Let the following sample of 8 observations be drawn from a n

Let the following sample of 8 observations be drawn from a normal population with unknown mean and standard deviation: 22, 18, 14, 25, 17, 28, 15, 21. Use Table 2.

Calculate the sample mean and the sample standard deviation. (Round intermediate calculations to 4 decimal places, \"sample mean\" to 3 decimal places and \"sample standard deviation\" to 2 decimal places.)

Construct the 80% confidence interval for the population mean. (Round \"t\" value to 3 decimal places, and final answers to 2 decimal places.)

Construct the 90% confidence interval for the population mean. (Round \"t\" value to 3 decimal places, and final answers to 2 decimal places.)

What happens to the margin of error as the confidence level increases from 80% to 90%?

Solution

a)
Mean(x)=20
Standard deviation( sd )=4.899
b)
Confidence Interval
CI = x ± t a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval

Sample Size(n)=8
Confidence Interval = [ 20 ± t a/2 ( 4.899/ Sqrt ( 8) ) ]
= [ 20 - 1.415 * (1.732) , 20 + 1.415 * (1.732) ]
= [ 17.549,22.451 ]
c)
Confidence Interval = [ 20 ± t a/2 ( 4.899/ Sqrt ( 8) ) ]
= [ 20 - 1.895 * (1.732) , 20 + 1.895 * (1.732) ]
= [ 16.718,23.282 ]

d)
As the confidence level increases, the interval becomes wider.