### Colonoscopy as a Screening Test for Colon CancerA colonoscopy is a screening test for colon cancer, recommended as a routine test for adults over age 50. One study¹ provides the best evidence yet that this test saves lives. The proportion of people with colon polyps expected to die from colon cancer is 0.01. A sample of 2602 people who had polyps removed during a colonoscopy were followed for 20 years, and 12 of them died from colon cancer. Does this provide evidence that the proportion of people who die from colon cancer after having polyps removed in a colonoscopy is significantly less than the expected proportion (without a colonoscopy) of 0.01?¹Zauber, et.al., "Colonscopic Polypectomy and Long-Term Prevention of Colorectal-Cancer Deaths," *New England Journal of Medicine*, 2012; 366: 687-696.### Hypothesis Testing**(a) What are the null and alternative hypotheses?**Below the text, there is a graphical tool for formulating hypotheses, allowing users to select various symbols and notation to build their hypotheses. Symbols and notations available include:- Equality (=)- Inequality (≠)- Less than (<)- Greater than (>)- Mean (μ)- Different population means (μ₁, μ₂)- Proportion (p)- Sample means (x̄₁, x̄₂)- Specific sample statistics (e.g., 0.01)- Correlation coefficient (ρ)- Sample proportions (p̂₁, p̂₂)- Linear correlation coefficient (r)**Hypotheses to test:**- Null Hypothesis (H₀): p = 0.01 (The proportion of people who die from colon cancer after having polyps removed in a colonoscopy is equal to 0.01)- Alternative Hypothesis (Hₐ): p < 0.01 (The proportion of people who die from colon cancer after having polyps removed in a colonoscopy is less than 0.01)By formulating these hypotheses, researchers can use statistical tests to determine whether the evidence supports that the colonoscopy procedure significantly reduces the risk of death from colon cancer compared to the expected proportion without the procedure.

Given values, p = 0.01n = 2602x = 12

A colonoscopy is a screening test for colon cancer, recommended as a routine test for adults over age 50. One study' provides the best evidence yet that this test saves lives. The proportion of people with colon polyps expected to die from colon cancer is 0.01. A sample of 2602 people who had polyps removed during a colonoscopy were followed for 20 years, and 12 of them died from colon cancer. Does this provide evidence that the proportion of people who die from colon cancer after having polyps removed in a colonoscopy is significantly less than the expected proportion (without a colonoscopy) of 0.01? 'Zauber, et.al., "Colonoscopic Polypectomy and Long-Term Prevention of Colorectal-Cancer Deaths," New England Journal of Medicine, 2012; 366: 687-696. (a) What are the null and alternative hypotheses? :: p : Pi : P2 :: 0.01 : Pi : P2 :: :: ::

A colonoscopy is a screening test for colon cancer, recommended as a routine test for adults over age 50. One study' provides the best evidence yet that this test saves lives. The proportion of people with colon polyps expected to die from colon cancer is 0.01. A sample of 2602 people who had polyps removed during a colonoscopy were followed for 20 years, and 12 of them died from colon cancer. Does this provide evidence that the proportion of people who die from colon cancer after having polyps removed in a colonoscopy is significantly less than the expected proportion (without a colonoscopy) of 0.01? 'Zauber, et.al., "Colonoscopic Polypectomy and Long-Term Prevention of Colorectal-Cancer Deaths," New England Journal of Medicine, 2012; 366: 687-696. (a) What are the null and alternative hypotheses? :: p : Pi : P2 :: 0.01 : Pi : P2 :: :: ::

