Is it a good idea to listen to music when studying for a big

Is it a good idea to listen to music when studying for a big test? In a study conducted by some Statistics students, 62 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. Here are the summary statistics:

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.

b) Create a 90% confidence interval for the mean difference in memory score between students who study to Mozart and those who listen to no music at all. Interpret your interval.

Solution

a.
Set Up Hypothesis
Null, There Is No-Significance between them Ho: u1 = u2
Alternate, better to study while listening to Mozart than to rap music - H1: u1 < u2
Test Statistic
X(Mean)=10.72
Standard Deviation(s.d1)=3.99 ; Number(n1)=29
Y(Mean)=12.77
Standard Deviation(s.d2)=4.73; Number(n2)=13
we use Test Statistic (t) = (X-Y)/Sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =10.72-12.77/Sqrt((15.9201/29)+(22.3729/13))
to =-1.36
| to | =1.36
Critical Value
The Value of |t | with Min (n1-1, n2-1) i.e 12 d.f is 1.782
We got |to| = 1.36064 & | t | = 1.782
Make Decision
Hence Value of |to | < | t | and Here we Do not Reject Ho
P-Value:Left Tail - Ha : ( P < -1.3606 ) = 0.09932
Hence Value of P0.05 < 0.09932,Here We Do not Reject Ho

We don\'t have evidence at 0.05 LOS, to indicate better to study while listening to Mozart than to rap music

b)
CI = x1 - x2 ± t a/2 * Sqrt ( sd1 ^2 / n1 + sd2 ^2 /n2 )
Where,
x1 = Mean of Sample 1, x2 = Mean of sample2
sd1 = SD of Sample 1, sd2 = SD of sample2
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
Mean(x1)=10
Standard deviation( sd1 )=3.19
Sample Size(n1)=10
Mean(x2)=12.77
Standard deviation( sd2 )=4.73
Sample Size(n1)=13
CI = [ ( 10-12.77) ±t a/2 * Sqrt( 10.1761/10+22.3729/13)]
= [ (-2.77) ± t a/2 * Sqrt( 2.74) ]
= [ (-2.77) ± 1.833 * Sqrt( 2.74) ]
= [-5.8 , 0.26]