The manufacturer of an MP3 player wanted to know whether a 1

The manufacturer of an MP3 player wanted to know whether a 10 percent reduction in price is enough to increase the sales of its product. To investigate, the owner randomly selected eight outlets and sold the MP3 player at the reduced price. At seven randomly selected outlets, the MP3 player was sold at the regular price. Reported below is the number of units sold last month at the sampled outlets.

  Regular price

138

123

88

118

144

125

96

  Reduced price

122

135

152

133

114

104

116

116

At the .050 significance level, can the manufacturer conclude that the price reduction resulted in an increase in sales? Hint: For the calculations, assume the \"Reduced price\" as the first sample.

Solution

Set Up Hypothesis
Null, There Is NoSignificance between them Ho: u1 = u2
Alternative, 10 percent reduction in price is enough to increase the sales of its product, H1: u1 < u2
Test Statistic
X (Mean)=118.857; Standard Deviation (s.d1)=20.53
Number(n1)=7
Y(Mean)=124; Standard Deviation(s.d2)=15.203
Number(n2)=8
Value Pooled variance S^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
S^2 = (6*421.4809 + 7*231.1312) / (15- 2 )
S^2 = 318.9849
we use Test Statistic (t) = (X-Y)/Sqrt(S^2(1/n1+1/n2))
to=118.857-124/Sqrt((318.9849( 1 /7+ 1/8 ))
to=-5.143/9.2435
to=-0.5564
| to | =0.5564
Critical Value
The Value of |t | with (n1+n2-2) i.e 13 d.f is 1.771
We got |to| = 0.5564 & | t | = 1.771
Make Decision
Hence Value of |to | < | t | and Here we Do not Reject Ho
P-Value: Left Tail - Ha : ( P < -0.5564 ) = 0.2937
Hence Value of P0.05 < 0.2937,Here We Do not Reject Ho

ANS:
Pooled variance = 318.98
to=-0.56
Do not Reject Ho