Statistical Inference

[Eugene99]

Q17:
Since sample size is large 50 (>30), we can assume that the sample distribution is normal. Hence

P(the mean to be at least $950 per month)

=P(m >=1950)

=P((m-1200)/(s/n1/2) > (1950-1200)/(s/n1/2))

= P((m-1200)/(s/n1/2) > -7.07) = 1- F(-7.07) =1 (approximation)

that is, the probability of the mean of 50 one-bedroom to be at least $950 is almost 1.

 

[Eugene99]

Q16:
The mean of the sample means is 75
The std of the sample means is 5/sqrt(40) = 0.7906
---------------------------
a. Less than 74.
t(74) = (74-75)/0.7906 = -1.2649
P(xbar < 74) = P(t < -1.2649 when df = 39) = 0.1067
----------------------------------------------------------
b. Between 74 and 76.
t(76) = (76-75)/0.7906 = +1.2649
P(74< xbar < 76) = P(-1.2649 < t < 1.2649 when df = 39) = 0.7866
------
c. Between 76 and 77.
d. Greater than 77.
----
Use the same procedure for "c" and for "d".
================================================

 

[Eugene99]

That's it!
CHEERS!

 

[deelaw007]

Do you know
probability density function?

 

[Eugene99]

Do you know
probability density function?

oops...of course I knew but seems like the time is up?

 

[deelaw007]

Yes

 

[Eugene99]

Having trouble in part (ii), somebody help!

 

[deelaw007]

Help yourself

 

[Eugene99]

Help yourself

Did I ask you?
Rizwan Javed could you help?

 

[Rizwan Javed]

Did I ask you?
Rizwan Javed could you help?

With what? :O

 

[Eugene99]

With what? :O

Q5 up there part ii

 

[Eugene99]

I end up having:
Φ(10-μ/4.58) = 0.2611
but there exists no value for z of 0.2611 in the table

 

[Rizwan Javed]

View attachment 59890

Having trouble in part (ii), somebody help!

P(X>10.0)
standardize X,
P(Z>10- μ / √21) = 0.7389
1 - P(Z < 10 - μ / √21) = 0.7389
P(Z < 10 - μ / √21) = 0.2611
ϕ(10 - μ / √21) = 0.2611 ---- (i)

Let (10 - μ / √21) = -v
ϕ(-v) = 0.2611
1 - ϕ(v) = 0.2611
ϕ(v) = 0.7389

use normal distribution tables,
v = 0.64
substitute this value of v back into (i)
10 - μ / √21 = -0.64
solving it will give you:
μ = 12.9

Is the answer correct?

 

[Eugene99]

P(X>10.0)
standardize X,
P(Z>10- μ / √21) = 0.7389
1 - P(Z < 10 - μ / √21) = 0.7389
P(Z < 10 - μ / √21) = 0.2611
ϕ(10 - μ / √21) = 0.2611 ---- (i)

Let (10 - μ / √21) = -v
ϕ(-v) = 0.2611
1 - ϕ(v) = 0.2611
ϕ(v) = 0.7389

use normal distribution tables,
v = 0.64
substitute this value of v back into (i)
10 - μ / √21 = -0.64
solving it will give you:
μ = 12.9

Is the answer correct?

yes the answer is correct but why did you let '(10-μ / √21) = -v', we don't normally do that!?

 

[Rizwan Javed]

yes the answer is correct but why did you let '(10-μ / √21) = -v', we don't normally do that!?

actually when ϕ with some value of z in it gives the probability less than 0.5, this shows that value of z is negative. So we put the value of z equal to -v to make our calculations easier because we don't know what the value of mean is.

 

[Eugene99]

actually when ϕ with some value of z in it gives the probability less than 0.5, this shows that value of z is negative. So we put the value of z equal to -v to make our calculations easier because we don't know what the value of mean is.

thank you
do you have any idea what degree of accuracy is required in our answer? like up to how many decimal places?

 

[Rizwan Javed]

thank you
do you have any idea what degree of accuracy is required in our answer? like up to how many decimal places?

3 significant figures.

 

[Eugene99]

3 significant figures.

Aren't May/June papers easier than Oct/Nov, or is it my myth?

 

[Rizwan Javed]

Aren't May/June papers easier than Oct/Nov, or is it my myth?

The difficulty level remains the same. No difference.

 

[Waleed001]

Aren't May/June papers easier than Oct/Nov, or is it my myth?

I hope you know how to calculate hypothesis