Question

1. If you were to flip a coin 16 times. A) What is the probability of getting 12 or fewer heads? B) What is the probability of getting 6 or fewer heads? 2. In a sample of N=36, what is the 90% confidence intervals of the mean, if the mean is 10 and the unbiased estimate of the standard deviation is 2? 3. What is the 99% confidence interval of the mean in a sample of N=20 with a mean of 100 and an unbiased estimate of the standard deviation of 30?

thank you !!

Answer 1

1.
BIONOMIAL DISTRIBUTION
pmf of B.D is = f ( k ) = ( n k ) p^k * ( 1- p) ^ n-k
where   
k = number of successes in trials
n = is the number of independent trials
p = probability of success on each trial
I.
mean = np
where
n = total number of repetitions experiment is executed
p = success probability
mean = 16 * 0.5
= 8
II.
variance = npq
where
n = total number of repetitions experiment is executed
p = success probability
q = failure probability
variance = 16 * 0.5 * 0.5
= 4
III.
standard deviation = sqrt( variance ) = sqrt(4)
=2
a.
probability that getting 12 or fewer heads
P( X < = 12) = P(X=12) + P(X=11) + P(X=10) + P(X=9) + P(X=8) + P(X=7) + P(X=6) + P(X=5) + P(X=4) + P(X=3) + P(X=2)
= ( 16 12 ) * 0.5^12 * ( 1- 0.5 ) ^4 + ( 16 11 ) * 0.5^11 * ( 1- 0.5 ) ^5 + ( 16 10 ) * 0.5^10 * ( 1- 0.5 ) ^6 + ( 16 9 ) * 0.5^9 * ( 1- 0.5 ) ^7 + ( 16 8 ) * 0.5^8 * ( 1- 0.5 ) ^8 + ( 16 7 ) * 0.5^7 * ( 1- 0.5 ) ^9 + ( 16 6 ) * 0.5^6 * ( 1- 0.5 ) ^10 + ( 16 5 ) * 0.5^5 * ( 1- 0.5 ) ^11 + ( 16 4 ) * 0.5^4 * ( 1- 0.5 ) ^12 + ( 16 3 ) * 0.5^3 * ( 1- 0.5 ) ^13 + ( 16 2 ) * 0.5^2 * ( 1- 0.5 ) ^14
= 0.9894
b.
probability that getting 6 or fewer heads
P( X < = 6) = P(X=6) + P(X=5) + P(X=4) + P(X=3) + P(X=2) + P(X=1) + P(X=0)
= ( 16 6 ) * 0.5^6 * ( 1- 0.5 ) ^10 + ( 16 5 ) * 0.5^5 * ( 1- 0.5 ) ^11 + ( 16 4 ) * 0.5^4 * ( 1- 0.5 ) ^12 + ( 16 3 ) * 0.5^3 * ( 1- 0.5 ) ^13 + ( 16 2 ) * 0.5^2 * ( 1- 0.5 ) ^14 + ( 16 1 ) * 0.5^1 * ( 1- 0.5 ) ^15 + ( 16 0 ) * 0.5^0 * ( 1- 0.5 ) ^16
= 0.2272
2.
TRADITIONAL METHOD
given that,
sample mean, x =10
standard deviation, s =2
sample size, n =36
I.
standard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 2/ sqrt ( 36) )
= 0.333
II.
margin of error = t α/2 * (standard error)
where,
ta/2 = t-table value
level of significance, α = 0.1
from standard normal table, two tailed value of |t α/2| with n-1 = 35 d.f is 1.69
margin of error = 1.69 * 0.333
= 0.563
III.
CI = x ± margin of error
confidence interval = [ 10 ± 0.563 ]
= [ 9.437 , 10.563 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =10
standard deviation, s =2
sample size, n =36
level of significance, α = 0.1
from standard normal table, two tailed value of |t α/2| with n-1 = 35 d.f is 1.69
we use CI = x ± t a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
ta/2 = t-table value
CI = confidence interval
confidence interval = [ 10 ± t a/2 ( 2/ Sqrt ( 36) ]
= [ 10-(1.69 * 0.333) , 10+(1.69 * 0.333) ]
= [ 9.437 , 10.563 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 90% sure that the interval [ 9.437 , 10.563 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 90% of these intervals will contains the true population mean
3.
TRADITIONAL METHOD
given that,
sample mean, x =100
standard deviation, s =30
sample size, n =20
I.
standard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 30/ sqrt ( 20) )
= 6.708
II.
margin of error = t α/2 * (standard error)
where,
ta/2 = t-table value
level of significance, α = 0.01
from standard normal table, two tailed value of |t α/2| with n-1 = 19 d.f is 2.861
margin of error = 2.861 * 6.708
= 19.192
III.
CI = x ± margin of error
confidence interval = [ 100 ± 19.192 ]
= [ 80.808 , 119.192 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =100
standard deviation, s =30
sample size, n =20
level of significance, α = 0.01
from standard normal table, two tailed value of |t α/2| with n-1 = 19 d.f is 2.861
we use CI = x ± t a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
ta/2 = t-table value
CI = confidence interval
confidence interval = [ 100 ± t a/2 ( 30/ Sqrt ( 20) ]
= [ 100-(2.861 * 6.708) , 100+(2.861 * 6.708) ]
= [ 80.808 , 119.192 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 99% sure that the interval [ 80.808 , 119.192 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 99% of these intervals will contains the true population mean