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Chapter 9: Problem 13

The article "Quantitative MRI and Electrophysiology of Preoperative Carpal Tunnel Syndrome in a Female Population" (Ergonomies, 1997: 642-649) reported that \((-473.13,1691.9)\) was a large-sample \(95 \%\) confidence interval for the difference between true average thenar muscle volume \(\left(\mathrm{mm}^{3}\right)\) for sufferers of carpal tunnel syndrome and true average volume for nonsufferers. Calculate and interpret a \(90 \%\) confidence interval for this difference.

Short Answer

Expert verified
90% CI: (-299.325, 1518.495)

Step by step solution

01

Understanding Confidence Interval Adjustment

The original confidence interval of \((-473.13, 1691.9)\) is given for a 95% confidence level. To adjust this to a 90% confidence level, we need a new, narrower interval.
02

Determine Z-score for 90% Confidence Level

The critical z-value for a 95% confidence level is approximately 1.96. For a 90% confidence level, this is about 1.645. We use these values to calculate the new margin of error.
03

Calculate Original Margin of Error

The original margin of error is half the width of the initial confidence interval. Calculate it: \[ \text{Original Margin} = \frac{1691.9 - (-473.13)}{2} = 1082.515 \]
04

Adjust Margin of Error for 90% Confidence

Since the z-value for 90% confidence is smaller, the margin of error will also be smaller. New margin: \[ \text{New Margin} = \frac{1.645}{1.96} \times 1082.515 \approx 908.71 \]
05

Calculate New Confidence Interval

Subtract and add the new margin of error from the midpoint of the original interval to get the 90% confidence interval. The midpoint is already given as half the sum of the interval bounds: \[ \text{Midpoint} = \frac{-473.13 + 1691.9}{2} = 609.385 \] Therefore, the 90% confidence interval is \( (609.385 - 908.71, 609.385 + 908.71) = (-299.325, 1518.495) \).
06

Interpret the Confidence Interval

A 90% confidence interval for the difference in true average thenar muscle volume between sufferers and nonsufferers of carpal tunnel syndrome is approximately \((-299.325, 1518.495)\). This means that we are 90% confident the true average volume difference falls within this range.

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Key Concepts

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

Carpal Tunnel Syndrome
Carpal tunnel syndrome (CTS) is a condition that affects the hand and arm. It is caused by compression of the median nerve as it travels through the carpal tunnel in the wrist. This condition can lead to symptoms like numbness, tingling, and weakness in the hand and fingers.

Understanding CTS is important, as it directly ties into medical statistics and studies, like the one discussed in quantitative assessments. By analyzing CTS through quantitative MRI and electrophysiology, research looks into aspects like average muscle volume, aiming to aid diagnosis and treatment planning.

CTS can affect various muscle groups in the hand, one of which is the thenar muscle. The thenar muscle is crucial for hand function, as it controls thumb movements. Disturbances in thenar muscle volume, as seen in CTS sufferers, can impact hand strength and dexterity.
  • Main symptoms: numbness, tingling, and weakness.
  • Affects the median nerve in the wrist.
  • Important for assessing changes in muscle volume.
Thenar Muscle Volume
The thenar muscles sit at the base of the thumb and are essential for movements like gripping and pinching. In studies of conditions like CTS, measuring the thenar muscle volume helps researchers understand the impact of the syndrome on muscle health.

In the provided study, the focus is on understanding the difference in average thenar muscle volume between sufferers and nonsufferers of carpal tunnel syndrome. Analyzing these differences using confidence intervals provides a statistically sound approach to understanding muscle atrophy linked with CTS.

The concept of muscle volume measurement:
  • A key indicator of muscle health and function.
  • Measured in cubic millimeters (\( ext{mm}^3 \)).
  • Used to assess the severity and impact of CTS.
Understanding variations in thenar muscle volume can shed light on the physiological implications of CTS, potentially guiding more effective treatment interventions.
Z-Value
In statistical analysis, the z-value is a critical number used to derive confidence intervals. It relates to the standard normal distribution and helps determine how data is distributed compared to the mean.

For confidence intervals, z-values translate the desired level of confidence into a specific multiplier of the standard error. Different confidence levels have corresponding z-values:
  • 95% confidence level: z-value of approximately 1.96.
  • 90% confidence level: z-value of approximately 1.645.

A smaller z-value, like the one for a 90% confidence level, means a narrower confidence interval. This reflects less certainty than the wider interval of a higher confidence level, like 95%

Key points to remember about z-values:
  • They help specify the precision of an estimate.
  • Are used to adjust the width of confidence intervals.
  • Essential for interpreting statistical significance in medical studies relating to CTS.

