Use the information in the previous exercise to answer the following questions.
- a. Construct a 90% confidence
interval estimate of the difference in mean Flesch reading ease score for health-related pages on Wikipedia and health-related pages on WebMD. - b. What does this confidence interval imply about the readability of health-related information from these two sources? Is this consistent with the conclusion in the hypothesis test of the previous exercise?
Many people now tum to the Internet to get information on health-related topics. The paper “An Examination of Health, Medical and Nutritional Information on the Internet: A Comparative Study of Wikipedia, WebMD and the Mayo Clinic Websites” (The International Journal of Communication and Health [2015]: 30–38) used Flesch reading ease scores (a measure of reading difficulty based on factors such as sentence length and number of syllables in the words used) to score pages on Wikipedia and on WebMD. Higher Flesch scores correspond to more difficult reading levels. The paper reported that for a representative sample of health-related pages on Wikipedia, the mean Flesch score was 26.7 and the standard deviation of the Flesch scores was 14.1. For a representative sample of pages from WebMD, the mean score was 43.9 and the standard deviation was 19.4. Suppose that these means and standard deviations were based on samples of 40 pages from each site. Is there convincing evidence that the mean reading level for health-related pages differs for Wikipedia and WebMD? Test the relevant hypotheses using a significance level of 0.05.
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Introduction To Statistics And Data Analysis
- A research group studying cell phone habits asked the question “Do you ever use your cell phone to make a payment at a convenience store?” to people selected from two random samples of cell phone users. One sample consisted of older adults, ages 35 years and older, and the other sample consisted of younger adults, ages 18 years to 34 years. The proportion of people who answered yes in each sample was used to create a 95 percent confidence interval of (0.097,0.125)(0.097,0.125) to estimate the difference (younger minus older) between the population proportions of people who would answer yes to the question. Which of the following is the best description of what is meant by 95 percent confidence? In repeated random sampling with the same sample size, approximately 95% of the sample proportions from the younger group will be between 0.097 and 0.125 greater than the sample proportion from the older group. A In repeated random sampling with the same sample size,…arrow_forwardThe following data were collected in a clinical trial to compare a new drug to a placebo for its effectiveness in lowering total serum cholesterol. New Drug (n=75) Placebo (n=75) Total Sample (n=150) Mean (SD) Total Serum Cholesterol 182.0 (24.5) 206.3 (21.8) 194.15 (23.2) % Patients with Total Cholesterol < 200 78.0% 65.0% 71.5% Generate the 95% confidence interval for the difference in mean total cholesterol levels between treatments Generate a 95% confidence interval for the difference in proportions of all patients with total cholesterol < 200. How many patients would be required to detect the difference in proportions observed in the current study with 80% power. A new investigation will be enrolling subjects to demonstrate the efficacy of the drug in individuals with high cholesterol at an early age. A two-sided test is planned at α = 0.05arrow_forward1. A recent survey showed that from a sample of 500 packages delivered by a Postal Service, 480were delivered on time. a) Construct a 95% confidence interval for the proportion of all packages that are deliveredon time by the Postal Service.arrow_forward
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- Glencoe Algebra 1, Student Edition, 9780079039897...AlgebraISBN:9780079039897Author:CarterPublisher:McGraw Hill