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Solid copper produced by sintering (heating without melting) a powder under specified environmental conditions is then measured for porosity (the volume fraction due to voids) in laboratory. A sample of \(n_{1}=4\) independent porosity measurements have mean \(\bar{y}_{1}=.22\) and variance \(s_{1}^{2}=.0010 .\) A second laboratory repeats the same process on solid copper formed from an identical powder and gets \(n_{2}=5\) independent porosity measurements with \(\bar{y}_{2}=.17\) and \(s_{2}^{2}=.0020\) Estimate the true difference between the population means \(\left(\mu_{1}-\mu_{2}\right)\) for these two laboratories, with confidence coefficient. 95

Short Answer

Expert verified
The 95% confidence interval for the difference is (-0.0311, 0.1311), not implying a significant difference.

Step by step solution

01

Calculate the Difference in Sample Means

First, subtract the mean porosity measurement from the second laboratory (\(\bar{y}_{2} = 0.17\)) from the mean porosity measurement from the first laboratory (\(\bar{y}_{1} = 0.22\)) to estimate the difference in sample means. \(\bar{y}_{1} - \bar{y}_{2} = 0.22 - 0.17 = 0.05\)
02

Calculate the Standard Error of the Difference

The standard error of the difference between two means is given by \[ SE = \sqrt{\frac{s_{1}^{2}}{n_{1}} + \frac{s_{2}^{2}}{n_{2}}} \] Substitute the given variances and sample sizes: \[ SE = \sqrt{\frac{0.0010}{4} + \frac{0.0020}{5}} = \sqrt{0.00025 + 0.0004} = \sqrt{0.00065} \approx 0.0255 \]
03

Determine the Appropriate t-Critical Value

To find the t-critical value for a 95% confidence interval with degrees of freedom approximated by the smaller sample size minus one (since they are small and may not qualify for additional simplification), use a t-distribution table or calculator:Degrees of freedom\( = \min(n_1 - 1, n_2 - 1) = 4 - 1 = 3 \).For a 95% confidence level and 3 degrees of freedom, the t-value approximately equals 3.182.
04

Calculate the Confidence Interval for the Difference

The formula for the confidence interval is given by: \[ (\bar{y}_{1} - \bar{y}_{2}) \pm t \times SE \]Substitute the values obtained: \[ 0.05 \pm 3.182 \times 0.0255 \]Calculate:\[ 0.05 \pm 0.0811 \]This results in the interval:\[-0.0311, 0.1311 \]
05

Interpret the Confidence Interval

The confidence interval for the difference in population means is \([-0.0311, 0.1311]\). This suggests that, with 95% confidence, the true difference in porosity means (\(\mu_{1} - \mu_{2}\)) could be as low as \(-0.0311\) or as high as \(0.1311\). We cannot conclude definitively that one laboratory's process results in higher porosity than the other, as the interval includes zero.

