Solution :
1. Let the sample proportion given be 0.5 since p is not given. The CI is 95% thus, we use the normal disribution with z value as 1.96
Given that the margin of error as 3% i.e 0.03
Now the estimated sample size be,
n= {Z/2
/( margin of error )} p*(1-p)
= (1.96 / 0.03)2 * 0.5 *(1-0.5)
=1067.111
Thus the estimated sample size will be 1068
Thus option d) is correct
2) Thus sample size is 20 ,we need to use exact test
Here they initially counted the headaches before treatment and then all these 20 adults are treated with chiropractic treatment. Again after the treatment they counted the number of times they experienced headache.
This is before and after case, thus we can see that, data is paired
Thus we have to use paired t test
Thus option a) is correct
3) To test whether chiropractic treatment is effective or not we need to test,
H0 : there is no difference after treatment i.e. treatment do not lower freq of headache
vs H1: there is no difference after treatment i.e. treatment lowers freq of headache
The type one error is reject H0 when it is true
This means, We rejecting the fact that treatment is not effective where in real it is not effective.
i.e we are saying treatment is effective actually its not.
Here b) you decide that chiropractic treatment significantly lowers the frequency of headache and in fact it does not
Thus option b) is correct
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Construct a confidence interval suitable for testing claim that
students taking non proctored tests get higher mean score than
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___<µ1 - µ2 < ____
Yes/No____ because the confidence interval contains only
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E.
Construct a confidence interval suitable for testing claim that
students taking non proctored tests get higher mean score than
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___<µ1 - µ2 < ____
Yes/No____ because the confidence interval contains only...
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