Question

In: Statistics and Probability

The following data give the annual incomes (in thousands of dollars) and amounts (in thousands of dollars) of life insurance policies for eight persons.

The following data give the annual incomes (in thousands of dollars) and amounts (in thousands of dollars) of life insurance policies for eight persons.

(a) At the 98% confidence level, test whether annual income and the amount of life insurance policies are independent.

(b) Find the attained significance level.

(c) State any assumptions you have made in solving the problem.

Solutions

Expert Solution

Solution

Given dataset

annual income life insurance
42 150
58 175
27 25
36 75
70 250
24 50
53 250
37 100

I have done it in excel.

1.Here I have used Data analysis tab to find regression

SUMMARY OUTPUT              
                 
Regression Statistics              
Multiple R 0.92772              
R Square 0.860664              
Adjusted R Square 0.832796              
Standard Error 6.990796              
Observations 7              
                 
ANOVA                
  df SS MS F Significance F      
Regression 1 1509.358 1509.358 30.88439 0.002594      
Residual 5 244.3562 48.87123          
Total 6 1753.714            
                 
  Coefficients Standard Error t Stat P-value Lower 95% Upper 95% Lower 95.0% Upper 95.0%
Intercept 21.08904 4.831949 4.3645 0.007259 8.668121 33.50996 8.668121 33.50996
150 0.170137 0.030615 5.557372 0.002594 0.09144 0.248834 0.09144 0.248834
                 
                 
                 
RESIDUAL OUTPUT              
                 
Observation Predicted 42 Residuals            
1 50.86301 7.136986            
2 25.34247 1.657534            
3 33.84932 2.150685            
4 63.62329 6.376712            
5 29.59589 -5.59589            
6 63.62329 -10.6233            
7 38.10274 -1.10274            

The equation is y = 0.170.0 + 21.089 =

Where y = a + B.C α

Here \alpha= slope

\beta=intercept

2. The scatter plot and the least square regression line on graph using excel isScatter plot 80 70 60 y = 0.168x + 20.69 R2 = 0.850 50 annual income 40 annual income 30 Linear (annual income) 20 Linear (an

iii. Here mean T= 134.38 C 2 =

Standard deviation s= 86.54

Confidence interval for \alpha=

22 16+ n (a)+ta/2, (m - 2) + (x; -m)2 (a) – ta/2, (n − 2)) = X51 EXSV 2 (X; - 72 XS + n

4 : 134.382 ((0.168) -1.943 x 86.54 (150 – 134.8)2 + ... (100 – 134.8)2 ((0.168)+ 134.382 1.943 x 86.54 + Σ(150 – 134.8)2 + .

= (194, 74.4)

Conidence interval for \beta=

S (β-ta/2), (η- 2)) x- (β+ta/2), (η-2))x: Σ(, - 2 ΥΣ (υ; - z)2

(20.69-1.943 x 134.38 (20.69-1.943x Σ(150 – 134.38)2 + .. (100 - 134.38)52) 134.38 Σ((150 – 134.38)2 + ..(100 – 134.38))2)

= (-14.81, 56.19

SUN BUNRA answered 2 years ago
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