Answer:
a-1.- location 1: =0.01- location 2: =0.113
a-2. F= 0.08849
b. A. yes.
c. t= -1.020
d. B. No.
Explanation:
Location1(%C) Location2(%C)
x x² y y²
1 10.40 108.16 10.10 102.01
2 10.20 104.04 10.90 118.81
3 10.30 106.09 10.20 104.04
4 10.40 108.16 10.70 114.49
5 10.20 104.04 10.40 108.16
∑ 51.5 530.49 52.3 547.51
x`= ∑x/n1 = 51.5/5= 10.3; y`= ∑y/n2= 52.3/5= 10.46
Sx² = 1/n1-1[ ∑x² - (∑x)²/n1] = 1/4[ 530.49 - (51.5)²/5]
= 1/4[ 530.49 - 530.45] = 0.01
Sy² = 1/n2-1[ ∑y² - (∑y)²/n2] = 1/4[ 547.51 - (52.3)²/5]
= 1/4[ 547.51 - 547.058] = 0.113
F= Sx²/Sy²= 0.01/0.113= 0.08849
The critical region is F ≥ F(0.025) (5,5)= 7.15
and F ≤ 1/F(0.025) (5,5) = 1/7.15= 0.13986
Since the calculated value of F =0.08849 falls in the critical region we conclude that the two standard deviations for the locations significantly different at the 95% confidence level.
c. t= x1`-x2`/ sqrt ( s₁²/n₁ + s₂²/n₂)
t= 10.3-10.46/ sqrt( 0.01/5 +0.113/5)
t=- 0.16/0.15684
t= -1.020
d. The calculated value t= -1.020 does not fall in the the critical region
t (0.025) (4) = ±2.776 we conclude that the means for the two different locations are not significantly different at the 95% confidence level.