INSTRUCTIONS: Provide (2) 150 words response for RESPONSES 1 AND 2 below. Responses may include direct questions. In your first peer posts, pick another confidence level, i.e. 90%, 99%, 97%, any other confidence level is fine. Have fun and be creative with it and calculate another T-confidence interval and interpret your results. Compare your results to that of the initial 95%, how much do they differ? How useful can this type of information be when you go to buy a new car, or even a house?
In your second peer posts, pick another confidence level, i.e. 90%, 99%, 97%, any other confidence level is fine. Have fun and be creative with it and calculate another proportion interval and interpret your results. Compare your results to that of the initial 95%, how much do they differ? How useful can this type of information be when you go to buy a new car, or even a house?
Attached are the excel docs for both responses to help with the post.
RESPONSE 1:
FIRST
1-0.95=0.05
0.05/2= 0.025
–
Change to less than equation for Excel function
1-0.025= 0.975.
α =0.975
Take N-1
10-1=9
=T.INV(0.975,9) = 2.26
T-critical value = 2.26
????̅= Mean (40,589)
????= .975
N-1= 9
????????= 15768.88432
T-crit= 2.26
Confidence Level @ 95%= 11280.3803 (Use your toolpack to get this number) OR wait for a detailed explanation below.
2.26 *(15768.88432/SQRT(10))= 11280.3803
=40,589± 11280.3803
(29308, 51869). Thus, I am 95% confident that the population mean car price for the type of cars I selected in Wk 1 is between $29,308 and $51,869. I realize this is quite the disparity, but I believe this is attributed to the wide range of cars I chose.
SECOND
P=0.7
Q=0.3
α =0.975
=NORM.S.INV(0.975)= 1.96
Z critical value=1.96
.70 ± 1.96(.1449137)
0.70 ± 0.28403=
(.41597, .98403). Thus, I am 95% confident that the population proportion of car prices that are less than the mean is between 41.6% and 98.4%. Please let me know if you got a different calculation here. I was confused as to why it was so similar to the example, when part 1 was not. Have a great week!
RESPONSE 2:
Step 1.
We are asked to determine the T- confidence interval for a sample of car population prices.
????̅± ???? ∗ ( ???????? √???? )
Step 2.
To calculate for 95% confidence
1-0.95=0.05
Step 3.
Calculate the confidence interval (NEEDS TO BE DIVIDED BY 2)
0.05/2= 0.025
Step 4: (IMPORTANT)
Change to less than equation for Excel function
1-0.025= 0.975. α =0.975
Step 5:
Calculate degrees of freedom (DF)
-Take N -1
10-1=9
Everything we have for Excel function!
N= 10
Excel Function:
=T.INV(0.975,9) = 2.26
T-critical value- 2.26
Equation should look like:
=2.26 *(33061.48742/SQRT(10))= 23650.76334
=56,813.70 ± 23650.76334
Equals (33,162.94; 80,464.46)
I am 95% confidence that the sample price of the cars will be between $33,162.94 and $80,464.46
Part 2:
Calculate the proportion confidence interval with the proportion of the number of vehicles that fall below the average.
????̂± ???? ∗ (√ ????̂∗????̂/ ???? )
Step 1:
calculate the success rate (p) and failure (q)
P=0.6
Q=0.4
Step 2:
Calculate Z critical value of 95% confidence
α =0.975
=NORM.S.INV(0.975)= 1.96
Z critical value=1.96
Step 3:
Put values in function
1.96 * ( √(.6*.4)/10)
0.6 ± 0.3036
(0.2963,0.9036)
I am 95% confident that the proportion of cars sampled that will fall below the average goes from 30% to 90%
>Sheet 0 8. 8 .5590608934
5768.8843 3733
8657712.677778
)
0.3802973 2
Price 19
20 6 Toyota 4 17 24 6 2020 00
20 25 8 15 16 8 28 4 19 26 6 15 8 26 4 22 28 4 QUAL QUAL QUAN QUAN QUAN QUAN Mean: 40,589 Median: 37,423 SD: 15768.8843193733 10 Sample Size: 10 0% )
e/Class
l
Qualitative Qualitative Quantitative Quantitative Quantitative Qualitative 00.00
20 21 20 All Wheel Drive 2 Wheel Drive-rear 16 2 Wheel Drive-Front 2020 22 All Wheel Drive 2 Wheel Drive-rear 15 21 All Wheel Drive 9,706.00
16 22 All Wheel Drive 7 All Wheel Drive :
19 :
UPPER CONFIDENCE LEVEL= LOWER CONFIDENCE LEVEL= 9 Mode ERROR:#N/A Minimum 167912
3
Price
Mean
4
5
8
7
Standard Error
4
9
6
Median
37423
Mode
ERROR:#N/A
Standard Deviation
1
19
Sample Variance
24
Kurtosis
-0.7745719499
Skewness
0.5873946372
Range
480
15
Minimum
19866
Maximum
67881
Sum
405887
Count
10
Confidence Level(95.
