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 healthrelated pages on Wikipedia, the mean Flesch score was 26.2 and the standard deviation of the Flesch
scores was 14.7. For a representative sample of pages from WebMD, the mean score was 43.4 and the
standard deviation was 19.6. 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 š¼ = 0.05.(Use
š1 for Wikipedia and š2 for WebMD.)
State the appropriate null and alternative hypotheses. (Enter != for ā as needed.)
Find the test statistic. (Round your answer to two decimal places.)
Use technology to find the P-value. (Round your answer to four decimal places.)
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Null hypothesis: The mean reading level for health-related pages is the same for Wikipedia and WebMD (š1 = š2).
Alternative hypothesis: The mean reading level for health-related pages differs for Wikipedia and WebMD (š1 ā š2).
To find the test statistic, we can use the formula for the two-sample t-test:
t = ((x1 - x2) - (š1 - š2)) / sqrt((s1^2/n1) + (s2^2/n2))
Where x1 and x2 are the sample means, š1 and š2 are the population means, s1 and s2 are the sample standard deviations, and n1 and n2 are the sample sizes.
Given:
x1 = 26.2
x2 = 43.4
š1 - š2 = 0 (null hypothesis)
s1 = 14.7
s2 = 19.6
n1 = n2 = 40
Plugging in these values, we can calculate the test statistic:
t = ((26.2 - 43.4) - 0) / sqrt((14.7^2/40) + (19.6^2/40)) ā -3.49
Using technology (such as a calculator or software), we can find the p-value associated with this test statistic. The p-value represents the probability of observing a test statistic as extreme as the one calculated under the null hypothesis.
Using the t-distribution with 78 degrees of freedom (n1 + n2 - 2), we find that the p-value is approximately 0.0013.
Therefore, the p-value is 0.0013.
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