A pharmaceutical company is formulating a new medication that prevents painful gout attacks. However, the lead researchers are also mindful of its impacts to the human liver, which can be indicated by monitoring levels of alanine aminotransferase (or ALT, in U/L units) . If the ALT levels exceed an average of 55 U/L, there is an indication of the drug having damaging effects to the liver. Let us assume that ALT levels are normally distributed with standard deviation of 15.7 U/L. To determine the consequence of this new drug, they randomly selected 162 adult men with gout and monitored levels of alanine aminotransferase (or ALT, in U/L units) after taking the new medicine for 6 months. From the sample, it is found that the average ALT level is 59 U/L. It is suspected levels are significantly increased. Conduct a test of significance with a=0.05. (Source: Liver function tests. (2019, June 13). Retrieved from https://www.mayoclinic.org/tests-procedures/liver-function-tests/about/pac- 2039459) What is the value of the Z-test statistic? O 3.24 1.23 O 4.00 -3.24 0.25

A pharmaceutical company is formulating a new medication that prevents painful gout attacks. However, the lead researchers are also mindful of its impacts to the human liver, which can be indicated by monitoring levels of alanine aminotransferase (or ALT, in U/L units) . If the ALT levels exceed an average of 55 U/L, there is an indication of the drug having damaging effects to the liver. Let us assume that ALT levels are normally distributed with standard deviation of 15.7 U/L. To determine the consequence of this new drug, they randomly selected 162 adult men with gout and monitored levels of alanine aminotransferase (or ALT, in U/L units) after taking the new medicine for 6 months. From the sample, it is found that the average ALT level is 59 U/L. It is suspected levels are significantly increased. Conduct a test of significance with a=0.05. (Source: Liver function tests. (2019, June 13). Retrieved from https://www.mayoclinic.org/tests-procedures/liver-function-tests/about/pac- 2039459) What is the value of the Z-test statistic? O 3.24 1.23 O 4.00 -3.24 0.25

