Do rock music CDs and country music CDs give the consumers the same amount of mu
ID: 3217568 • Letter: D
Question
Do rock music CDs and country music CDs give the consumers the same amount of music listening time? A sample of 12 randomly selected single rock musicCDs and a sample of 14 randomly selected single country music CDs have the following total lenghts (in minutes)
Country
music
Assume that the two populations are normally distributed with equal standard deviations.
a) Manualy calculate the value of the test statistic t for testing the null hypothesis that the mean lenghts of the rock and country music single CDs are the same against the alternative hyphotesis that these meanlenghts are not the same. Use the value of this t statisticto compute the ( approximate) p-value.
b) Manualy calculate test statistic F for performing the test of equality of the mean lenghts of the rock and country music single CDs and use it to find the (approximate) p-value.
c) How do the test statistic in parts a and b compare? How do the p-values computed in parts a and b compare? Do you think that this is a coinsidence , or will this always happen?
Rock MusicCountry
music
43 45.3 44.3 40.2 63.8 42.8 32.8 33 54.2 33.5 51.3 37.7 64.8 36.8 36.1 34.6 33.9 33.4 51.7 36.5 36.5 43.3 59.7 31.7 44.0 42.7Explanation / Answer
Given that,
mean(x)=47.675
standard deviation , s.d1=11.5716
number(n1)=12
y(mean)=38.25
standard deviation, s.d2 =4.7061
number(n2)=14
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, = 0.05
from standard normal table, two tailed t /2 =2.201
since our test is two-tailed
reject Ho, if to < -2.201 OR if to > 2.201
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =47.675-38.25/sqrt((133.90193/12)+(22.14738/14))
to =2.64
| to | =2.64
critical value
the value of |t | with min (n1-1, n2-1) i.e 11 d.f is 2.201
we got |to| = 2.64052 & | t | = 2.201
make decision
hence value of | to | > | t | and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 2.6405 ) = 0.023
hence value of p0.05 > 0.023,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: 2.64
critical value: -2.201 , 2.201
decision: reject Ho
p-value: 0.023
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