/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 11 A geologist collected 20 differe... [FREE SOLUTION] | 91Ó°ÊÓ

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A geologist collected 20 different ore samples, all of the same weight, and randomly divided them into two groups. The titanium contents of the samples, found using two different methods, are listed in the table.\begin{tabular}{ccccc|ccccc} \multicolumn{5}{c|} { Method 1 } & \multicolumn{5}{c} { Method 2 } \\ \hline 011 & .013 & 013 & .015 & .014 & .011 & 016 & .013 & .012 & .015 \\ .013 & .010 & .013 & .011 & .012 & .012 & .017 & .013 & .014 & .015 \end{tabular} a. Use an appropriate method to test for a significant difference in the average titanium contents using the two different methods. b. Determine a \(95 \%\) confidence interval estimate for \(\left(\mu_{1}-\mu_{2}\right) .\) Does your interval estimate substantiate your conclusion in part a? Explain.

Short Answer

Expert verified
Answer: ______ [Please refer to Steps 3 and 5 of the provided solution for the conclusion and confidence interval interpretation, respectively. You will need to perform calculations using the given data to fill in the answer.]

Step by step solution

01

Calculate the summary statistics

Calculate the means, sample standard deviations, and sample sizes for both methods: Method 1: \(n_1 = 10\) samples, \(\bar{x}_1\) = mean, \(s_1\) = standard deviation Method 2: \(n_2 = 10\) samples, \(\bar{x}_2\) = mean, \(s_2\) = standard deviation Using the given data in the table, compute the means and standard deviations for Method 1 and Method 2.
02

Perform a two-sample t-test

To test for a significant difference in the average titanium contents, we will perform a two-sample t-test. The null hypothesis is that there is no difference in means, i.e. \(\mu_1 = \mu_2\). The alternative hypothesis is that there is a difference in means, i.e. \(\mu_1 \neq \mu_2\). Calculate the test statistic \(t\), and the degrees of freedom \(df\): \(t = \frac{(\bar{x}_1-\bar{x}_2)-0}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}\) Calculate the critical value using a 95% confidence level (\(\alpha=0.05\)) and degrees of freedom \(df = min(n_1 - 1, n_2 - 1)\), and find the corresponding p-value.
03

Make a conclusion

Compare the test statistic \(t\) and p-value to the critical value. If the absolute value of \(t\) is greater than the critical value or the p-value is less than \(\alpha\), reject the null hypothesis and conclude that there is a significant difference in the average titanium contents using the two methods. Otherwise, fail to reject the null hypothesis and conclude that there is no significant difference.
04

Determine the 95% confidence interval estimate for \(\left(\mu_{1}-\mu_{2}\right)\)

The formula for the 95% confidence interval for \(\left(\mu_{1}-\mu_{2}\right)\) is: \(\left(\bar{x}_1 - \bar{x}_2\right) \pm t_{\alpha/2} \cdot \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}\) Use the calculated means, standard deviations, and sample sizes to find the 95% confidence interval.
05

Interpret the confidence interval

If the confidence interval contains 0, it suggests that there is not enough evidence to say that there is a significant difference in the average titanium contents between the two methods. If the confidence interval does not contain 0, it supports our conclusion from the t-test that there is a significant difference. Compare the confidence interval to the conclusion made in Step 3, and check whether it substantiates the conclusion by analyzing if the confidence interval contains 0 or not.

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Key Concepts

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

Confidence Interval
Estimating a range where the true difference in means might lie can be done with a confidence interval. Essentially, it gives us a range of values, based on our sample data, in which we are fairly certain the true difference between the population means (\( \mu_{1} - \mu_{2} \)) lies. When constructing a confidence interval, we use point estimates from our data—such as the sample means and standard deviations—along with a multiplier that accounts for our desired confidence level (often 95%).
  • The sample difference (\( \bar{x}_1 - \bar{x}_2 \)) is our best guess for the true difference.
  • We add and subtract a margin of error, determined by a critical t-value and the standard error, to find this interval.
Mathematically, for the 95% confidence interval for the difference of means, the formula is:\[(\bar{x}_1 - \bar{x}_2) \pm t_{\alpha/2} \cdot \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}\]Where:
  • \( t_{\alpha/2} \) is the t-value for a 95% confidence level.
  • \( s_1 \, \text{and} \, s_2 \) are the standard deviations from each group.
A confidence interval that includes 0 suggests there might not be a significant difference between group means.
Hypothesis Testing
Hypothesis testing is like a detective's tool to determine whether observed data reveals a significant effect or is just due to chance. In this case, we want to know if the two methods that measure titanium content yield results that are genuinely different. To start, we form two hypotheses:
  • The null hypothesis (\( H_0 \)): Assumes no effect, stating \( \mu_1 = \mu_2 \), meaning the two methods have the same result on average.
  • The alternative hypothesis (\( H_a \)): Suggests a true difference exists, \( \mu_1 eq \mu_2 \).
To test these hypotheses, we utilize the two-sample t-test. This test compares the means of two independent groups to see if the differences observed could realistically occur under the null hypothesis.
  • The test statistic \( t \), calculated from the sample data, tells us how much the sample means deviate from the null hypothesis.
  • A p-value is derived from this test which helps make the decision: if it is less than our significance level (usually 0.05), we reject the null hypothesis, suggesting a significant difference.
In hypothesis testing, it's important to decide in advance which level of significance we'll use. This decision helps in understanding whether observed differences are likely due to true effects or random variations.
Summary Statistics
Summary statistics are the building blocks of data analysis and give us useful snapshots of the data in our samples. When dealing with two-sample t-tests, these statistics are crucial for calculating the test statistic and confidence intervals. The summary statistics include:
  • Mean ( \( \bar{x} \) ): It is the average value of the data set and provides a central point of reference.
  • Standard deviation ( \( s \) ): Measures the variability within the sample, indicating how spread out the observations are from the mean.
  • Sample size ( \( n \) ): The number of observations in each sample, impacting the reliability of our estimates.

In this two-method experiment, we first calculate these statistics for each group. We find the average titanium content for each method (method 1 and method 2) and determine their respective variability through the standard deviation. These statistics are then plugged into our formulas to explore hypotheses about the two methods. By calculating and understanding these values, we prepare ourselves for the further statistical analysis needed in hypothesis testing and confidence interval estimation. An accurate assessment of these values is vital for the two-sample t-test to be valid and reliable.

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Most popular questions from this chapter

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