Question

In the population of students who own video games prequarantine, they played on average 2 hours a night with a standard deviation of 3.

After the quarantine, a random sample of 6 students in our class claimed to play: 4, 8, 7, 0, 0, 8 hours a day. Are they significantly different than the population of students who own video games pre-quarantine?

what is the:
UCB:
LCB:
Accept or Reject the null hypotheses:

Answer 1

a.
TRADITIONAL METHOD
given that,
standard deviation, σ =3
sample mean, x =4.5
population size (n)=6
I.
standard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
stanadard error = ( 3/ sqrt ( 6) )
= 1.22
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.05
from standard normal table, two tailed z α/2 =1.96
since our test is two-tailed
value of z table is 1.96
margin of error = 1.96 * 1.22
= 2.4
III.
CI = x ± margin of error
confidence interval = [ 4.5 ± 2.4 ]
= [ 2.1,6.9 ]
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DIRECT METHOD
given that,
standard deviation, σ =3
sample mean, x =4.5
population size (n)=6
level of significance, α = 0.05
from standard normal table, two tailed z α/2 =1.96
since our test is two-tailed
value of z table is 1.96
we use CI = x ± Z a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
Za/2 = Z-table value
CI = confidence interval
confidence interval = [ 4.5 ± Z a/2 ( 3/ Sqrt ( 6) ) ]
= [ 4.5 - 1.96 * (1.22) , 4.5 + 1.96 * (1.22) ]
= [ 2.1,6.9 ]
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interpretations:
1. we are 95% sure that the interval [2.1 , 6.9 ] contains the true population mean
2. if a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population mean

b.
Given that,
population mean(u)=2
standard deviation, σ =3
sample mean, x =4.5
number (n)=6
null, Ho: μ=2
alternate, H1: μ!=2
level of significance, α = 0.05
from standard normal table, two tailed z α/2 =1.96
since our test is two-tailed
reject Ho, if zo < -1.96 OR if zo > 1.96
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 4.5-2/(3/sqrt(6)
zo = 2.041
| zo | = 2.041
critical value
the value of |z α| at los 5% is 1.96
we got |zo| =2.041 & | z α | = 1.96
make decision
hence value of | zo | > | z α| and here we reject Ho
p-value : two tailed ( double the one tail ) - ha : ( p != 2.041 ) = 0.041
hence value of p0.05 > 0.041, here we reject Ho
ANSWERS
---------------
null, Ho: μ=2
alternate, H1: μ!=2
test statistic: 2.041
critical value: -1.96 , 1.96
decision: reject Ho
p-value: 0.041
we have enough evidence to support the claim that significantly different than the population of students who own video games pre-quarantine.