One of the authors came across an article (USA Today, 2008) that said that on average Americans have visited 16 states in the United States. In a survey of 50 students in her introductory statistics class, she found the average number of states the students had visited to be 9.48 and the standard deviation to be 7.13. The data were not strongly skewed.1. Identify the observational unit for this study. a. studentsb. Americansc. number of states2. Identify the variable of interest and whether it is categorical or quantitative. a. number of U. S. states visited, categorialb. number of students, quantitativec. number of students, categoriald. number of U. S. states visited, quantitative3. Regardless of your answer to part (c), state the null and the alternative hypotheses in symbols, to test whether the average number of states all students at the author’s school have visited is different from 16.a. H0: mu = 16, Ha: mu ≠ 16b. H0: mu = 16, Ha: mu > 16c. H0: mu = 16, Ha: mu < 164. Using the 2SD approach to find a 95% confidence interval for the average number of states all students at the author’s school have visited. Round to two decimal places

1. a. Students

2. d. number of U.S. states visited, quantitative

3. a. H o: mu = 16, Ha: mu ≠ 16

4. (7.46,11.50)

Step-by-step explanation:

1. The researcher is interested in finding out the average number of states in US that the students have visited.

The unit that is being observed to arrive at that conclusion is the Students. The observational unit for this study is  a. Students.

2. The variable of interest from the information above is the number of U.S. states visited. The observations recorded for this variable are in numbers which are numeric in nature. This implies that the variable is quantitative in nature.

Therefore, the answe  is d. number of U.S. states visited, quantitative

3. To test whether the average number of states all students at the author’s school have visited,

The null hypothesis (H o) is always given as the negation statement/null of what we wish to test.

So, we wish to test that the average is different from 16. Therefore, the null would be that the average is 16.

Answer  is a. H o: mu = 16, Ha: mu ≠ 16

4. A  95% confidence interval for the average number of states all students at the author’s school have visited can be calculated using the formula:

x  bar + or - 2 √s/n

=9.48 + or - 2 × 7.13/ √50

= (7.46,11.50)

The 95% confidence interval for the average number of states all students at the author’s school have visited is (7.46,11.50). The average falls betw 7.64 and 11.50.

answer: f = 32.2m

step-by-step explanation: that is the answer

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).