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Chapter 5: Problem 22

The U.S. Department of Agriculture provided homogenized baby food samples to three labs for analysis. \({ }^{3}\) Results agreed well for protein, fat, zinc, riboflavin, and palmitic acid. Results for iron were questionable: Lab A, \(1.59 \pm 0.14\) (13); Lab B, \(1.65 \pm 0.56\) (8); Lab C, \(2.68 \pm 0.78\) (3) \(\mathrm{mg} / 100 \mathrm{~g}\). Uncertainty is the standard deviation, with the number of replicate analyses in parentheses. Use two separate \(t\) tests to compare results from Lab C with those from Lab A and Lab B at the \(95 \%\) confidence level. Comment on the sensibility of the \(t\) test results. Offer your own conclusions.

Short Answer

Expert verified
The t-tests show significant differences between Lab C and the other labs, indicating inconsistency in iron measurements.

Step by step solution

01

Identify the Given Information

Lab A reports a mean iron content of 1.59 with a standard deviation of 0.14 and 13 samples. Lab B reports 1.65 with a standard deviation of 0.56 and 8 samples. Lab C reports 2.68 with a standard deviation of 0.78 and 3 samples.
02

State the Null and Alternative Hypotheses

For Lab A vs Lab C: The null hypothesis (\(H_0\)) is that there is no significant difference between the iron content measured by Lab A and Lab C. The alternative hypothesis (\(H_1\)) is that there is a significant difference. Similarly, set up hypotheses for Lab B vs Lab C.
03

Calculate the T-test Statistic for Lab A vs Lab C

Use the formula for the t-statistic: \[ t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \] where \(\bar{x}_1 = 1.59\), \(\bar{x}_2 = 2.68\), \(s_1 = 0.14\), \(s_2 = 0.78\), \(n_1 = 13\), and \(n_2 = 3\). Substitute these values to find the t-statistic.
04

Calculate the Degrees of Freedom for A vs C

Using the formula for degrees of freedom with unequal variance: \[ df = \frac{\left( \frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} \right)^2}{\frac{\left( \frac{s_1^2}{n_1} \right)^2}{n_1 - 1} + \frac{\left( \frac{s_2^2}{n_2} \right)^2}{n_2 - 1}} \] Calculate \(df\) for Labs A and C.
05

Determine Critical T-value

Using the degrees of freedom calculated and a standard t-table, find the critical t-value for a 95% confidence level.
06

Compare T-statistic to Critical Value for A vs C

If the calculated t-statistic is greater than the critical t-value, reject the null hypothesis. Otherwise, fail to reject the null hypothesis. Determine if there is a statistically significant difference between Labs A and C.
07

Repeat Steps 3-6 for Lab B vs Lab C

Calculate the t-statistic, degrees of freedom, and critical t-value for Lab B vs Lab C. Compare the t-statistic with the critical value to determine significance.
08

Interpret the Results

If either comparison rejects the null hypothesis, there is a significant difference between lab results. Analyze and conclude based on the findings from both comparisons.

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Key Concepts

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

Statistical Hypothesis Testing
Statistical hypothesis testing is a method used to make inferences or draw conclusions about a population based on sample data. It involves setting up two contrasting hypotheses: the null hypothesis (\(H_0\)) and the alternative hypothesis (\(H_1\)).
  • The null hypothesis generally states that there is no effect or no difference, meaning any observed effect in the data is due to sampling error or chance.
  • The alternative hypothesis posits that there is indeed a significant effect or difference.

In this exercise, for both Lab A vs Lab C and Lab B vs Lab C, the null hypothesis will say that the iron content measurements are the same across the labs, whereas the alternative suggests they are different.
Confidence Interval
A confidence interval provides a range of values that is likely to contain the population parameter of interest. For a 95% confidence interval, we are saying we are 95% sure that the true mean lies within this range.
In practical terms, if we were to take many samples and build an interval estimate for each sample, about 95% of those intervals would capture the true parameter value. Confidence intervals help us understand the uncertainty around a sample estimate, such as the mean iron level measured by each laboratory.
Using a confidence interval alongside hypothesis testing can give further insight into not only whether there is a significant difference but how large and meaningful that difference is likely to be.
Degrees of Freedom
The concept of "degrees of freedom" in statistics refers to the number of values in a calculation that are free to vary while still conforming to a given constraint.
In the context of a two-sample t-test, the degrees of freedom are used when determining the critical value from the t-distribution, which we then compare our t-statistic to.
Calculating degrees of freedom when comparing two samples with unequal variances involves using this formula:
\[df = \frac{\left( \frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} \right)^2}{\frac{\left( \frac{s_1^2}{n_1} \right)^2}{n_1 - 1} + \frac{\left( \frac{s_2^2}{n_2} \right)^2}{n_2 - 1}}\]
This formula accounts for the variability in both samples and is crucial for finding the appropriate critical value in hypothesis testing. It ensures the test's validity, especially when sample sizes vary as in this lab comparison.
Laboratory Comparison
Comparing laboratory results is important when assessing consistency and reliability in analysis. In this exercise, labs have been compared based on their measurement of iron content in baby food.
The t-test helps determine if observed differences between laboratory results are due to random variation or are statistically significant.
  • Lab A reported \(1.59 \pm 0.14\) mg, Lab B reported \(1.65 \pm 0.56\) mg, and Lab C reported \(2.68 \pm 0.78\) mg of iron per 100 g.

One must proceed with caution when handling laboratory comparisons, as it's crucial to consider not only the means but also the variances and sample sizes. Variability in reported results can arise from differences in instruments, techniques, or even the samples themselves.
By using statistical analyses such as the t-test, analysts can assess if differences are meaningful or simply a result of random noise.

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Most popular questions from this chapter

Correcting for matrix effects with an internal standard. The appearance of pharmaceuticals in municipal wastewater (sewage) is an increasing problem that is likely to have adverse effects on our drinking water supply. Sewage is a complex matrix. When the drug carbamazepine was spiked into sewage at a concentration of \(5 \mathrm{ppb}\), chromatographic analysis gave an apparent spike recovery of \(154 \%{5}^{15}\) When deuterated carbamazepine was used as an internal standard for the analysis, the apparent recovery was \(98 \%\). Explain how the internal standard is used in this analysis and rationalize why it works so well to correct for matrix effects. Experimental Design

Detection limit. Low concentrations of \(\mathrm{Ni}^{2+}\)-EDTA near the detection limit gave the following counts in a mass spectral measurement: \(175,104,164,193,131,189,155,133,151,176\). Ten measurements of a blank had a mean of 45 counts. A sample containing \(1.00 \mu \mathrm{M} \mathrm{Ni}^{2+}\)-EDTA gave 1797 counts. Estimate the detection limit for Ni- EDTA.

Verifying constant response for an internal sfandard. When we develop a method using an internal standard, it is important to verify that the response factor is constant over the calitration range. Data are shown below for a chromstographic analysis of naphthalene \(\left(\mathrm{C}_{10} \mathrm{H}_{2}\right)\), using deuterated naphthalene \(\left(\mathrm{C}_{10} \mathrm{D}_{\mathrm{s}}\right.\) in which \(\mathrm{D}\) is the isotope \({ }^{2} \mathrm{H}\) ) as an internal standard. The two compounds emerge from the column at almost identical times and are measured by a mass spectrometer, which distinguishes them by molecular mass. From the definition of response factor in Equation \(5-11\), we can write $$ \frac{\text { Area of analyte signal }}{\text { Area of standard signal }}-F\left(\frac{\text { concentration of analyte }}{\text { concentration of standard }}\right) $$ Prepare a graph of peak area ratio \(\left(\mathrm{C}_{10} \mathrm{H}_{\mathrm{s}} / \mathrm{C}_{10} \mathrm{D}_{\mathrm{k}}\right)\) versus concentration ratio \(\left(\left[\mathrm{C}_{10} \mathrm{H}_{\mathrm{k}}\right]\left[\mathrm{C}_{10} \mathrm{D}_{\mathrm{k}}\right]\right)\) and find the slope, which is the response factor. Evaluate \(F\) for each of the three samples and find the standard deviation of \(F\) to see how "constant" it is. \begin{tabular}{ccccc} Sample & \(C_{10} \mathrm{H}_{\mathrm{s}}\) \((\mathrm{ppm})\) & \(C_{10} D_{\mathrm{s}}\) \((\mathrm{ppm})\) & \(C_{10} H_{\mathrm{k}}\) peak area & \(C_{10} D_{\mathrm{s}}\) peak area \\ \hline 1 & \(1.0\) & \(10.0\) & 303 & 2992 \\ 2 & \(5.0\) & \(10.0\) & 3519 & 6141 \\ 3 & \(10.0\) & \(10.0\) & 3023 & 2819 \\ \hline \end{tabular}

What is the difference between a false positive and a false negative?

imes Siandard addition graph. Students performed an experiment like that in Figure \(5.7\) in which each flask contained \(25.00 \mathrm{~mL}\) of serum, varying additions of \(2.640 \mathrm{M} \mathrm{NaCl}\) standard, and a total volume of \(50.00 \mathrm{~mL}\). \begin{tabular}{ccc} Flask & Volume of standard (mL) & \(\mathrm{Na}^{*}\) atomic emission sagnal (mV) \\ \hline 1 & 0 & \(3.13\) \\ 2 & \(1.000\) & \(5.40\) \\ 3 & \(2.000\) & \(7.89\) \\ 4 & \(3.000\) & \(10.30\) \\ 5 & \(4.000\) & \(12.48\) \\ \hline \end{tabular} (a) Prepare a standard addition graph and find \(\left[\mathrm{Na}^{+}\right]\)in the serum. (b) Find the standard deviation and \(95 \%\) confidence interval for \(\left[\mathrm{Na}^{+}\right]\)
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