Use a one-sample t ‑test, based on the data below, to test the null hypothesis H0:µ=100.63 against the alternative hypothesis H1:µ>100.63 . The sample has a mean of x⎯⎯⎯=101.09 and a standard deviation of s=0.4887 . 100.68,101.23,100.82,101.15,100.96, 100.70,102.09 Calculate the standard error (SE) and the t ‑statistic for this test. Give the standard error to four decimal places and t to three decimal places.

The standard error (SE) is 0.1847.

The t-statistic for this test is 2.490.

Step-by-step explanation:

We are given that the sample has a mean of = 101.09 and a standard deviation of s = 0.4887 .

Also, the 7 sample values are also given.

Let = population mean.

So, Null Hypothesis, : = 100.63  

Alternate Hypothesis, : > 100.63  

The test statistics that would be used here One-sample t test statistics as we don't know about population standard deviation;

                         T.S. =    ~

where,  = sample mean = 101.09

            s = sample standard deviation = 0.4887

            n = sample values = 7

The Standard Error (SE) is given by =    =   = 0.1847

So, test statistics  =    ~

                                =  2.490

The value of t test statistics is 2.490.

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).

answer: the answer is ann