The U. S. Department of Agriculture claims that the mean consumption of coffee by a person in the United States is 24.2 gallons per year. A random sample of 101 people in the United States shows that the mean coffee consumption is 23.5 gallons per year with a standard deviation of 3.2 gallons. At α=0.05, can you reject the claim? Using a P-value, evaluate the claim. Using Rejection Regions for a z-test, evaluate the claim.

The decision rule is  

Reject the null hypothesis  

The conclusion is  

There is sufficient evidence to reject the claim  

Step-by-step explanation:

From the question we are told that

   The population mean is  

    The sample size is  n  =  101

     The sample mean is  

     The standard deviation is  

The null hypothesis is  

 The alternative hypothesis is

Generally the test statistics is mathematically represented as    

=>              

=>            

From the z table  the area under the normal curve to the left corresponding to    -2.198  is  

Gnerally the  p-value is mathematically represented as

=>      

From the value obtained we have that   hence  

The decision rule is  

Reject the null hypothesis  

The conclusion is  

There is sufficient evidence to reject the claim  

Using the Rejection Regions for a z-test method

Generally from the z table  the critical value  of  is  

Not we are using    because it is a two - tailed test

   Now comparing the critical value and the test statistics we see that the

region covered by  the test statistics (i.e )   is greater than the region covered by  the critical value  (i.e )

Hence

The decision rule is  

Reject the null hypothesis  

The conclusion is  

There is sufficient evidence to reject the claim  

this shows a reflection over the x-axis and is translated one unit down and 6 units to the right (x + 6, y - 1)

hope this !

~just a girl in love with shawn mendes

part a

answer: a reasonable domain is where n is a real number. so n can be between 0 and 10. both endpoints are included.

work shown:

n is the number of days after the experiment starts. the smallest n can be is n = 0 which means that 0 days have gone by, and we're at the start. to find out how large n should be, then replace f(n) with 16.13 and solve for n. use logarithms to isolate the exponent.

f(n) = 12*(1.03)^n

16.13 = 12*(1.03)^n

16.13/12 = (1.03)^n

1.34416666666667 = (1.03)^n

(1.03)^n = 1.34416666666667

log[ (1.03)^n ] = log[ 1.34416666666667 ]

n*log[ 1.03 ] = log[ 1.34416666666667 ]

n = log[ 1.34416666666667 ]/log[ 1.03 ]

n = 10.0062999823929

this rounds to n = 10 which is fairly close but not 100% perfect. so this is the largest n can be.

note: if you plug n = 10 into f(n), you'll get roughly 16.127 which rounds to 16.13 (this comes up again in part c)

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part b

answer: the y intercept is 12. it represents the starting height of the plant in cm.

work shown:

plug n = 0 into the f(n) function. simplify

f(n) = 12*(1.03)^n

f(0) = 12*(1.03)^0

f(0) = 12*(1)

f(0) = 12

on day n = 0, aka the starting point, the height f(n) is 12 cm

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part c

answer: the average rate of change is approximately 0.43061036458991 (round however you need to). this represents the average growth rate from day n = 3 to day n = 10. so the plant grew roughly 0.43 cm per day during this timespan, assuming you round to 2 decimal places.

work shown:

compute f(3)

f(n) = 12*(1.03)^n

f(3) = 12*(1.03)^3

f(3) = 13.112724 < we'll use this later

compute f(10)

f(n) = 12*(1.03)^n

f(10) = 12*(1.03)^10

f(10) = 16.1269965521294 < we'll use this later

now use the formula below with a = 3 and b = 10

aroc = average rate of change

aroc = [ f(b) - f(a) ]/[ b - a ]

aroc = [ f(10) - f(3) ]/[ 10 - 3 ]

aroc = (16.1269965521294 - 13.112724)/(10 - 3)

aroc = 3.0142725521294/7

aroc = 0.43061036458991

round this however you need to

note: the plant grew approximately 3.01 cm over 7 days, so roughly 0.43 cm per day is the average growth rate (if you were to round to 2 decimal places).