15 pts + brainliest to right/best answer
1 answer:
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Fun. I prefer Oxymetazoline.
For the control group we have a headache probability of
c = 368/1671 = .220
For the experimental group we have a headache probability of
e = 494/2013 = .245
The observed difference is
d = e - c = .025
The variance of the difference is
s² = c(1-c)/n₁ + e(1-e)/n₂
so the standard deviation is
We get a t statistic on the difference of
t = d/s = .025/.0139 = 1.79
We're interested in the one sided test, P(d > 0). We have enough dfs to assume normality. We look up in the standard normal table
P(z < 1.79) = .96327
so
p = P(z > 1.79) = 1 - .96327 = 0.037 = 3.7%
Answer: That's less that 10% so we have evidence to conclude that headaches are significantly greater in the experimental group.
Given:
Number is 96.45.
To find:
The expression that has a value of 96.45.
Solution:
In option a,
In option b,
In option c,
In option d,
Therefore, the correct option is a.