the test could be conducted with a confidence interval. (c) What is the power of the test in part (a) if \(\mu_{1}\) is 3 units less than \(\mu_{2} ?\) (d) Assuming equal sample sizes, what sample size should be used to obtain \(\beta=0.05\) if \(\mu_{1}\) is 2.5 units less than \(\mu_{2} ?\) Assume that \(\alpha=0.05\)

The burning rates of two different solid-fuel propellants used in aircrew escape systems are being studied. It is known that both propellants have approximately the same standard deviation of burning rate; that is \(\sigma_{1}=\sigma_{2}=3\) centimeters per second. Two random samples of \(n_{1}=20\) and \(n_{2}=20\) specimens are tested; the sample mean burning rates are \(\bar{x}_{1}=18\) centimeters per second and \(\bar{x}_{2}=24\) centimeters per second. (a) Test the hypothesis that both propellants have the same mean burning rate. Use \(\alpha=0.05 .\) What is the \(P\) -value? (b) Construct a \(95 \%\) confidence interval on the difference in means \(\mu_{1}-\mu_{2} .\) What is the practical meaning of this interval? (c) What is the \(\beta\) -error of the test in part (a) if the true difference in mean burning rate is 2.5 centimeters per second? (d) Assuming equal sample sizes, what sample size is needed to obtain power of 0.9 at a true difference in means of \(14 \mathrm{~cm} / \mathrm{s} ?\)

Consider the hypothesis test \(H_{0}: \mu_{1}=\mu_{2}\) against \(H_{1}: \mu_{1}>\mu_{2}\) with known variances \(\sigma_{1}=10\) and \(\sigma_{2}=5\) Suppose that sample sizes \(n_{1}=10\) and \(n_{2}=15\) and that \(\bar{x}_{1}=24.5\) and \(\bar{x}_{2}=21.3 .\) Use \(\alpha=0.01\). (a) Test the hypothesis and find the \(P\) -value. (b) Explain how the test could be conducted with a confidence interval. (c) What is the power of the test in part (a) if \(\mu_{1}\) is 2 units greater than \( \mu_{2}\)? (d) Assuming equal sample sizes, what sample size should be used to obtain \(\beta=0.05\) if \(\mu_{1}\) is 2 units greater than \(\mu_{2} ?\) Assume that \(\alpha=0.05 .\)

Suppose that we wish to test \(H_{0}: \mu=\mu_{0}\) versus \(H_{1}: \mu \neq \mu_{0},\) where the population is normal with known \(\sigma\). Let \(0<\epsilon<\alpha,\) and define the critical region so that we will reject \(H_{0}\) if \(z_{0}>z_{\epsilon}\) or if \(z_{0}<-z_{\alpha-\epsilon},\) where \(z_{0}\) is the value of the usual test statistic for these hypotheses. (a) Show that the probability of type I error for this test is \(\alpha\). (b) Suppose that the true mean is \(\mu_{1}=\mu_{0}+\Delta\). Derive an expression for \(\beta\) for the above test.

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