3. Netflix Addiction. Researchers are concerned about the impact of Netflix on s

Question

3. Netflix Addiction. Researchers are concerned about the impact of Netflix on student's focus. lu particular, they believe ihai io0 llcli Nei llix (zul (YIaiively unpaci a sindeti s a" alcuni performance. First, the researchers need to find out, on average, how many hours a week students spend watching or playing a Netflix video. A random sample of 300 students resulted in a sample mean of 10.7 hours watched per week. They know from previous studies that the population standard deviation of this variable is 4.2 hours

Explanation / Answer

TRADITIONAL METHOD
given that,
standard deviation, =4.2
sample mean, x =10.7
population size (n)=300
I.
stanadard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
stanadard error = ( 4.2/ sqrt ( 300) )
= 0.242
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.38
from standard normal table, two tailed z /2 =0.878
since our test is two-tailed
value of z table is 0.878
margin of error = 0.878 * 0.242
= 0.213
III.
CI = x ± margin of error
confidence interval = [ 10.7 ± 0.213 ]
= [ 10.487,10.913 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
standard deviation, =4.2
sample mean, x =10.7
population size (n)=300
level of significance, = 0.38
from standard normal table, two tailed z /2 =0.878
since our test is two-tailed
value of z table is 0.878
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 = [ 10.7 ± Z a/2 ( 4.2/ Sqrt ( 300) ) ]
= [ 10.7 - 0.878 * (0.242) , 10.7 + 0.878 * (0.242) ]
= [ 10.487,10.913 ]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 62% sure that the interval [10.487 , 10.913 ] contains the true population mean
2. if a large number of samples are collected, and a confidence interval is created
for each sample, 62% of these intervals will contains the true population mean
[ANSWERS]
best point of estimate = mean = 10.7
standard error =0.242
z table value = 0.878
margin of error = 0.213
confidence interval = [ 10.487 , 10.913 ] ~  [ 10.49 , 10.91 ]
no, we don't relay on result if not normally distributed

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