Question

Solve this confidence interval problem-- Since we are currently conducting a study on millennials and their TV time during the week, I decided to estimate the population mean number of hours that millennials watch more than 5 hours a day and if so, how many hours did they spend watching TV. According to the survey, I pulled 10 individuals randomly and these were the results—

6, 10, 3, 0, 4, 8, 2, 1, 1, 8

Calculate a confidence level of 95%.

Answer 1

Solution:

Given that,

x	x2
6	36
10	100
3	9
0	0
4	16
8	64
2	4
1	1
1	1
8	64
$\sum$ x = 43	$\sum$ x2 = 295

The sample mean is $\bar x$

Mean $\bar x$  =  $\sum$ x / n

= 43 / 10

= 4.3

Mean $\bar x$  = 4.3

The sample standard deviation is S

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

= $\sqrt{}$ ( 295 - ( 43 )² / 10 ) 9

=  $\sqrt{}$ ( 295 - 184 / 9 )

   =  $\sqrt{}$110.1 / 9

   =  $\sqrt{}$12.2333

= 3.4976

The sample standard deviation = 3.5

$\bar x$ = 4.3

s = 3.5

n = 10

Degrees of freedom = df = n - 1 = 10- 1 = 9

At 95% confidence level the z is ,

$\alpha$ = 1 - 95% = 1 - 0.95 = 0.05

$\alpha$ / 2 = 0.05 / 2 = 0.025

t_{$\alpha$ /2,df} = t_0.025,9 =2.262

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

= 2.262 * (3.5 / $\sqrt$10)

= 2.504

The 95% confidence interval estimate of the population mean is,

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

4.3 - 2.504 < $\mu$ < 4.3 + 2.504

1.796 < $\mu$ < 6.804