Question

The following data show the number of hours per day 12 adults spent in front of screens watching​ television-related content.

1.7, 4.6, 4.1, 4.7, 7.1, 6.6, 5.9, 2.6, 5.3, 1.5, 2.3, 8.6

Construct a 95​% confidence interval to estimate the average number of hours per day adults spend in front of screens watching​ television-related content.

Estimate the average number of hours per day adults spend in front of screens watching​ television-related content from min to max hours

Answer 1

Solution:

x	x2
1.7	2.89
4.6	21.16
4.1	16.81
4.7	22.09
7.1	50.41
6.6	43.56
5.9	34.81
2.6	6.76
5.3	28.09
1.5	2.25
2.3	5.29
8.6	73.96
x = 55	x2 = 308.08

The sample mean is

Mean   = (x / n) )

= (1.7+ 4.6+ 4.1+ 4.7+ 7.1+ 6.6+ 5.9+ 2.6+ 5.3+ 1.5+ 2.3+ 8.6 / 12 )

= 55 / 12

= 4.5833

Mean   = 4.58

The sample standard is S

  S = ( x² ) - (( x)² / n ) n -1

= (308.08( (55 )² / 12 ) 11

   = ( 308.08 - 252.0833 / 11)

= (55.9967 / 11)

= 5.0906

= 2.0962

The sample standard is 2.10

Degrees of freedom = df = n - 1 = 12 - 1 = 11

At 95% confidence level the t is ,

  = 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t _/2,df = t_0.025,11 = 2.201

Margin of error = E = t_/2,df * (s /n)

= 2.201 * (2.10 / 12)

= 1.33

Margin of error =1.33

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

- E < < + E

4.58 - 1.33 < < 4.58 + 1.33

3.24 < < 5.91

min = 3.24  hours

max = 5.91  hours