A researcher wanted to determine the effectiveness of a new cream in the treatment of warts. She identified 145 individuals who had two warts. She applied cream A on one wart and cream B on the second wart. Test whether the proportion of successes with cream A is different from cream B at the alpha equals 0.05 level of significance. Treatment ATreatment B Success Failure Success 61 11 Failure 21 52(a) What type of test should be used? A. A hypothesis test regarding the difference of two means using a matched-pairs design B. A hypothesis test regarding the difference between two population proportions from dependent samples C. A hypothesis test regarding two population standard deviations D. A hypothesis test regarding the difference between two population proportions from independent samples (b) Determine the null and alternative hypotheses. A. H0 : p1 = p2, H1 : p1 >p2B. H0 : p1 < p2, H1 : p1 >p2C. H0 : p1 = p2, H1 : p1

a. Option D

b. H₀ : p₁ = p₂, H₁: p1 ≠ p₂

Step-by-step explanation:

a. A hypothesis test regarding the difference between two population proportions from independent samples. This type of test is used to compare the proportions of two populations. This test is usually appropriate under the conditions that:

The sampling method for each population is simple random sampling.

The samples are independent.

Each sample includes at least 10 successes and 10 failures.

Each population is at least 20 times as big as its sample.

b. Null hypothesis: H₀ : p₁ = p₂

   Alternative hypothesis: H₁ : p₁ ≠ p₂

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).

cant see photo

step-by-step explanation: