Question

Class data (hours on electronics) are below.. please use data fto answer question...

1. A national survey conducted by Common Sense Media in early 2017 found that the average amount of time in front of various electronic devices was 6 hours and 40 minutes. I believe that this actual number of hours is less than this. Using our class data, at α=.05 is there enough evidence to support my claim?

Hours on electronics 6,6.5,4,1,8,8,0.5,4,4,9,4,3,3,2,4,4,3,3,2,2,3,4,1,6,3,5,8,3,2,4,6,4,4,5,6,4,1,1,13,0.75,10,2,9,6,9,6,5,2,3.5,8,4,3,4,2,1.5,6,4,2,3,6,10,8,4,2,4,3,4,7,6,8,4.5,5,2,7,12,3,4.5,11,8,6,5,8,7,8,2,6,2,7,5,4,2,2,1,3,4,4,2,2,1,6,4,6,7,9,5,2,7,11,10,3,10,2.5,4.5,5,5,4,8,9,2,14,12,6,18,6,14,12,8,10,15,16,16,15,16,10,15,17,8,17,17,5,3,2,3,4,5,3,4,2,4,6,6,4,9,5,10,5,9,2,8,12,5,1,5,6,8,5,10,8,10,6,5,10,2,3,10,2,3.5,2.5,2,5,4,0,4,2,2,6,7,1,3.

Answer 1

1.

Given that,
population mean(u)=6.66 hours
sample mean, x =5.744
standard deviation, s =3.835
number (n)=189
null, Ho: μ=6.66
alternate, H1: μ<6.66
level of significance, α = 0.05
from standard normal table,left tailed t α/2 =1.653
since our test is left-tailed
reject Ho, if to < -1.653
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =5.744-6.66/(3.835/sqrt(189))
to =-3.2837
| to | =3.2837
critical value
the value of |t α| with n-1 = 188 d.f is 1.653
we got |to| =3.2837 & | t α | =1.653
make decision
hence value of | to | > | t α| and here we reject Ho
p-value :left tail - Ha : ( p < -3.2837 ) = 0.00061
hence value of p0.05 > 0.00061,here we reject Ho
ANSWERS
---------------
null, Ho: μ=6.66
alternate, H1: μ<6.66
test statistic: -3.2837
critical value: -1.653
decision: reject Ho
p-value: 0.00061
we have enough evidence to support the claim that I believe that this actual number of hours is less than this mean.