Question

Many people now turn to the internet to get information on health-related topics. A ressarchatice used Flesch reading case scores a measure of reading difficulty based on factors such as sentence length and number of sylables in the words used to score pages on Wikipedia and on WebMD. Higher resch scores correspond to more difficult reading levels. The paper reported that for a representative sample of health related pages on Wikipedia, the mean schscore was 26.4 and the standard deviation of the Flesch scores was 14. For a representative sample of pages from Wet, the mean score was 13.1 and the standard devion was 19.5 Suppose that these means and standard deviations were based on samples of 40 pages from each te (a) Construct a confidence interval estimate of the difference in mean Fisch reading ease score for health related pages on Wikipedia and health related pages on WMD. (red. Use technology to calculate the critical value Round your answers to three decimal places) () What does this confidence interval imply about the readability of health related information from these two sources This werval indicates that, on avera, Wipedia pages have a more difficult reading level This interval indicates that, on average Wikipedia pages and WMD pages have a difficult reading levels This interval indicates that, on were Web pages have a more reading level Is this consistent with the conclusion of apothesis test that there is convincing evidence that the mean reading level for health related pages differs for Wikipedia and WebMD? The best it for this test was This is consistent with the conclusion to content with the test statistic This is not consistent with the conclusion and consistent with the test This is consistent with the conclusion and content with the test This is not condent with the content with the rest

Answer 1

a.
TRADITIONAL METHOD
given that,
mean(x)=26.4
standard deviation , s.d1=14.9
number(n1)=40
y(mean)=43.1
standard deviation, s.d2 =19.6
number(n2)=40
I.
standard error = sqrt(s.d1^2/n1)+(s.d2^2/n2)
where,
sd1, sd2 = standard deviation of both
n1, n2 = sample size
standard error = sqrt((222.01/40)+(384.16/40))
= 3.893
II.
margin of error = t a/2 * (standard error)
where,
t a/2 = t -table value
level of significance, α = 0.1
from standard normal table, two tailed and
value of |t α| with min (n1-1, n2-1) i.e 39 d.f is 1.685
margin of error = 1.685 * 3.893
= 6.559
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (26.4-43.1) ± 6.559 ]
= [-23.259 , -10.141]
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DIRECT METHOD
given that,
mean(x)=26.4
standard deviation , s.d1=14.9
sample size, n1=40
y(mean)=43.1
standard deviation, s.d2 =19.6
sample size,n2 =40
CI = x1 - x2 ± t a/2 * Sqrt ( sd1 ^2 / n1 + sd2 ^2 /n2 )
where,
x1,x2 = mean of populations
sd1,sd2 = standard deviations
n1,n2 = size of both
a = 1 - (confidence Level/100)
ta/2 = t-table value
CI = confidence interval
CI = [( 26.4-43.1) ± t a/2 * sqrt((222.01/40)+(384.16/40)]
= [ (-16.7) ± t a/2 * 3.893]
= [-23.259 , -10.141]
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interpretations:
1. we are 90% sure that the interval [-23.259 , -10.141] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 90% of these intervals will contains the true population proportion
b.
Given that,
mean(x)=26.4
standard deviation , s.d1=14.9
number(n1)=40
y(mean)=43.1
standard deviation, s.d2 =19.6
number(n2)=40
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, α = 0.1
from standard normal table, two tailed t α/2 =1.685
since our test is two-tailed
reject Ho, if to < -1.685 OR if to > 1.685
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =26.4-43.1/sqrt((222.01/40)+(384.16/40))
to =-4.29
| to | =4.29
critical value
the value of |t α| with min (n1-1, n2-1) i.e 39 d.f is 1.685
we got |to| = 4.28992 & | t α | = 1.685
make decision
hence value of | to | > | t α| and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != -4.2899 ) = 0
hence value of p0.1 > 0,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: -4.29
critical value: -1.685 , 1.685
decision: reject Ho
p-value: 0
we have enough evidence to support the claim that difference of means between wikipedia pages and webmd pages.
yes,
it is consistent and t test is used.