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Key Concepts

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

Sample Means
In statistics, sample means play a crucial role, especially when conducting experiments or making estimates about a population. A sample mean is simply the average of a set of observations. It summarizes the characteristics of a sample, helping researchers make inferences about the population as a whole.
Consider a scenario where different labs are testing copper samples for porosity. For each lab, they take several independent porosity measurements and calculate the mean of these measurements. This mean, known as the sample mean, provides an estimate of the true population mean for the porosity of the copper being tested.
When the problem mentions
  • first laboratory's mean as \(ar{y}_{1} = 0.22\)
  • second laboratory's mean as \(ar{y}_{2} = 0.17\)
it's these sample means that allow statisticians to estimate the difference in population means, aiming to determine if one laboratory process results in different porosity levels compared to the other.
T-Distribution
The t-distribution is a critical concept in statistical analysis, particularly useful when working with small sample sizes. This distribution is used when the sample size is small and the population standard deviation is unknown.
A t-distribution looks similar to a normal distribution but has heavier tails. This means it tends to produce values that fall further from the mean, which provides a more accurate reflection of variability when sample sizes are small.
In the context of estimating the confidence interval for the difference in porosity means between two laboratories, one uses the t-distribution to find the critical t-value.
This t-value helps scale the standard error to create the confidence interval, accounting for the possibility of extreme values due to the small sample size.
Standard Error
Standard error is an essential part of estimating how sample means are expected to differ from the actual population mean. It provides a measure of the variability of a sample mean.
  • A smaller standard error indicates that the sample mean is a more accurate reflection of the population mean.
  • A larger standard error suggests more variability and less certainty in the estimate.
In the problem, when calculating the confidence interval for the difference in porosities, we compute the standard error of the difference between the two sample means.
The formula used here combines variances from both samples and their respective sizes:
\[ SE = \sqrt{\frac{s_{1}^{2}}{n_{1}} + \frac{s_{2}^{2}}{n_{2}}} \]
By substituting the given values, you determine how much uncertainty to expect in the mean difference estimate. This value is then used to construct the confidence interval around the estimated mean difference.
Degrees of Freedom
Degrees of freedom are used in statistical calculations, such as estimating population parameters, in order to ensure that calculations are as unbiased and accurate as possible. When determining the t-critical value in a small sample, the degrees of freedom play a significant role.
In the context of this problem, degrees of freedom are determined by the smaller of the sample sizes minus one. Since we're working with sample sizes of 4 and 5, the degrees of freedom here are calculated as:
  • For the smaller sample: \(n_1 - 1 = 4 - 1 = 3\).
These degrees of freedom help in identifying the appropriate t-critical value from the t-distribution table.
This value ensures that when the confidence interval is constructed, it properly accounts for the small sample size, thus making our interval estimate more reliable.

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Most popular questions from this chapter

Sometimes surveys provide interesting information about issues that did not seem to be the focus of survey initially. Results from two CNN/USA Today/Gallup polls, one conducted in March 2003 and one in November \(2003,\) were recently presented online. \(^{\star}\) Both polls involved samples of 1001 adults, aged 18 years and older. In the March sample, \(45 \%\) of those sampled claimed to be fans of professional baseball whereas \(51 \%\) of those polled in November claimed to be fans. a. Give a point estimate for the difference in the proportions of Americans who claim to be baseball fans in March (at the beginning of the season) and November (after the World Series). Provide a bound for the error of estimation. b. Is there sufficient evidence to conclude that fan support is greater at the end of the season? Explain.

Suppose that two independent random samples of \(n_{1}\) and \(n_{2}\) observations are selected from normal populations. Further, assume that the populations possess a common variance \(\sigma^{2}\). Let $$S_{i}^{2}=\frac{\sum_{j=1}^{n_{i}}\left(Y_{i j}-\bar{Y}_{i}\right)^{2}}{n_{i}-1}, \quad i=1,2$$ a. Show that \(S_{p}^{2}\), the pooled estimator of \(\sigma^{2}\) (which follows), is unbiased: $$S_{p}^{2}=\frac{\left(n_{1}-1\right) S_{1}^{2}+\left(n_{2}-1\right) S_{2}^{2}}{n_{1}+n_{2}-2}$$ b. Find \(V\left(S_{p}^{2}\right)\)

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A survey of 415 corporate, government, and accounting executives of the Financial Accounting Foundation found that 278 rated cash flow (as opposed to earnings per share, etc.) as the most important indicator of a company's financial health. Assume that these 415 executives constitute a random sample from the population of all executives. Use the data to find a \(95 \%\) confidence interval for the fraction of all corporate executives who consider cash flow the most important measure of a company's financial health.

The Mars twin rovers, Spirit and Opportunity, which roamed the surface of Mars in the winter of 2004, found evidence that there was once water on Mars, raising the possibility that there was once life on the plant. Do you think that the United States should pursue a program to send humans to Mars? An opinion poll \(^{\star}\) indicated that \(49 \%\) of the 1093 adults surveyed think that we should pursue such a program. a. Estimate the proportion of all Americans who think that the United States should pursue a program to send humans to Mars. Find a bound on the error of estimation. b. The poll actually asked several questions. If we wanted to report an error of estimation that would be valid for all of the questions on the poll, what value should we use? [Hint: What is the maximum possible value for \(p \times q ?]\)

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