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)