On a plant reserach farm 2 fields of watermelons were grown

On a plant reserach farm 2 fields of watermelons were grown, w/ newer formula fertilizer in Field 1. In Field 1 14 watermelons were randomly sampled weight. At the same time 12 watermelons were randomly selected on field 2. You want to show that watermelons are significantly larger on the average in field 1 compared w/ field 2. Asume as an approximation that the weight of watermelons follows a normal distribution & the standard Dev is the same for the population in both fields.(use alpha=.05) (there is an obvious approximation in the the weight cannot be negative so the real weight distibution are going to be a little skewed)

Field 1: 23.1, 26.1, 18.3, 25.1, 20, 19, 30.1, 24.8, 25.6, 21.1, 22.2, 21, 21.3, 27.1

Field 2: 20, 17.9, 21.8, 25.1, 24.8, 26.5, 15.1, 20.3, 24.5, 17..5, 18.8, 14.9

A) write the null & alternative hypothesis

B) sketch the distribution diagram & show tail(s) w/ numerical critical valu(s) on it

C) Compute the sample means & standard dev that you will need. Preferably use an automated calculator method.

D) compute the tes value---shoe formula & ubstitue appropriate #\'s

E) what is your decision about the null hypothesis/

Solution

Set Up Hypothesis
Null, Ho: u1 = u2
Alternate, watermelons are significantly larger on the average in field 1 - H1: u1 > u2
Test Statistic
X(Mean)=23.2
Standard Deviation(s.d1)=3.3846 ; Number(n1)=14
Y(Mean)=20.8818
Standard Deviation(s.d2)=4.0274; Number(n2)=11
we use Test Statistic (t) = (X-Y)/Sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =23.2-20.8818/Sqrt((11.45552/14)+(16.21995/11))
to =1.531
| to | =1.531
Critical Value
The Value of |t | with Min (n1-1, n2-1) i.e 10 d.f is 1.812
We got |to| = 1.53098 & | t | = 1.812
Make Decision
Hence Value of |to | < | t | and Here we Do not Reject Ho
P-Value:Right Tail -Ha : ( P > 1.531 ) = 0.07839
Hence Value of P0.05 < 0.07839,Here We Do not Reject Ho

[ANSWERS]
A.
Null: Ho: u1 = u2 , Alternate H1: u1 > u2
C.
field 1
X(Mean)=23.2
Standard Deviation(s.d1)=3.3846 ; Number(n1)=14
field 2
Y(Mean)=20.8818
Standard Deviation(s.d2)=4.0274; Number(n2)=11
D.
to =1.531
E.
Do not Reject Ho