A community aasciation is interested to estimating the average number of matemit
ID: 3149972 • Letter: A
Question
A community aasciation is interested to estimating the average number of matemity days woman stay in the local hospital. A random sample is taken of 36 women who had babies in the hospital during the past year. The following numbers of maternity days each woman was in the hospital are rounded to the nearest day. Use these data and a population standard deviation of 1.13 to construct a 98% confidence interval to estimate the average maternity stay in the hospital for all woman who have babies in this hospital. (Round the intermediate values to 3 decimal places. Round your answers to 3 decimal places.) muExplanation / Answer
Note that
Lower Bound = X - z(alpha/2) * s / sqrt(n)
Upper Bound = X + z(alpha/2) * s / sqrt(n)
where
alpha/2 = (1 - confidence level)/2 = 0.01
X = sample mean = 3.333333333
z(alpha/2) = critical z for the confidence interval = 2.326347874
s = sample standard deviation = 1.13
n = sample size = 36
Thus,
Lower bound = 2.895204484
Upper bound = 3.771462183
Thus, the confidence interval is
( 2.895204484 , 3.771462183 ) [ANSWER]
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If you use tables instead of technology, we round z = 2.33.
Note that
Lower Bound = X - z(alpha/2) * s / sqrt(n)
Upper Bound = X + z(alpha/2) * s / sqrt(n)
where
alpha/2 = (1 - confidence level)/2 = 0.01
X = sample mean = 3.333333333
z(alpha/2) = critical z for the confidence interval = 2.33
s = sample standard deviation = 1.13
n = sample size = 36
Thus,
Lower bound = 2.894516667
Upper bound = 3.77215
Thus, the confidence interval is
( 2.894516667 , 3.77215 ) [ALTERNATE ANSWER FOR TABLE USE]
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