Homework Soursehe sample data below have been collected based on a simple r
he sample data below have been collected based on a simple random sample from a normally distributed population. Complete parts a and b
5 6 0 9 3 2 9 3 8 6
a)Compute a 95 % confidence interval estimate for the population mean.
B) Show what the impact would be if the confidence level is increased to 98 %. Discuss why this occurs. Select the correct choice below and fill in the answer boxes to complete your choice.
Solution
a)
Note that
Lower Bound = X - t(alpha/2) * s / sqrt(n)
Upper Bound = X + t(alpha/2) * s / sqrt(n)
where
alpha/2 = (1 - confidence level)/2 = 0.025
X = sample mean = 5.1
t(alpha/2) = critical t for the confidence interval = 2.262157163
s = sample standard deviation = 3.0713732
n = sample size = 10
df = n - 1 = 9
Thus,
Lower bound = 2.902871971
Upper bound = 7.297128029
Thus, the confidence interval is
( 2.902871971 , 7.297128029 ) [ANSWER]
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b)
Note that
Lower Bound = X - t(alpha/2) * s / sqrt(n)
Upper Bound = X + t(alpha/2) * s / sqrt(n)
where
alpha/2 = (1 - confidence level)/2 = 0.01
X = sample mean = 5.1
t(alpha/2) = critical t for the confidence interval = 2.821437925
s = sample standard deviation = 3.0713732
n = sample size = 10
df = n - 1 = 9
Thus,
Lower bound = 2.359668581
Upper bound = 7.840331419
Thus, the confidence interval is
( 2.359668581 , 7.840331419 )
As we can see, the interval became wider, and the margin of error became larger.
This is so because the critical t value becomes larger with larger confidence level.
This makes sense because you need to enclose more values to be \"more confident\" that you have the true mean.
