Problems Analysis and Statistics 2022

Running head: PROBLEM ANALYSIS 1
Problem Analysis and Statistics
Student Name
Institution

PROBLEM ANALYSIS 2
Question 1
a)
Estimated linear regression model is given as:
Price = -259.626 +3721.025*weight
b)
Step 1: calculating correlation coefficient
Coefficient of determination (R square) = correlation coefficient(r )2
r2 = 0.9783
r = √0.9783 =0.9783
There exists 97.83% correlation between price and weight of diamond rings. The price and
weight of diamond rings are positively and strongly related.
Step 2: hypothesis statement
H0 : β0=0 (the slope is equal to zero)
H1 : β1 ≠0 (the slope is not equal to zero)
Step 3: calculation of test statistic for the regression model
To test the significance of linear relations, r, we calculate the test statistic, at α=0.02
t= b1
SE
Where:
b1 is the slope of the regression model
SE is the standard error of the slope
t=−259.626
17.319 = -14.991
Step 4: calculating the critical value at α =0.02

PROBLEM ANALYSIS 3
Degrees of freedom, DF = n-1 = 48-1=47
t α
2 ,df =47 =t0.01 ,df =47 = ±2.408
Step 5: Decision rule
Decision rule: Reject null hypothesis if the test statistic is less than t critical value
Step 6: interpretation of test results
Reject null hypothesis since the test statistic (-14.991) is less than -2.408. Therefore, we
conclude that there is statistically significant linear relationship between price and weight at α
=0.02
c)
When weight = 0.2
Price = -259.626 +3721.025(0.2) = 484.579
When weight = 0.3
Price = -259.626 +3721.025(0.3) = 856.682
Pooled sample variance: SP
2 = (n¿ ¿ x−1) Sx
2 + (n¿¿ y−1)S y
2
(n¿¿ x +n y−2)¿ ¿ ¿
SP
2 = ( 44−1 ) 213.6432+(44−1) 0.0572
44+ 44−2 = 22821.667
Formula : ¿- y) ± t(n¿¿ x+n y−2 , α
2 )+ √ SP
2
nx
+ SP
2
n y
¿
Critical value, t(86,0.02) =2.3705
80% CI = (856.682-484.579)± 2.3705*
√ 22821.667
44 + 22821.667
44
= 372.103±76.3488
= (295.754, 448.452)
Therefore, 80% Confidence interval is (295.754, 448.452)

PROBLEM ANALYSIS 4
d)
When weight = 0.18
Price = -259.626 +3721.025(0.18) = 410.1585
Z= x−μ
σ
0.01= x−325
213.63
0.01*213.63 = x−325
x = 327.136
e)
Answers (ii) and (v) are false statements
Question 2
a)
First, we construct a probability table
X P(x)
0.2 0.6
-0.15 0.4
E(x) = X.P(x)
= 0.2(0.6)+(-0.15)(0.4)
= 0.06
The expected price
E(price) = $5*0.06 = $0.3

PROBLEM ANALYSIS 5
The expected price is $0.3 per day
The expected price after 5 days = $0.3*5 = $1.5
Var(X) = X2P(x)
= 0.22 (0.6)+(−0.15¿¿ 2)(0.4)¿
=0.033
Var (price) = 0.033*$5
The variance of price is $ 0.165
The variance price after 5 days = $ 0.165*5 = $0.825
b)
After 100 days, the expected price = $0.3*100 = $30
Variance after 100 days = $ 0.165*100= $16.5
P(x≥ 9.60) =1- P(Z< x−μ
σ )
P(Z< x−μ
σ )= 9.60−30
16.5 = 0.108
P(x≥ 9.60) =1- P(Z< x−μ
σ ) = 1-0.108 = 0.892
The probability that after 100 days, the price of twocoin is at least $9.60 is 0.892
c)
Probability in a month = 0.9726
Probability in a day = 0.9726
30
Probability of getting consecutive increment in 12 days = 0.9726
30 ∗12 = 0.389
Question 3
a)

PROBLEM ANALYSIS 6
σ =5
μ =30
Z= x−μ
σ
P( x>38) = P(Z> 38−30
5 ) = 1.6
P(Z≤ 1.6 ¿ = 0.945
P(Z>1.6 ¿ =1- P(Z≤ 1.6 ¿ =1-0.945 = 0.055
The probability that on a given day the oil yield is greater than 38 barrels is 0.055
b)
P(X=28)
P( x <28) = P(Z¿ 28−30
5 ) = -0.4
P(Z←0.4 ¿ = 0.345
P(X=32)
P( x <32) = P(Z¿ 32−30
5 ) = 0.4
P(Z¿ 0.4 ¿ = 0.655
Therefore,
P(X=32) - P(X=28) = 0.655-0.345=0.31
The probability that the sample mean of a yield is between 28 and 32 barrels is 0.31

PROBLEM ANALYSIS 7