Question

Many people now turn to the Internet to get information on health-related topics. A research article 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 26.6 and the standard deviation of the Flesch scores was 14.2. For a representative sample of pages from WebMD, the mean score was 43.8 and the standard deviation was 19.9. Suppose that these means and standard deviations were based on samples of 40 pages from each site. (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 - MWebMD. Use technology. Round your answers to two decimal places.) WebMD. (Use M Wikipedia ( -18.10 -16.30 * )

Answer 1

solution:-

(a) given that

wikipedia

mean x1 = 26.6 , standard deviation s1 = 14.2 , n1 = 40

webMD

mean x2 = 43.8 , standard deviation s2 = 19.9 , n2 = 40

degree of freedom df = (n1+n2)-2 = (40+40)-2 = 78

we look into t table with df and with 90% confidence level

critical value t = 1.665

confidence interval formula

=> (x1-x2) +/- t * sqrt(s1^2/n1 + s2^2/n2)

=> (26.6-43.8) +/- 1.665 * sqrt((14.2^2/40) + (19.9^2/40))

=> (-23.64 , -10.76)