In a clinical study of an allergy drug, 108 of the 200 subjects reported experiencing significant relief from their symptoms. Test the claim that more than half of all those using the drug experience relief. Using a 0.01 significance level, what is your decision

The decision is to not reject the null hypothesis.

At a significance level of 0.01, there is not enough evidence to support the claim that the proportion of all those using the drug that experience relief is significantly higher than 50% (P-value = 0.1443).

Step-by-step explanation:

This is a hypothesis test for a proportion.

The claim is that the proportion of all those using the drug that experience relief is significantly higher than 50%.

Then, the null and alternative hypothesis are:

The significance level is 0.01.

The sample has a size n=200.

The sample proportion is p=0.54.

The standard error of the proportion is:

Then, we can calculate the z-statistic as:

This test is a right-tailed test, so the P-value for this test is calculated as:

As the P-value (0.1443) is greater than the significance level (0.01), the effect is  not significant.

The null hypothesis failed to be rejected.

At a significance level of 0.01, there is not enough evidence to support the claim that the proportion of all those using the drug that experience relief is significantly higher than 50%.

b

step-by-step explanation:

calculate the scale factor as the ratio of corresponding sides, image to original, that is

scale factor = = = = 1

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).