5. [2/4 Points] DETAILS PREVIOUS ANSWERS PECKSTAT2 13.3.033.S. MY NOTES ASK YOUR TEACHER 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 14.5. 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. LA USE SALT (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. (Use "Wikipedia " /WebMD. Use SALT. Round your answers to two decimal places.) (b) What does this confidence interval imply about the readability of health-related information from these two sources? This interval indicates that, on average, Wikipedia pages and WebMD pages have equally difficult reading levels. This interval indicates that, on average, Wikipedia pages have a more difficult reading level. This interval indicates that, on average, WebMD pages have a more difficult reading level. Is this consistent with the conclusion of a hypothesis test that there is convincing evidence that the mean reading level for health-related pages differs for Wikipedia and WebMD? The t test statistic for this test was -4.4. This is not consistent with the conclusion and not consistent with the t test statistic. This is consistent with the conclusion but not consistent with the t test statistic. This is consistent with the conclusion and consistent with the t test statistic. This is not consistent with the conclusion but is consistent with the t test statistic. Need Help? Read It Watch It