0%
11
28
16
Sheet5
Sheet1
Vehicle type/Class
Year
Make
Model
MPG (City)
MPG (Hwy)
Quan (cylinders)
1
SUV
20
Toyota
4Runner
34,565
17
2 SUV
2020
Rav4
28,517
26
35
3 SUV
2019
Ford
Expidition
60,247
4
Truck
Ram
Ram
25
40,437
5 Truck 2019 Ford
F-250
40,281
6 SUV
2018
Kia
Sorento
19,866
21
7 SUV 2019
Dodge
Durango
32,946
8 SUV 2020
GMC
Yukon
55,349
22
9 SUV 2020
Chevy
Equinox
25,798
31
10 SUV 2020
Land Rover
Range Rover
67,881
QUAL
QUAN
Mean:
40,589
Median:
37,423
SD:
15768.8843193733
Sample Size:
Wk4
New SD
7052.0594534899
1
P(X<40,089)
0.4717381647
47.17%
2
P(X>41,589)
0.4436179857
44.36%
3
P(X=40,589)
0.000056571
4
P(
39,089
0.16844201468267
16.88%
39,089
42,089
Wk5
2.2621571628
<---------- T Interval
Z Crit Value——>
1.9599639845
.70 ± 1.96(.1449137)
Column1
. 70 ± .28403
(.41597, .98403)
Confidence Level(95.0%)
11280.3802973162
equals:41.6% to 98.4%
Mean+CL
51,869
Mean-CL
29,308
Sheet1
Vehicles Ty
p
Year
Make
Mode
Price
MPG (City)
MPG (Highway)
Drive Type
Qualitative
Quantitative
SUV
20
21
Hyundai
Genesis GV80
$ 48,
9
All Wheel Drive
Hybrid/SUV
2021
Lexus
GX
$ 58,665.00
16
Coupe
201
7
Honda
Accord EX
$ 16,791.00
27
36
2 Wheel Drive-Front
Coupe
2014
Chevrolet
Corvette
$ 38,990.00
15
23
2 Wheel Drive-rear
Coupe
2020
Toyota
Supra
$ 52,777.00
24
31
SUV 2021
Volkswagen
Atlas
$ 43,320.00
22
Sedan
BMW
M340i
$ 50,998.00
30
Coupe 2020 Chevrolet
Camaro LT
$ 27,996.00
19
29
Coupe 2020 BMW
M8
$ 119,994.00
Coupe 2020
Nissan
GT-R
$
10
Sports Car
2006
Maserati
BirdCage 75th
$ 3,000,000.00
12
Before Adding Outlier DATA
Mean
$ 56,813.70
25.5
Median
$ 49,949.00
17.5
22.5
STD:
33061.4874137736
4.18993503
5.5226805086
After Adding Outlier Data
Mean:
$ 324,376.09
17.9090909091
24.2727272727
Median: $ 50,998.00 16 22
STD:
887958.174169195
5.375026427
6.6346199453
56,813,70
Average under 6
p
0.6
q
0.4
P(x=4)
11.15%
Exactly 4
P(x<5) = P (x ≤ 5-1) = P(x ≤ 4)
16.62%
Fewer than 5
P (x > 6)=1-p(x≤ 6)=1-p(x≤ 6)
38.23%
More than 6
P(x≥4)=1-P(x≤ 4-1)=1-P(x≤ 3)
94.52%
At least 4
Week 4
New SD
16530.7437068868
1 P(X< 56,313) mean minus $500
0.4879351629
(= 48.79%)
2 P(X>57,813) mean Add $1000
0.4758982201
(=47.58%)
3 P(X=56,813)
0.0000241334
(= 0%)
4 P(55,313 0.0723008289
(=7%)
$ 55,313.70
(mean minus 1500)
$ 58,313.70
(mean add 1500)
MEAN = 56,813.70
T CONFIDENCE INTERVAL
PROPORTIONAL CONFIDENCE INTERVAL
SD = 33061.48742
T=
2.2621571628
Z- critical values
1.9599639845
n = 10
SD/√n=
10454.9603060514
p-hat * q-hat=
0.24
P = 0.6
95% CONFIDENCE LEVEL=
23650.7633431052
/n
0.024
Q = 0.4
UPPER CONFIDENCE LEVEL=
$ 80,464.46
SQRT (0.024)
0.1549193338
LOWER CONFIDENCE LEVEL=
$ 33,162.94
Z =
0.3036363149
ALPHA = α (1-0.25=0.975)
0.9036363149
90%
degrees of freedom (DF) (10-1=9)
0.2963636851
30%
Sheet 2 (W_O Supercar)
W/O SuperCar
Mean
57693
Standard Error
11647.587009286
Median
50998
Mode
ERROR:#N/A
Standard Deviation
34942.7610278581
Sample Variance
1220996548.25
Kurtosis
0.1230751271
Skewness
1.0471267667
Range
103203
Minimum
16791
Maximum
119994
Sum
519237
Count
Confidence Level(95.0%)
26859.3838085639
Sheet 3 (With SuperCar)
With Supercar
Mean
351923.7
Standard Error
294415.077482457
Median
51887.5
Standard Deviation
931022.222339516
Sample Variance
866802378490.011
Kurtosis
9.965500215
Skewness
3.1549042033
Range
2983209
Maximum
3000000
Sum
3519237
Count 10
Confidence Level(95.0%)
666013.176362728
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