A random sample of 100 observations produced a sample proportion of 0.25. An approximate 95% confidence interval for the population proportion p is between? a. 0.165 and 0.335
b. 0.165 and 0.321
c. 0.179 and 0.321
d. 0.207 and 0.293
e. 0.152 and 0.348

The confidence interval for the population proportion p is between ( 0.165 and 0.335 )

Hence, Option a) 0.165 and 0.335 is the correct answer.

Step-by-step explanation:

Given the data in the question;

sample size n = 100

sample proportion p₀ = 0.25

q₀ = 1 - p₀ = 1 - 0.25 = 0.75

confidence interval = 95%

First we calculate our Standard Error

Standard Error = √( ( p₀ × q₀ ) / n )

we substitute

Standard Error = √( ( 0.25 × 0.75 ) / 100 )

Standard Error = √( 0.1875 / 100 )

Standard Error = √0.001875

Standard Error = 0.0433

Now, For 95% confidence interval; z = 1.96

so confidence interval for the population proportion p will be;

⇒ p₀ ± ( z × Standard Error )

we substitute

⇒ 0.25 ± ( 1.96 × 0.0433 )

⇒ 0.25 ± 0.084868

⇒ (  0.25 - 0.084868 ), (0.25 + 0.084868 )

⇒ (  0.25 - 0.084868 ), (0.25 + 0.084868 )

⇒ ( 0.165, 0.335 )

Therefore, The confidence interval for the population proportion p is between ( 0.165 and 0.335 )

Hence, Option a) 0.165 and 0.335 is the correct answer.

answer: it is 1: 47 pm

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).