Trials in an experiment with a polygraph include 97 results that include 23 cases of wrong results and 74 cases of correct results. Use a 0.05 signifcance level to test the claim that such polygraph results are correct less than 80% of the time. Based on the results should polygraph test results be prohibited as evidence in trials? identify the null hypothesis, alternative hypothesis, test statistic, P-value conclusion about the null hypothesis and final conclusion that addresses the original claim. Use the P-value method. Use the normal distribution as an approximation of the binomial distribution a. H0:p= 0.80 b. H0:p=0.20
H1:p>0.80 H1: p =/ 0.20
c. H0:p=0.20 d. H0:p= 0.80
H1:p< 0.20 H1: p< 0.80
e. H0: p= 0.20 f. H0: p= 0.80
H1: p>0.20 H1:p =/ 0.80
the test statistic is z=
the P-value is ?

H0:p= 0.80                 H1: p< 0.80  one tailed test

Step-by-step explanation:

We state the null and alternative hypotheses as that the results are 80 % against the claim that the results are less than 80%.

H0:p= 0.80                 H1: p< 0.80  one tailed test

p2= 0.8 , p1= 74/97= 0.763

q1= 1-0.763= 0.237    q2= 0.2

The level of significance is 0.05 .

The Z∝= ±1.645 for ∝= 0.05

The test statistic used here is

Z= p1-p2/ √pq/n

Putting the values:

Z= 0.763 -0.8 / √ 0.8*0.2/97

z= -0.037/ 0.0406

z= -0.9113

The Z∝ = ±1.645 for ∝= 0.05 for one tailed test.

As the calculated value  does not fall in the critical region  we fail to reject the null hypothesis. There is not sufficient evidence to support  the claim that such polygraph results are correct less than 80% of the time.

Using the normal probability table.

P (Z <  -0.9113)= 1- P(z= 0.9311) = 1- 0.8238= 0.1762

If P- value is smaller than the significance level reject H0.

0.1762> 0.005  Fail to reject H0.

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).

1/2*qr*rs

step-by-step explanation: