Answer:
0.0623 ± ( 2.056 )( 0.0224 ) can be used to compute a 95% confidence interval for the slope of the population regression line of y on x
Step-by-step explanation:
Given the data in the question;
sample size n = 28
slope of the least squares regression line of y on x or sample estimate = 0.0623
standard error = 0.0224
95% confidence interval
level of significance ∝ = 1 - 95% = 1 - 0.95 = 0.05
degree of freedom df = n - 2 = 28 - 2 = 26
∴ the equation will be;
⇒ sample estimate ± ( t-test) ( standard error )
⇒ sample estimate ± (
) ( standard error )
⇒ sample estimate ± (
) ( standard error )
⇒ sample estimate ± (
) ( standard error )
{ from t table; (
) = 2.055529 = 2.056
so we substitute
⇒ 0.0623 ± ( 2.056 )( 0.0224 )
Therefore, 0.0623 ± ( 2.056 )( 0.0224 ) can be used to compute a 95% confidence interval for the slope of the population regression line of y on x
Answer:
D |-11 | >-25
Step-by-step explanation:
|-11 | = 11
11 > -25
Ist amount paid = $1500
Making $350 for 10 months = 10*350 = $3500
Total amount paid = 1500 + 3500 = 5000
So an amount of $5000 was paid to cover the cost of $4500 within the ten month period.
I = PRT
Interest, I = 5000 - 4500 = 500Time, t = 10 months = 10/12 = (5/6) year.Principal P = 4500
R = I /(PT) R = 500 / (4500*5/6)
R = (500*6) / (4500*5)
R = 0.1333..
R ≈ 13.33 % per annum.
Answer:
Slope = 46.000/2.000 = 23.000
x-intercept = 6/23 = 0.26087
y-intercept = -6/1 = -6.00000