A researcher is comparing the reaction times of 10-year-olds who play video game

Question

A researcher is comparing the reaction times of 10-year-olds who play video games and 10-year-olds who do not. Random samples from each group provided the results shown below.

Sample 1:

n=15 , mean=0.50 , standard deviation=0.01

Sample 2:

n=10 , mean=0.52 , standard deviation=0.02

* Assume Normally distributed with the same standard deviation

a. Construct a 98% confidence interval for the difference between the population means of reaction times. Note: Express your answer to 5 decimal places of accuracy

b. Using a 2.5% significance level can the researcher conclude that the mean reaction time of players of video games is less than the mean reaction time of non-players of video games? formulate and test the approprate hypothesis. use critical value approach

Explanation / Answer

PART A.
TRADITIONAL METHOD
given that,
mean(x)=0.5
standard deviation , s.d1=0.01
number(n1)=15
y(mean)=0.52
standard deviation, s.d2 =0.02
number(n2)=10
I.
stanadard error = sqrt(s.d1^2/n1)+(s.d2^2/n2)
where,
sd1, sd2 = standard deviation of both
n1, n2 = sample size
stanadard error = sqrt((0/15)+(0/10))
= 0.01
II.
margin of error = t a/2 * (stanadard error)
where,
t a/2 = t -table value
level of significance, =
from standard normal table, two tailedand
value of |t | with min (n1-1, n2-1) i.e 9 d.f is 2.82
margin of error = 2.821 * 0.01
= 0.02
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (0.5-0.52) ± 0.02 ]
= [-0.04 , 0]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
mean(x)=0.5
standard deviation , s.d1=0.01
sample size, n1=15
y(mean)=0.52
standard deviation, s.d2 =0.02
sample size,n2 =10
CI = x1 - x2 ± t a/2 * Sqrt ( sd1 ^2 / n1 + sd2 ^2 /n2 )
where,
x1,x2 = mean of populations
sd1,sd2 = standard deviations
n1,n2 = size of both
a = 1 - (confidence Level/100)
ta/2 = t-table value
CI = confidence interval
CI = [( 0.5-0.52) ± t a/2 * sqrt((0/15)+(0/10)]
= [ (-0.02) ± t a/2 * 0.01]
= [-0.04 , 0]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 98% sure that the interval [-0.04 , 0] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 98% of these intervals will contains the true population proportion

PART B.
Given that,
mean(x)=0.5
standard deviation , s.d1=0.01
number(n1)=15
y(mean)=0.52
standard deviation, s.d2 =0.02
number(n2)=10
null, Ho: u1 = u2
alternate, H1: u1 < u2
level of significance, = 0.025
from standard normal table,left tailed t /2 =2.262
since our test is left-tailed
reject Ho, if to < -2.262
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =0.5-0.52/sqrt((0.0001/15)+(0.0004/10))
to =-2.9277
| to | =2.9277
critical value
the value of |t | with min (n1-1, n2-1) i.e 9 d.f is 2.262
we got |to| = 2.9277 & | t | = 2.262
make decision
hence value of | to | > | t | and here we reject Ho
p-value:left tail - Ha : ( p < -2.9277 ) = 0.00841
hence value of p0.025 > 0.00841,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 < u2
test statistic: -2.9277
critical value: -2.262
decision: reject Ho
p-value: 0.00841

the mean reaction time of players of video games is less than the mean reaction time

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