13 People were asked how much money on average they spend on

13.) People were asked how much money, on average, they spend on coffee daily. 120 men said they spend an average of $2.35 with a standard deviation of $O.75. 200 women said they spend, on average, $3.21 with a standard deviation of $1.03. At = 0.02, test claim that men spend less than women on coffee. the

Solution

a.
Set Up Hypothesis
Null , Ho: u1 > u2
Alternate, H1: u1 <= u2
Test Statistic
X(Mean)=2.35
Standard Deviation(s.d1)=0.75
Number(n1)=120
Y(Mean)=3.21
Standard Deviation(s.d2)=1.03
Number(n2)=200
we use Test Statistic (Z) = (X-Y)/Sqrt(s.d1^2/n1)+(s.d2^2/n2)
Zo=2.35-3.21/Sqrt((0.5625/120)+(1.0609/200))
Zo =-8.6
| Zo | =8.6
Critical Value
The Value of |Z | at LOS 0.02% is 2.054
We got |Zo | =8.603 & | Z | =2.054
Make Decision
Hence Value of | Zo | > | Z | and Here we Reject Ho
P-Value: Left Tail - Ha : ( P < -8.6 ) = 0
Hence Value of P0.02 > 0,Here we Reject Ho

b.
CI = x1 - x2 ± Z 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)
Za/2 = Z-table value
CI = Confidence Interval
Mean(x1)=2.35
Standard deviation( sd1 )=0.75
Sample Size(n1)=120
Mean(x2)=3.21
Standard deviation( sd2 )=1.03
Sample Size(n12=200
CI = [ ( 2.35-3.21) ±Z a/2 * Sqrt( 0.5625/120+1.0609/200)]
= [ (-0.86) ± Z a/2 * Sqrt( 0.01) ]
= [ (-0.86) ± 2.326 * Sqrt( 0.01) ]
= [-1.0925 , -0.6275]

interval does n\'t fall in Zero, There is a diffrence