confidenece interval????????`?

Question

when consumers apply for credit, their credit is rated using FICO (Fair, Issac, and company) scores. Credit ratings are given below for a sample of applicants for car loans. Use sample data to construct a 99% confidence interval for the standard deviation of FICO scores for all applicants for credit.

661 595 548 730 791 678 672 491 492 583 762 624 769 729 734 706


*****please kindly show how u got the standard deviation thanks:-)

Answer 1

I haven't learned the Confidence Interval yet, but I just learned the S.Dev. again yesterday.
661 595 548 730
791 678 672 491
492 583 762 624
769 729 734 706
Sixteen numbers. Find average.
661 + 595 + 548 + 730 + 791 + 678 + 672 + 491 + 492 + 583 + 762 + 624 + 769 + 729 + 734 + 706 = 10 565
10565/16 = 660.3125 average.
Now get all sixteen
differences squared.
(660.3125 - 661)^2 = 0.47265625
(660.3125 - 595)^2 = 4265.72266
(660.3125 - 548)^2 = 12614.0977
(660.3125 - 730)^2 = 4856.34766
(660.3125 - 791)^2 = 17079.2227
(660.3125 - 678)^2 = 312.847656
(660.3125 - 672)^2 = 136.597656
(660.3125 - 491)^2 = 28666.7227
(660.3125 - 492)^2 = 28329.0977
(660.3125 - 583)^2 = 5977.22266
(660.3125 - 762)^2 = 10340.3477
(660.3125 - 624)^2 = 1318.59766
(660.3125 - 769)^2 = 11812.9727
(660.3125 - 729)^2 = 4717.97266
(660.3125 - 734)^2 = 5429.84766
(660.3125 - 706)^2 = 2087.34766
Now add these 16 #'s up and divide by 16.
0.47265625 + 4 265.72266 + 12 614.0977 + 4 856.34766 + 17 079.2227 + 312.847656 + 136.597656 + 28 666.7227 + 28 329.0977 + 5 977.22266 + 10 340.3477 + 1 318.59766 + 11 812.9727 + 4 717.97266 + 5 429.84766 + 2 087.34766 = 137945.438
Divide that sum by 16 to get Variance.
137 945.438 / 16 = 8621.58987
Square root that to get Standard Deviation.
8621.58987^0.5 = 92.8525168
So Standard Deviation is 92.8525 I guess.

Okay I learned a little.
For 99% z=2.576 or something according to google. So 92.8525 times 2.57600 = + or -239.188 from average 660.3125

660.3125 + 239.18800 = 899.5005
660.3125 - 239.18800 = 421.1245
So 421.12 is lower limit
And 899.50 is upper limit
for 99% confidence I guess.

Answer 2

Using a calculator in statistics mode I get
m = 660.3125
σ = 95.89767
Z ≡ (X - m) / σ
99% = P(- 2.5758 < Z < 2.5758) =
P(- 2.5758 < (X - m) / σ < 2.5758) =
P(- 2.5758σ < (X - m) < 2.5758σ ) =
P(m - 2.5758σ < X < m + 2.5758σ ) =
P(660.3125 - 2.5758*95.89767 < X < 660.3125 + 2.5758*95.89767 ) =
P(413.2993 < X < 907.3257) = 99%
or
P(413 < X < 907) = 99%
m = (∑x)/n
σ = √((∑(x - m)^2))/n)
which can be restated as
σ = √((∑x^2)/n - m^2)
(derived below)
which is the way standard deviation is calculated by a calculator. Calculating "by hand", of course, can use either method. Usually I use a spreadsheet for this as it has built-in functions, or can be set up for the more detailed display.

σ = √(∑(x^2 - 2mx + m^2))/n)
σ = √(∑x^2 - ∑2mx + ∑m^2)/n)
σ = √(∑x^2 - 2m∑x + nm^2)/n)
σ = √(∑x^2 - 2((∑x)/n)∑x + n((∑x)/n)^2)/n)
σ = √(∑x^2 - 2((∑x)^2/n) + ((∑x)^2/n))/n)
σ = √(∑x^2 - ((∑x)^2/n)/n)
σ = √((∑x^2)/n - ((∑x)^2/n^2))
σ = √((∑x^2)/n - ((∑x)/n)^2)
σ = √((∑x^2)/n - m^2)

Answer 3

Because the sample size is < 30, a t-table is appropriate here.
The general formula for this confidence interval is:
Xbar ± t(sub α/2)[s/√n]; Where t(sub α/2) = 2.947 (from the t-tables with df = 15)
s = √[(n(Σx²)-(Σx)²)/(n(n-1))] = 95.9; Xbar = 660.3; n = 16
(Σx²) means square all the data first and then sum the squares.
(Σx)² means sum the data first and then square the sum.

99% confidence interval is:
660.3 ± 2.947(95.9/√16)
660.3 ± 70.65
or 589.7 < µ < 730.9