Question

The random sample shown below was selected from a normal distribution.

3,6,8,3,8,8

Complete parts a and b.

a. Construct a 99% confidence interval for the population mean μ.

(1.971,10.03)

​(Round to two decimal places as​ needed.)

b. Assume that sample mean x overbar x and sample standard deviation s remain exactly the same as those you just calculated but that are based on a sample of n=25 observations. Repeat part a. What is the effect of increasing the sample size on the width of the confidence​ intervals?The confidence interval is (______,_______)​(Round to two decimal places as​ needed.)

Answer 1

Solution:

Given that

x	x2
3	9
6	36
8	64
3	9
8	64
8	64
$\sum$x = 36	$\sum$x2 = 246

The sample mean is $\bar x$

$\bar x$ = $\sum$ x / n

= (3 + 6 + 8+ 3 + 8 + 8 / n )

= 36 /6

= 6

The sample mean is = 6

  The sample standard is S

S = $\sqrt{}$ ( $\sum$ x² ) - (( $\sum$ x )² / n ) / 1 -n )

= $\sqrt{}$ ( ( 246 -( 36 )² / 6 ) 5

=$\sqrt{}$ (246 -216 / 5 )

= $\sqrt{}$ 30 / 5

= $\sqrt{}$ 6

= 2.4

The sample standard is 2.4

a ) n = 6

$\bar x$ = 6

s = 2.4

n = 6

Degrees of freedom = df = n - 1 = 6 - 1 = 5

At 99% confidence level the z is ,

$\alpha$ = 1 - 99% = 1 - 0.99 = 0.01

$\alpha$ / 2 = 0.01 / 2 = 0.005

t_{$\alpha$ /2,df} = t_0.005,5 = 4.031

Margin of error = E = t_{$\alpha$/2,df} * (s /$\sqrt$n)

= 4.031 * (2.4 / $\sqrt$ 6)

= 395

Margin of error = 3.95

The 99% confidence interval estimate of the population mean is,

$\bar x$ - E < $\mu$ < $\bar x$ + E

6 - 3.95 < $\mu$ < 6 + 3.95

2.05 < $\mu$ < 9.95

(2.05,9.95 )

b ) n = 25

$\bar x$ = 6

s = 2.4

n = 25

Degrees of freedom = df = n - 1 = 25- 1 = 24

At 99% confidence level the z is ,

$\alpha$ = 1 - 99% = 1 - 0.99 = 0.01

$\alpha$ / 2 = 0.01 / 2 = 0.005

t_{$\alpha$ /2,df} = t_0.005,24 = 2.797

Margin of error = E = t_{$\alpha$/2,df} * (s /$\sqrt$n)

= 2.797 * (2.4 / $\sqrt$ 25)

= 1.34

Margin of error = 1.34

The 99% confidence interval estimate of the population mean is,

$\bar x$ - E < $\mu$ < $\bar x$ + E

6 - 1.34 < $\mu$ < 6 + 1.34

4..66 < $\mu$ < 7.34

(4.66, 7.34 )

The effect of increasing the sample size on the width of the confidence​ intervals is