Aconsumer protection group is concerned that a ketchup manufacturer is filling its 20-ounce family-size containers with less than 20 ounces of ketchup. the group purchases 10 family-size bottles of this ketchup, weighs the contents of each, and finds that the mean weight is equal to 19.86 ounces, and the standard deviation is equal to 0.22 ounce. do the data provide sufficient evidence for the consumer group to conclude that the mean fill per family-size bottle is less than 20 ounces? use α = 0.01

We accept H₀, we dont have evidence to say that the family size container

of ketchup has smaller quantity

Step-by-step explanation:

Population mean  μ₀ =  20 ounces

sample size  n  =  10   df = n -1  df = 10-1   df= 9

n < 30   use of t-student distribution

sample mean  μ  = 19.86

sample standard deviation  s  =  0,22

One tail-test ( left tail)

1.-Test Hypothesis

H₀     null hypothesis                μ₀ =  20

Hₐ  Alternative hypothesis       μ₀ < 20

2.- α  =  0,01   and one test tail

3.- Compute

t(s)  =  [ ( μ  -   μ₀ ) ] / s/√n           t(s)  = [( 19.86  -  20 )* √10 ] / 0.22

t(s)  =  - ( 0,14 * 3.16) / 0,22  

t(s)  =  - 2.01

4.- We go to  t-student table  t(c) for df = 9  and 0,01 = α  

and find t(c) =  - 2.821

5.-Compare   t (s)   and  t (c)

t (c)  <   t (s)      -  2.821  < - 2.01

6. t(s)  is inside de acceptance region  we accept H₀

it would more likely be bald eagle because its has more in the box so about 75%.

step-by-step explanation:

126.66 m²

step-by-step explanation:

the pertinent area formula is a = πr², where r is the radius.

here, a = π(6.35 m)² = 40.32π, or approximately 126.66 m²