A simple random sample of size n is drawn from a population that is normally distributed. The sample mean x bar is found to be
109,
and the sample standard deviation, s, is found to be 10.
a. Construct 95% confidence interval about miu, if the sample size n is 28. Find lower and upper bound.
b. Construct 95% confidence interval about miu, if the sample size n is 17. Find lower and upper bound.
c.Construct 80% confidence interval about miu, if the sample size n is 28. Find lower and upper bound.
d. Could we have computed the confidence intervals in part a to c if the population had not been normally distributed?
Solution:
here, n=sample size.
Part(a):
Here, sample mean = .
and sample standard deviation = s = 10.
We are given, n = 28.
The 95% confidence interval wil be:(109 - 3.7, 109 + 3.7) = (105.3, 112.7) .
Upper Bound = 105.3
Lower Bound = 112.7
Part(b):
Now, are given that n = 17.
The 95% confidence interval wil be:(109 - 4.75, 109 + 4.75) = (104.25, 113.75) .
Upper Bound = 104.25
Lower Bound = 113.75
Part(c):
Now, are given that n = 28.
The 80% confidence interval wil be:(109 - 3.11, 109 + 3.11) = (105.89, 112.11)
Upper Bound = 105.89
Lower Bound = 112.11
Part(d):
If the population was not Normal then we could not compute the confidence interval as the distribution is unknown.
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