
Answer:
A)
Note that       Â
      Â
Lower Bound = X - z(alpha/2) * s / sqrt(n) Â Â Â Â Â Â Â
Upper Bound = X + z(alpha/2) * s / sqrt(n) Â Â Â Â Â Â Â
      Â
where       Â
alpha/2 = (1 - confidence level)/2 = Â Â 0.025 Â Â Â Â Â
X = sample mean = Â Â 1723.4 Â Â Â Â Â
z(alpha/2) = critical z for the confidence interval = Â Â 1.959963985 Â Â Â Â Â
s = sample standard deviation = Â Â 89.55083319 Â Â Â Â Â
n = sample size = Â Â 30 Â Â Â Â Â
      Â
Thus, Â Â Â Â Â Â Â
      Â
Lower bound = Â Â 1691.355235 Â Â Â Â Â
Upper bound = Â Â 1755.444765 Â Â Â Â Â
      Â
Thus, the confidence interval is       Â
      Â
( Â 1691.355235 Â , Â 1755.444765 Â ) [ANSWER]
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b)
We assumed taht the distirbution of these observations is approximately normally distributed.
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c)
Yes, because the values are not far away from each other.