A random sample of patient records from a large medical facility were taken andsubjects were classified by smoking status and cholesterol levels. The results aresummarized in the table with observed counts next to expected values inparentheses followed by the Chi-square component for each cell in brackets.Cholesterol Level Never Smoked80 (68.86) [1.80]49 (48.55) [0.00]28 (39.59) [3.40]Totals 157LowModerateHighSmoking StatusFormer Smoker63 (63.16) [0.00]37 (44.53) [1.27]44 (36.32) [1.63]144Current Smoker57 (67.98) [1.77]55 (47.93) [104]43 (39.09) [0.39]155Discuss briefly if the conditions for this test have been met.Totals200141115456

In the provided data table, the given variables are "smoking status" (which is categorized into…

Answered: A random sample of patient records from…

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 healthcare provider monitors the number of CAT scans perfomed each month in each of its clinics. The most recent year of data for a particular clinics follows (the reported variable is the number of CAT scans each month expressed as the number of CAT scans per thousand members of the health plan): 2.31.2.09. 2.36, 1.95, 1.98, 2.25. 2.16, 2.07. 1.88, 1.94, 1.97, 2.02. Find a two sided 95% confidence interval for the standard deviation. What should you do to address any reservations about this confidence interval? Round your answers to two decimal places (e.g. 98.76). Sos
A randomly sampled group of patients at a major U.S. regional hospital became part of a nutrition study on dietary habits. Part of the study consisted of a 50-question survey asking about types of foods consumed. Each question was scored on a scale from one (most unhealthy behavior) to five (most healthy behavior). The answers were summed and averaged. The population of interest is the patients at the regional hospital. A prior survey of patients had found the mean score for the population of patients to be μ=2.9 After careful review of these data, the hospital nutritionist decided that patients could benefit from nutrition education. The current survey was implemented after patients were subjected to this education, and it produced the following sample statistics for the 15 patients sampled: x¯=3.5 and s=1.2 We would like to know if the education improved nutrition behavior. We test the hypotheses H0:μ=2.9 versus Ha:μ>2.9. The tt test to be used has the test statistic: Answer:
A study was conducted to examine the impact of speaking in public on college students. A class of 15 statistics students participated in the study. At the beginning of a lecture the students recorded their systolic blood pressure. During the lecture the instructor called on each student to stand and answer questions about topics in the lecture. After speaking the students once again recorded their blood pressure. The resulting values are given below along with summary statistics. before 100 123 128 125 136 108 130 113 105 136 146 138 108 149 132 after 141 142 146 136 144 111 132 114 103 133 140 130 100 133 113 difference -41 -19 -18 -11 -8 -3 -2 -1 2 3 6 8 8 16 19 before after difference mean 125.1 127.9 -2.733 st dev 15.26 15.45 15.23 Does this information indicate that blood pressure increased because of speaking? Conduct an appropriate hypothesis test and show appropriate steps. Let μ1 mean systolic blood pressure before speaking in public, and μ2 mean…
An Internet retailer would like to investigate the relationship between the amount of time in minutes a purchaser spends on its Web site and the amount of money he or she spends on an order. The table to the right shows the data from a random sample of
A healthcare provider monitors the number of CAT scans perfomed each month in each of its clinics. The most recent year of data for a particular clinics follows (the reported variable is the number of CAT scans each month expressed as the number of CAT scans per thousand members of the health plan): 2.31, 2.09, 2.36, 1.95, 1.98, 2.25, 2.16, 2.07, 1.88, 1.94, 1.97, 2.02. Find a two-sided 95% confidence interval for the standard deviation. What should you do to address any reservations about this confidence interval? Round your answers to two decimal places (e.g. 98.76).
The mean birthweight of an infant is thought to be associated with the smoking status of the mother during the first trimester of the pregnancy. Consider the following data from mothers divided into four categories according to smoking habits, and the corresponding birthweights of their children. Group 1: Mother is a non smoker (NON): 7.5, 6.2, 6.9, 7.4, 9.2, 8.3, 7.6 Group 2: Mother is an ex-smoker (EX): 5.8, 7.3, 8.2, 7.1, 7.8 Group 3: Mother is a current smoker, smokes less than 1 pack per day (CUR < 1) 5.9, 6.2, 5.8, 4.7, 8.3, 7.2, 6.2 Group 4: Mother is a current smoker, smokes more than or equal to 1 pack per day (Cur ≥ 1) 6.2, 6.8, 5.7, 4.9, 6.2, 7.1, 5.8, 5.4 Using Stata test whether or not the mean birthweights are equal across all four groups at the alpha = 0.05 level. What are your null and alternative hypothesis? Ο HO: μ_1== μ_2 HA: μ_1=\= μ_2 O HO: μ_i== u_j for all i and j HA: μ_i=\= μ_j for all i and j O HO: μ_i == μ_j for all i and j HA: μ_i=\= μ_j for at least one i,j…
The age of cars in the staff parking lot of a suburban college is uniformly distributed from six months (0.5) to 9.5 years. Find the 35th percentile
A certain virus affects 0.7% of the population. A test used to detect the virus in a person is positive 87% of the time if the person has the virus (true positive) and 14% of the time if the person does not have the virus (false positive). Fill out the remainder of the following table and use it to answer the two questions below based on a total sample of 100,000 people. Virus No Virus TotalPositive Test Negative Test Total 100,000a) Find the probability that a person has the virus given that they have tested positive. Round your answer to the nearest hundredth of a percent and do not include a percent sign. % b) Find the probability that a person does not have the virus given that they test negative. Round your answer to the nearest hundredth of a percent and do not include a percent sign. %
I need help filling in the following blanks based on the attached table: The median frequency of attendance is [Blank]. This means [Blank] percent of people in this sample attend [Blank] or less. The 36th percentile is [Blank]. That means [Blank] percent of people in the same area attend [Blank] or less frequently. The 65th percentile is [Blank]. That means [Blank] percent of people in the same area attend [Blank] or less frequently. frequently.

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