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

See similar textbooks

Similar questions

The average American consumes 86 liters of alcohol per year. Does the average college student consume more alcohol per year? A researcher surveyed 10 randomly selected college students and found that they averaged 88.8 liters of alcohol consumed per year with a standard deviation of 17 liters. What can be concluded at the the αα = 0.10 level of significance?
It takes an average of 9.8 minutes for blood to begin clotting after an injury. An EMT wants to see if the average will change if the patient is immediately told the truth about the injury. The EMT randomly selected 52 injured patients to immediately tell the truth about the injury and noticed that they averaged 10.7 minutes for their blood to begin clotting after their injury. Their standard deviation was 1.98 minutes. What can be concluded at the the αα = 0.05 level of significance? For this study, we should use Select an answer t-test for a population mean, z-test for a population proportion The null and alternative hypotheses would be: H0:H0: ? μ p Select an answer = ≠ < > H1:H1: ? p μ Select an answer < ≠ > = The test statistic ? t z = (please show your answer to 3 decimal places.) The p-value = (Please show your answer to 4 decimal places.) The p-value is ? > ≤ αα Based on this, we should Select an answer reject fail to reject accept…
Overproduction of uric acid in the body can be an indication of cell breakdown. This may be an advance indication of illness such as gout , leukemia or lymphoma. Over a period of months an adult male patient has taken even blood tests for uric acid. The mean concentration was x =5.30mh/dL. The distribution of uric acid in healthy adult males can be assumed to be normal, with standard deviation =1.95mg/dL. Find a 95% confidence interval for the population mean concentration of uric acid in this patients blood What is margin of error? Lower limit? Upper limit? And margin of error?
The mean life expectancy in the US pop. is currently 81.1 years with a standard deviation of 15.9. A researcher examined the age of death of 70 people in Florida hospitals and the mean was 85.2 yrs. He runs a 2tailed test with a=.05 to see if FL has significantly different life expectancy to the overall US pop. Calculate the standard error, z statistic and critical z values from this information.
It takes an average of 14.5 minutes for blood to begin clotting after an injury. An EMT wants to see if the average will increase if the patient is immediately told the truth about the injury. The EMT randomly selected 65 injured patients to immediately tell the truth about the injury and noticed that they averaged 15.4 minutes for their blood to begin clotting after their injury. Their standard deviation was 3.05 minutes. What can be concluded at the the αα = 0.05 level of significance? For this study, we should use The null and alternative hypotheses would be: H0:H0: H1:H1: The test statistic = (please show your answer to 3 decimal places.) The p-value = (Please show your answer to 4 decimal places.) The p-value is αα Based on this, we should the null hypothesis. Thus, the final conclusion is that ... The data suggest the populaton mean is significantly greater than 14.5 at αα = 0.05, so there is statistically significant…
It takes an average of 10.8 minutes for blood to begin clotting after an injury. An EMT wants to see if the average will increase if the patient is immediately told the truth about the injury. The EMT randomly selected 46 injured patients to immediately tell the truth about the injury and noticed that they averaged 11.7 minutes for their blood to begin clotting after their injury. Their standard deviation was 2.98 minutes. What can be concluded at the the αα = 0.05 level of significance? The null and alternative hypotheses would be: Please provide number H0:H0: mean =Correct __?__ H1:H1: mean > Correct_ __?__ The test statistic t = __?__ (please show your answer to 3 decimal places.) The p-value = __?__ (Please show your answer to 4 decimal places.)
Someone once dropped a 'mint imperial", a type of sweet, from the top of a multi-story car park and it landed on my grandmother's head. The average speed of a falling mint imperial is 4 m/s and the velocity is a Gaussian distribution with standard deviation 0.25 m/s. If a mint travelling faster than 5 m/s causes injury, what is the chance my grandmother was injured? In fact she was fine, but very annoyed. O(i- erf (2/2)/2 O [1- erf (v2)/2 O [1- erf (2) O [1- erf (4)/2
Albumin is a liver protein that helps circulate vitamins throughout the body. Reduced Albumin has been associated with severe illness with Covid-19 patients. Suppose we wanted to conduct a study to compare the mean Albumin concentration (g/L) between a sample of adult patients hospitalized for Covid 19 (Sample Mean= 32.8, s=6.0) to the mean albumin concentration (g/L) of healthy adults (u=34.1, o=7.13). How many patients would need to enroll from the hospital to conduct this study with 99% confidence, and 80% power? A. 267 B. 120 C. 352 D. 225
As of 2021 research is being conducted to study the effects of space travel on human body. For example, astronauts stay on the international Space Station (ISS) on missions and the amount of radiation they are exposed to is measured. Suppose the radiation they are exposed to on a regular six month mission follows a Normal distribution with mean 85 millisieverts(mSv) and standard deviation of 30 mSv. If the risk of cancer increases when the astranauts are exposed to more than 100mSv, what is the probability that an astranaut's cancer risk will increase during a regular mission?
The nicotine content in cigarettes of a certain brand is normally distributed with mean (in milligrams) u and standard deviation o = 0.1. The brand advertises that the mean nicotine content of their cigarettes is 1.5 mg. Now, suppose a reporter wants to test whether the mean nicotine content is actually higher than advertised. He takes measurements from a SRS of 15 cigarettes of this brand. The sample yields an average of 1.6 mg of nicotine. Conduct a test using a significance level of a = 0.05. (a) The standardized test statistic (b) The critical value (endpoint of Rejection Region) is z* = (c) The final conclusion is OA. The nicotine content is probably higher than advertised. B. There is not sufficient evidence to show that the ad is misleading.
Does chronic hepatitis C infection impact bone mineral density? In the general population of healthy young adults, bone mineral density is Normally distributed, and density scores are standardized to have a mean u of 0 and a standard deviation o of 1. A study examined the bone densities of a random sample of 28 adult men suffering from chronic hepatitis C infection. Their mean standardized bone density was -0.61. Does this provide strong evidence that the mean standardized bone density of adult men suffering from chronic hepatitis C infection is less than 0?
It takes an average of 12.7 minutes for blood to begin clotting after an injury. An EMT wants to see if the average will decline if the patient is immediately told the truth about the injury. The EMT randomly selected 51 injured patients to immediately tell the truth about the injury and noticed that they averaged 12.4 minutes for their blood to begin clotting after their injury. Their standard deviation was 3.77 minutes. What can be concluded at the the αα = 0.10 level of significance? For this study, we should use Select an answer t-test for a population mean z-test for a population proportion The null and alternative hypotheses would be: H0:H0: ? p μ Select an answer > < ≠ = H1:H1: ? p μ Select an answer = < ≠ > The test statistic ? z t = (please show your answer to 3 decimal places.) The p-value = (Please show your answer to 4 decimal places.) The p-value is ? > ≤ αα Based on this, we should Select an answer reject accept fail to reject…