Answered

Discover a wealth of knowledge at Westonci.ca, where experts provide answers to your most pressing questions. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform.

To determine the organic material in a dried lake ted, the percent carbon by mass is measured at two different locations. In order to compare the means of the two different locations, it must first be determined whether the standard deviations of the two locations are different.
Location1(%C) Location2(%C)
1 10.40 10.10
2 10.20 10.90
3 10.30 10.20
4 10.40 10.70
5 10.20 10.40
a-1. Far each location, calculate the standard deviation and report it.
- location 1: _____
- location 2: _____
a-2. What is the calculated F value for comparing the standard deviation?
b. Are the two standard deviations for the locations significantly different at the 95% confidence level?
A. yes.
B. No.
c. What is the calculated t value used to compare the means of these two locations?
d. Are the means for the two different locations significantly different at the 95% confidence level?
A. yes.
B. No.

Sagot :

akiran007

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.

Step-by-step 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.

Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. We're glad you chose Westonci.ca. Revisit us for updated answers from our knowledgeable team.