**Vitamin D Intoxication Case Study****Background:**Eight individuals in Massachusetts experienced unexplained vitamin D intoxication, leading to hospitalization. It was hypothesized that excessive dairy milk supplementation might be the cause. Below are the blood levels of calcium and albumin for each individual at the time of hospital admission.**Blood Levels:**| Calcium (mmol/l) | Albumin (g/l) ||------------------|---------------|| 2.99 | 40 || 3.79 | 41 || 3.17 | 36 || 2.48 | 41 || 3.25 | 46 || 3.47 | 42 || 2.39 | 39 || 3.21 | 40 |**Questions:**(a) **Construct a One-Sided 95% Confidence Interval:** - Find the lower bound for the true mean calcium level of individuals experiencing vitamin D intoxication.(b) **Compute a 95% Lower Confidence Bound:** - Determine the true mean albumin level for this group.(c) **Comparison with Normal Ranges:** - For healthy individuals, normal calcium values range from 2.12 to 2.74 mmol/l and normal albumin levels range from 32 to 55 g/l. - Assess whether patients with vitamin D intoxication have normal blood levels of calcium and albumin.**Analysis:**Use the given blood levels to perform statistical analysis and compare with standard health ranges to assess the extent of deviation, if any, in these patients potentially due to vitamin D intoxication.

Given data calcium Albumin2.99 403.79…

Answered: 10. When eight persons in Massachusetts…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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A random sample of n1 = 10 regions in New England gave the following violent crime rates (per million population). x1: New England Crime Rate 3.3 3.9 4.2 4.1 3.3 4.1 1.8 4.8 2.9 3.1 Another random sample of n2 = 12 regions in the Rocky Mountain states gave the following violent crime rates (per million population). x2: Rocky Mountain Crime Rate 3.7 4.1 4.7 5.5 3.3 4.8 3.5 2.4 3.1 3.5 5.2 2.8 Assume that the crime rate distribution is approximately normal in both regions. Do the data indicate that the violent crime rate in the Rocky Mountain region is higher than in New England? Use ? = 0.01. Solve the problem using both the traditional method and the P-value method. (Test the difference ?1 − ?2. Round the test statistic and critical value to three decimal places.) test statistic critical value Find (or estimate) the P-value. P-value > 0.250 0.125 < P-value < 0.250 0.050 < P-value < 0.125 0.025 < P-value < 0.050 0.005 <…
A random sample of n1 = 10 regions in New England gave the following violent crime rates (per million population). x1: New England Crime Rate 3.3 3.9 4.2 4.1 3.3 4.1 1.8 4.8 2.9 3.1 Another random sample of n2 = 12 regions in the Rocky Mountain states gave the following violent crime rates (per million population). x2: Rocky Mountain Crime Rate 3.7 4.1 4.7 5.5 3.3 4.8 3.5 2.4 3.1 3.5 5.2 2.8 Assume that the crime rate distribution is approximately normal in both regions. Do the data indicate that the violent crime rate in the Rocky Mountain region is higher than in New England? Use ? = 0.01. Solve the problem using both the traditional method and the P-value method. (Test the difference ?1 − ?2. Round the test statistic and critical value to three decimal places.) test statistic critical value Find (or estimate) the P-value. P-value > 0.2500.125 < P-value < 0.250 0.050 < P-value < 0.1250.025 < P-value < 0.0500.005 <…
The price to earnings ratio (P/E) is an important tool in financial work. A random sample of 14 large U.S. banks (J. P. Morgan, Bank of America, and others) gave the following P/E ratios.† 24 16 22 14 12 13 17 22 15 19 23 13 11 18 The sample mean is x ≈ 17.1. Generally speaking, a low P/E ratio indicates a "value" or bargain stock. Suppose a recent copy of a magazine indicated that the P/E ratio of a certain stock index is μ = 19. Let x be a random variable representing the P/E ratio of all large U.S. bank stocks. We assume that x has a normal distribution and σ = 4.6. Do these data indicate that the P/E ratio of all U.S. bank stocks is less than 19? Use α = 0.01. (a) What is the level of significance? Compute the z value of the sample test statistic. (Round your answer to two decimal places.)(c) Find (or estimate) the P-value. (Round your answer to four decimal places.)
A random sample of n1 = 10 regions in New England gave the following violent crime rates (per million population). x1: New England Crime Rate 3.3 3.7 4.2 3.9 3.3 4.1 1.8 4.8 2.9 3.1 Another random sample of n2 = 12 regions in the Rocky Mountain states gave the following violent crime rates (per million population). x2: Rocky Mountain Crime Rate 3.5 4.1 4.7 5.5 3.3 4.8 3.5 2.4 3.1 3.5 5.2 2.8 Assume that the crime rate distribution is approximately normal in both regions. Do the data indicate that the violent crime rate in the Rocky Mountain region is higher than in New England? Use ? = 0.01. Solve the problem using both the traditional method and the P-value method. (Test the difference ?1 − ?2. Round the test statistic and critical value to three decimal places.) test statistic critical value Find (or estimate) the P-value. A. P-value > 0.250 B. 0.125 < P-value < 0.250 C. 0.050 < P-value < 0.125 D. 0.025 < P-value <…
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A random sample of n1 = 10 regions in New England gave the following violent crime rates (per million population). x1: New England Crime Rate 3.3 3.9 4.2 4.1 3.3 4.1 1.8 4.8 2.9 3.1 Another random sample of n2 = 12 regions in the Rocky Mountain states gave the following violent crime rates (per million population). x2: Rocky Mountain Crime Rate 3.5 4.1 4.5 5.5 3.3 4.8 3.5 2.4 3.1 3.5 5.2 2.8 Assume that the crime rate distribution is approximately normal in both regions. Do the data indicate that the violent crime rate in the Rocky Mountain region is higher than in New England? Use ? = 0.01. Solve the problem using both the traditional method and the P-value method. (Test the difference ?1 − ?2. Round the test statistic and critical value to three decimal places.)
In a 1868 paper, German physician Carl Wunderlich reported based on over a million body temperature readings that the mean body temperature for healthy adults is 98.6° F. However, it is now commonly believed that the mean body temperature of a healthy adult is less than what was reported in that paper. To test this hypothesis a researcher measures the following body temperatures from a random sample of healthy adults. 98.2, 98.6, 98.5, 98.6, 97.6, 98.2, 98.0 (a) Find the value of the test statistic. (b) Find the 10% critical value.
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The age-specific death rate for the over-65 age group is Age-Group Population Number of Deaths 65 6,000 155 a. 155 b. 25.8 per 1,000 O c. 1.55 per 10,000 d. 25.8 per 10,000
Refer to the following morbidity and mortality statistics for Community X for 2008. Total One-Year Population: 250,000 Population of Women 15-44: 75,000 Number of live births: 6,000 Total Deaths: 2,260 Number of infant deaths: 120 Deaths from heart disease: 222 Deaths from Cancer: 115 Deaths from Stroke: 90 Number of persons diagnosed with cancer 10,500 Calculate the following: Infant Mortality Rate _____________________________________ Prevalence of cancer for 2008 _____________________________ Crude Death Rate ______________________________________ Cause-specific mortality rate for stroke…
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