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

In a study conducted by the U.S. Department of Health and Human Services, a sample of 546 boys aged 6–11 was weighed, and it was determined that 87 of them were overweight. A sample of 508 girls aged 6–11 was also weighed, and 74 of them were overweight. Can you conclude that the proportion of boys who are overweight differs from the proportion of girls who are overweight? Please show calcul
Imagine you are a statistical consultant and your client, a weightlifting supplement company, is looking to see if their product is effective in increasing muscle mass gained over a training cycle. You randomly sampled 20 female weightlifters and assigned them to take the supplement and 20 to take a placebo. You have both groups run the same controlled training program over 12 weeks and calculated their whole body lean mass difference in pounds from the beginning to the end of the program. Can you infer that the supplement group gained more muscle mass on average than the placebo group? Use a = 0.05 for any hypothesis test. પ્રમ 1) State the research question in one sentence and use the question to determine the parameter we are using to conduct statistical inference. Describe this parameter in one sentence using symbols and a description. manu Aren
Part 1 of 4 The restaurants of a fast food chain are primarily located in three regions of Arizona. The CEO of the chain wants to know if the average monthly revenue of the restaurants in each region are equal. The monthly revenues (in thousads of dollars) for this past month of the restaurants in each of the three regions are given below. Conduct an ANOVA test using a 5% level of significance to determine whether or not the average monthly revenues in the three regions are equal. Monthly Revenues of Restaurants in Region 1: 179, 183.3, 156.8, 155.1, 253, 105.4, 219, 134.3, 119.9, 193.1, 188.5, 188.4 Monthly Revenuse of Restaurants in Region 2: 158.7, 175.2, 187.8, 187.1, 118.1, 179.5, 195.1, 199.7, 349.6, 226.5, 253.7, 287.4, 172.8, 290.5, 149 Monthly Revenuse of Restaurants in Region 3: 215.4, 199.8, 360.7, 273, 250, 214.2, 144.5, 165, 137, 187.6 Step 1: State the null and alternative hypotheses. Ho: P1 = P2 = H3 Ha: At least one mean isn't equal to the other means v Part 2 of 4 Step…
PLEASE ANSWER ASAP: A case-control study was conducted to evaluate the relationship between physical activity and coronary heart disease (CHD) in men. A total of 406 men newly diagnosed with CHD were included together with 406 men of similar ages who did not have CHD. The risk of CHD (measured by the OR) was higher among men who were inactive or only moderately active (collectively called ‘inactive’) than among those who were physically active: (See attached image) Odds ratio for inactive men compared to active men= (299 x 136)/(270 x 107)=1.41Suppose there is no misclassification in these data.What would the observed OR have been... 1. If 20% of all the inactive men had been misclassified as active? 2. If 10 % of inactive cases had been misclassified as active?
A colleague of yours has made a decision to decrease sample size because of limited funding. What should they expect to happen to the probability of uncovering statistically significant findings in their research? Why?
Left ventricular mass (LVM) is an important risk factor for subsequent cardiovascular disease. A study is proposed to assess the relationship between childhood blood-pressure levels and LVM in children as determined from echocardiograms. The goal is to stratify children into a normal bp group (< 90th percentile for their age, gender, and height) and an elevated bp group (≥ 90th percentile for their age, gender, and height) and compare change in LVM between the two groups. Before this can be done, one needs to demonstrate that LVM actually changes in children over a 4-year period. To help plan the main study, a pilot study is conducted where echocardiograms are obtained from 10 random children from the Bogalusa Heart Study at baseline and after 4 years of follow-up. The data are given in Table 1 . Table 1 Pilot data on left ventricular mass (LVM) in children from the Bogalusa Heart Study ID Baseline LVM (g) 4- year LVM (g) Change (g)* 1 139 163 24 2 134 126 -8 3 86 142 56…
A company claims that their product has an average lifespan of 6 years. An independent company tested the product and claimed that was false. What type of statistical error did the original company make? Explain your reasoning.
please answer questions B,C,F
Postpartum depression and anxiety (PPD), is a common medical condition affecting mothers and their families after the birth of a baby. The CDC estimated that 18% of women who have recently given birth suffer from PPD. However, this research only reflected self-reported cases. Therefore, one group dedicated to helping women and their families with PPD believes that the true percentage of women who suffer from PPD is much higher. The group conducts a simple random sample of 112 women who had given birth in the last year and discovers that 26 of them report having PPD. Based on this evidence, can the group claim that the true percentage of women who have PPD is greater than 18%? Use a 0.05 level of significance. Step 1 of 3 : State the null and alternative hypotheses for the test. Fill in the blank below. H0: p=0.18 Ha: p⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯0.18
Postpartum depression and anxiety (PPD), is a common medical condition affecting mothers and their families after the birth of a baby. The CDC estimated that 18% of women who have recently given birth suffer from PPD. However, this research only reflected self-reported cases. Therefore, one group dedicated to helping women and their families with PPD believes that the true percentage of women who suffer from PPD is much higher. The group conducts a simple random sample of 112 women who had given birth in the last year and discovers that 26 of them report having PPD. Based on this evidence, can the group claim that the true percentage of women who have PPD is greater than 18%? Use a 0.05 level of significance. Step 2 of 3: Compute the value of the test statistic. Round your answer to two decimal places.
You conducted a study to assess the effect of smoking on systolic blood pressure. The results showed that there was a significant association between the two variables, with an odds ratio of 1.5. Explain this finding to someone who has no background in epidemiologic research.
Please answer these questions thanks