Question

Let the following sample of 8 observations be drawn from a normal population with unknown mean and standard deviation: 21, 20, 25, 18, 28, 19, 13, 22. [You may find it useful to reference the t table.]

a. Calculate the sample mean and the sample standard deviation. (Round intermediate calculations to at least 4 decimal places. Round "Sample mean" to 3 decimal places and "Sample standard deviation" to 2 decimal places.)

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

c. Construct the 99% confidence interval for the population mean. (Round "t" value to 3 decimal places and final answers to 2 decimal places.)

d. What happens to the margin of error as the confidence level increases from 90% to 99%? As the confidence level increases, the margin of error becomes larger. As the confidence level increases, the margin of error becomes smaller.

Answer 1

Answer:

n= 8

a)

sample standard deviation (s) = 4.53

sample mean ($\bar x$) = 20.75

b)

c= 90%

formula for confidence interval is

1 tc*

where tc is the t critical value for c= 90% with df= n-1 =7

we get

tc= 1.895

$20.75 \pm 1.895 * \frac{4.53}{\sqrt{8}}$

20.75−3.034 <  $\mu$ < 20.75+3.034

confidence interval is = (17.72 , 23.78)

c)

c= 99%

formula for confidence interval is

1 tc*

where tc is the t critical value for c= 90% with df= n-1 =7

we get

tc= 3.499

$20.75 \pm 3.499 * \frac{4.53}{\sqrt{8}}$

20.75−5.605 <  $\mu$ < 20.75+5.605

confidence interval is = ( 15.15 , 26.36 )

d)

As the confidence level increases, the margin of error becomes larger