Is it a good idea to listen to music when studying for a big test? In a study co

Question

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.

Explanation / Answer

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]

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