No Music Rap Mozart In a study conducted by some Statistics students, 65 people

Question

No Music Rap Mozart In a study conducted by some Statistics students, 65 people were randomly assigned to listen to rap music, music by Mozart, or no music while attempting to memorize objects pictured on a page. They were then asked to list all the objects they could remember. The summary statistics for each group are shown in the table. Complete parts a and b 29 18 18 oun ean10.01 SD 3.79 3.39 4.28 8.29 14.41 a) Does it appear that it is better to study while listening to Mozart than to rap music? Test an appropriate hypothesis and state your conclusion. Let group M correspond to Mozart listeners and group R correspond to rap listeners. Write the null and alternative hypotheses. Choose the correct answer below. Test the hypothesis. t(Round to two decimal places as needed.) P(Round to four decimal places as needed.) State your conclusion. Use -0.05 | HO. There | evidence that the mean number of objects remembered by those who listen to Mozart is higher than the mean number of objects remembered by those who listen to rap music.

Explanation / Answer

a.
Given that,
mean(x)=8.29
standard deviation , s.d1=3.39
number(n1)=18
y(mean)=10.01
standard deviation, s.d2 =3.79
number(n2)=29
null, Ho: u1 = u2
alternate, H1: u1 > u2
level of significance, = 0.05
from standard normal table,right tailed t /2 =1.74
since our test is right-tailed
reject Ho, if to > 1.74
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =8.29-10.01/sqrt((11.4921/18)+(14.3641/29))
to =-1.62
| to | =1.62
critical value
the value of |t | with min (n1-1, n2-1) i.e 17 d.f is 1.74
we got |to| = 1.61535 & | t | = 1.74
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value:right tail - Ha : ( p > -1.6154 ) = 0.93768
hence value of p0.05 < 0.93768,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 > u2
test statistic: -1.62
critical value: 1.74
decision: do not reject Ho
p-value: 0.93768
we do not have enough evidence to support the claim that while listening mozart is more than rap music

b.
TRADITIONAL METHOD
given that,
mean(x)=8.29
standard deviation , s.d1=3.39
number(n1)=18
y(mean)=14.41
standard deviation, s.d2 =4.28
number(n2)=18
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((11.492/18)+(18.318/18))
= 1.287
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 17 d.f is 1.74
margin of error = 1.74 * 1.287
= 2.239
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (8.29-14.41) ± 2.239 ]
= [-8.359 , -3.881]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
mean(x)=8.29
standard deviation , s.d1=3.39
sample size, n1=18
y(mean)=14.41
standard deviation, s.d2 =4.28
sample size,n2 =18
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 = [( 8.29-14.41) ± t a/2 * sqrt((11.492/18)+(18.318/18)]
= [ (-6.12) ± t a/2 * 1.287]
= [-8.359 , -3.881]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 90% sure that the interval [-8.359 , -3.881] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 90% of these intervals will contains the true population proportion

option :C

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