In: Math
Assume that a sample is used to estimate a population mean μ . Find the margin of error M.E. that corresponds to a sample of size 18 with a mean of 47.4 and a standard deviation of 16.9 at a confidence level of 90%. Report ME accurate to one decimal place because the sample statistics are presented with this accuracy. M.E. _________
he effectiveness of a blood-pressure drug is being investigated.
An experimenter finds that, on average, the reduction in systolic
blood pressure is 32.7 for a sample of size 288 and standard
deviation 11.5.
Estimate how much the drug will lower a typical patient's systolic
blood pressure (using a 80% confidence level).
Enter your answer as a tri-linear inequality accurate to one
decimal place (because the sample statistics are reported accurate
to one decimal place).
______< μ< ______
SAT scores are distributed with a mean of 1,500 and a standard deviation of 300. You are interested in estimating the average SAT score of first year students at your college. If you would like to limit the margin of error of your 95% confidence interval to 25 points, how many students should you sample?
Make sure to give a whole number answer.
(1)
n = Sample Size = 18
= Sample Mean = 47.4
s = Sample SD = 16.9
SE = s/
= 16.9/
= 3.9834
= 0.10
ndf = n - 1 = 18 - 1 = 17
From Table, critical values of t = 1.7396
So,
Margin of Error = ME = t X SE
= 1.7396 X 3.9834
= 6.9
So,
Answer is:
6.9
(2)
n = Sample Size = 288
= Sample Mean = 32.7
s = Sample SD = 11.5
SE = s/
= 11.5/
= 0.6776
= 0.20
ndf = n - 1 = 288 - 1 = 287
From Table, critical values of t = 1.2845
So,
Margin of Error = ME = t X SE
= 1.2845 X 0.6776
= 0.8704
So,
Confidence Interval:
32.7 0.8704
= ( 31.8 ,33.6)
Answer is:
31.8 < < 33.6
(3)
Number of students (n) is given by:
Given:
= 0.05
From Table, critical values of Z = 1.96
= 300
e = 25
Substituting, we get:
So,
Answer is:
554