Thread: Probability of type I and II errors!!!!!!!!

1. Probability of type I and II errors!!!!!!!!

Hi, I have problems calculating the errors i have answers could some confirm if these are correct. thanks a lot guys.

i)A botanist has a plant which she thinks is a ephermeralis, although it could also be a semperflorens. The flower of the ephermeralis opens for a random length of time, exponentially distributed with mean 9: for the semperflorens the distribuition i salso exponential , but with mean 25 hrs. The next times the plabndt flowers the botanist will measure how long the flower is open and se is to decide on the species of the plant.
As a statician, you regard this as a hypothesis-testing problem, where the null hypotheses is that plant is an ephermeralis, the alternatice that it is a semperflorens.

b) the botanist decides that she will classify the plant as an ephermeralis if the flower lasts less than 15, as a semperflorens otherwise. Calculate the probabilities of he two types of errors.

ii) The Botanists handbook states that the ephemeralis has average leaf length of 13cm , the semperflorens of 17.5cm, and that in both cases the variance of the leave length is 5cm^2. The botanist finds that her plant has 8 mature leaves, with sample mean 14.25cm.
a) construct a 95% confindence interval for the mean leaf lenghth of this specimen. State assumptions
b) does your result shed any light on whether the plant is a ephermeralis or a semperflorens.

My answer for the type 1 error is 0.79014 and the type II is 0.95949
for iia) 15.7995 lower and upper 12.7005

2. Originally Posted by rishul
Hi, I have problems calculating the errors i have answers could some confirm if these are correct. thanks a lot guys.

i)A botanist has a plant which she thinks is a ephermeralis, although it could also be a semperflorens. The flower of the ephermeralis opens for a random length of time, exponentially distributed with mean 9: for the semperflorens the distribuition i salso exponential , but with mean 25 hrs. The next times the plabndt flowers the botanist will measure how long the flower is open and se is to decide on the species of the plant.
As a statician, you regard this as a hypothesis-testing problem, where the null hypotheses is that plant is an ephermeralis, the alternatice that it is a semperflorens.

b) the botanist decides that she will classify the plant as an ephermeralis if the flower lasts less than 15, as a semperflorens otherwise. Calculate the probabilities of he two types of errors.

ii) The Botanists handbook states that the ephemeralis has average leaf length of 13cm , the semperflorens of 17.5cm, and that in both cases the variance of the leave length is 5cm^2. The botanist finds that her plant has 8 mature leaves, with sample mean 14.25cm.
a) construct a 95% confindence interval for the mean leaf lenghth of this specimen. State assumptions
b) does your result shed any light on whether the plant is a ephermeralis or a semperflorens.

My answer for the type 1 error is 0.79014 and the type II is 0.95949
for iia) 15.7995 lower and upper 12.7005
$\displaystyle \alpha = \Pr(T > 15 \, | \, \text{ephemeral}) = \int_{15}^{+\infty} \frac{1}{9} \, e^{-x/9} \, dx \neq 0.79014$.

$\displaystyle \beta = \Pr(T < 15 \, | \, \text{semperflorens}) = \int_{0}^{15} \frac{1}{25} \, e^{-y/25} \, dy \neq 0.95949$.

I haven't checked bothered to check the rest. But showing all of your working will make checking a lot easier.

3. for the type 1 error alpha i did it like this

Type I error

if P(X = x) = (Lander) exp^(-lander * x)
and P(X is less than or equal to x)= 1 - exp^(-lander * X)

hence P(mean < 15)= P(mean < or = 15) - P(mean > or = 15)
hence alpha=(1 -Exp^(-0.1111... * 15)) -( 0.111... * exp^-0.111*15)
hence alpha=0.79014

Type II error
i used the same method but uising the central limit theroem i obtained x bar which was 29.805.
Putting this value in to
beta=(1 -Exp^(-0.1111... * 29.805)) -( 0.111... * exp^-0.111*29.805)
beta= 0.95949

the 0.111111.. is (1/9)

4. Originally Posted by rishul
for the type 1 error alpha i did it like this

Type I error

if P(X = x) = (Lander) exp^(-lander * x)
and P(X is less than or equal to x)= 1 - exp^(-lander * X)

hence P(mean < 15)= P(mean < or = 15) - P(mean > or = 15)
hence alpha=(1 -Exp^(-0.1111... * 15)) -( 0.111... * exp^-0.111*15)
hence alpha=0.79014

Type II error
i used the same method but uising the central limit theroem i obtained x bar which was 29.805.
Putting this value in to
beta=(1 -Exp^(-0.1111... * 29.805)) -( 0.111... * exp^-0.111*29.805)
beta= 0.95949

the 0.111111.. is (1/9)

My previous post shows what to do.

5. yo man i did the integrals

for the type 1 error when i intergrated i got
0.18888(5dp)

for the type 2 error when i integrated i got 0.45119(5dp)

are these probabiliites correct?

thank you

6. Originally Posted by rishul
yo man i did the integrals

for the type 1 error when i intergrated i got
0.18888(5dp)

for the type 2 error when i integrated i got 0.45119(5dp)

are these probabiliites correct?

thank you
yo bro quite so.