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Most popular questions from this chapter

Sometimes experiments involving success or failure responses are run in a paired or before/after manner. Suppose that before a major policy speech by a political candidate, \(n\) individuals are selected and asked whether \((S)\) or not \((F)\) they favor the candidate. Then after the speech the same \(n\) people are asked the same question. The responses can be entered in a table as follows: where \(x_{1}+x_{2}+x_{3}+x_{4}=n\). Let \(p_{1}, p_{2}, p_{3}\), and \(p_{4}\) denote the four cell probabilities, so that \(p_{1}=P(S\) before and \(S\) after), and so on. We wish to test the hypothesis that the true proportion of supporters \((S)\) after the speech has not increased against the alternative that it has increased. a. State the two hypotheses of interest in terms of \(p_{1}, p_{2}\), \(p_{3}\), and \(p_{4}\). b. Construct an estimator for the after/before difference in success probabilities. c. When \(n\) is large, it can be shown that the rv \(\left(X_{i}-X_{j}\right) / n\) has approximately a normal distribution with variance given by \(\left[p_{i}+p_{j}-\left(p_{i}-p_{j}\right)^{2}\right] / n\). Use this to construct a test statistic with approximately a standard normal distribution when \(H_{0}\) is true (the result is called McNemar's test). d. If \(x_{1}=350, x_{2}=150, x_{3}=200\), and \(x_{4}=300\), what do you conclude?

The article "Lrban Battery Litter" cited in Example \(8.14\) gave the following summary data on zinc mass \((\mathrm{g})\) for two different brands of size D batteries: \begin{tabular}{lccc} Brand & Sample Size & Sample Mean & Sample SD \\ \hline Duracell & 15 & \(138.52\) & \(7.76\) \\ Energizer & 20 & \(149.07\) & \(1.52\) \\ \hline \end{tabular} Assuming that both zinc mass distributions are at least approximately normal, carry out a test at significance level \(.05\) to decide whether true average zinc mass is different for the two types of batteries.

Give as much information as you can about the \(P\)-value of the \(F\) test in each of the following situations: a. \(v_{1}=5, v_{2}=10\), upper-tailed test, \(f=4.75\) b. \(v_{1}=5, v_{2}=10\), upper-tailed test, \(f=2.00\) c. \(v_{1}=5, v_{2}=10\), two-tailed test, \(f=5.64\) d. \(v_{1}=5, v_{2}=10\), lower-tailed test, \(f=.200\) e. \(v_{1}=35, v_{2}=20\), upper-tailed test, \(f=3.24\)

The accompanying summary data on total cholesterol level (mmol/l) was obtained from a sample of Asian postmenopausal women who were vegans and another sample of such women who were omnivores ("Vegetarianism, Bone Loss, and Vitamin D: A Longitudinal Study in Asian Vegans and Non-Vegans," European \(J\). of Clinical Nutr., 2012: 75-82). \begin{tabular}{lccc} Diet & Sample Size & Sample Mean & Sample SD \\ \hline Vegan & 88 & \(5.10\) & \(1.07\) \\ Omnivore & 93 & \(5.55\) & \(1.10\) \\ \hline \end{tabular} Calculate and interpret a \(99 \%\) CI for the difference between population mean total cholesterol level for vegans and population mean total cholesterol level for omnivores (the cited article included a 95\% CI). [Note: The article described a more sophisticated statistical analysis for investigating bone density loss taking into account other characteristics ("covariates") such as age, body weight, and various nutritional factors; the resulting CI included 0 , suggesting no diet effect.]

Reliance on solid biomass fuel for cooking and heating exposes many children from developing countries to high levels of indoor air pollution. The article "Domestic Fuels, Indoor Air Pollution, and Children's Health" (Annals of the N.Y. Academy of Sciences, 2008: 209-217) presented information on various pulmonary characteristics in samples of children whose households in India used either biomass fuel or liquefied petroleum gas (LPG). For the 755 children in biomass households, the sample mean peak expiratory flow (a person's maximum speed of expiration) was \(3.30 \mathrm{~L} / \mathrm{s}\), and the sample standard deviation was \(1.20\). For the 750 children whose households used liquefied petroleum gas, the sample mean PEF was \(4.25\) and the sample standard deviation was 1.75. a. Calculate a confidence interval at the \(95 \%\) confidence level for the population mean PEF for children in biomass households and then do likewise for children in LPG households. What is the simultaneous confidence level for the two intervals? b. Carry out a test of hypotheses at significance level \(.01\) to decide whether true average PEF is lower for children in biomass households than it is for children in LPG households (the cited article included a \(P\)-value for this test). c. \(\mathrm{FEV}_{1}\), the forced expiratory volume in 1 second, is another measure of pulmonary function. The cited article reported that for the biomass households the sample mean \(\mathrm{FEV}_{1}\) was \(2.3 \mathrm{~L} / \mathrm{s}\) and the sample standard deviation was \(.5 \mathrm{~L} / \mathrm{s}\). If this information is used to compute a \(95 \%\) CI for population mean \(\mathrm{FEV}_{1}\), would the simultaneous confidence level for this interval and the first interval calculated in (a) be the same as the simultaneous confidence level determined there? Explain.
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