Question

Construct 90%, 95%, and 99% confidence intervals to estimate μ from the following data. State the point estimate. Assume the data come from a normally distributed population. 13.3 11.6 11.9 12.2 12.5 11.4 12.0 11.7 11.8 13.3

(Round the intermediate values to 4 decimal places. Round your answers to 2 decimal places.)

90% confidence interval: enter the lower limit of the 90% confidence interval  ≤ μ ≤ enter the upper limit of the 90% confidence interval
95% confidence interval: enter the lower limit of the 95% confidence interval  ≤ μ ≤ enter the upper limit of the 95% confidence interval
99% confidence interval: enter the lower limit of the 99% confidence interval  ≤ μ ≤ enter the upper limit of the 99% confidence interval

The point estimate is ..............

Answer 1

First of all we need to calculate mean and standard deviation of given data as following type

sample mean will be point estimate of population mean.

point estimate is 12.17

95% confidence interval will be

M = 12.17
t = 2.26
sM = √(0.672/10) = 0.21

μ = M ± t(sM)
μ = 12.17 ± 2.26*0.21
μ = 12.17 ± 0.4793

[11.6907, 12.6493]

90% confidence interval will be

M = 12.17
t = 1.83
sM = √(0.672/10) = 0.21

μ = M ± t(sM)
μ = 12.17 ± 1.83*0.21
μ = 12.17 ± 0.3884

[11.7816, 12.5584]

99% confidence will be

M = 12.17
t = 3.25
sM = √(0.672/10) = 0.21

μ = M ± t(sM)
μ = 12.17 ± 3.25*0.21
μ = 12.17 ± 0.6886

[11.4814, 12.8586]

point estimate is 12.17

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