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
To calculate for 95% confidence
1-0.95=0.05
Calculate the confidence interval (NEEDS TO BE DIVIDED BY 2)
0.05/2= 0.025
***IMPORTANT*** (I missed this the first calculation around, you will get a very wrong negative number if you do not remember the more than L)
-Change to less than equation for Excel function
1-0.025= 0.975. α =0.975
Calculate degrees of freedom (DF)
Take N-1
10-1=9
Calculate for T-critical value:
=T.INV(0.975,9) = 2.26
T-critical value = 2.26
At this point, we have everything we need to plug it into excel – YAY!
????̅= Mean (40,589)
????= .975
N-1= 9
????????= 15768.88432
N= 10
T-crit= 2.26
Confidence Level @ 95%= 11280.3803 (Use your toolpack to get this number) OR wait for a detailed explanation below.
FINALLY (use the given equation from given pdf).
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
calculate the success rate (p) and failure (q)
P=0.7
Q=0.3
Calculate Z critical value of 95% confidence
α =0.975
=NORM.S.INV(0.975)= 1.96
Z critical value=1.96
Now, you have everything necessary to plug into the given equation. Go for it!
.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!
????̅= Mean (56,813.70)
????= T.INV( .975, 9)
????????= 33061.48742
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%
Sheet3
Price
Mean 40588.7
Standard Error 4986.5590608934
Median 37423
Mode ERROR:#N/A
Standard Deviation 15768.8843193733
Sample Variance 248657712.677778
Kurtosis -0.7745719499
Skewness 0.5873946372
Range 48015
Minimum 19866
Maximum 67881
Sum 405887
Count 10
Confidence Level(95.0%) 11280.3802973162
Sheet5
Sheet1
Vehicle type/Class Year Make Model Price MPG (City) MPG (Hwy) Quan (cylinders)
1 SUV 2019 Toyota 4Runner 34,565 17 20 6
2 SUV 2020 Toyota Rav4 28,517 26 35 4
3 SUV 2019 Ford Expidition 60,247 17 24 6
4 Truck 2020 Ram Ram 2500 40,437 20 25 8
5 Truck 2019 Ford F-250 40,281 15 16 8
6 SUV 2018 Kia Sorento 19,866 21 28 4
7 SUV 2019 Dodge Durango 32,946 19 26 6
8 SUV 2020 GMC Yukon 55,349 15 22 8
9 SUV 2020 Chevy Equinox 25,798 26 31 4
10 SUV 2020 Land Rover Range Rover 67,881 22 28 4
QUAL QUAN QUAL QUAL QUAN QUAN QUAN QUAN
Mean: 40,589 Mean: 40,589
Median: 37,423 Median: 37,423
SD: 15768.8843193733 SD: 15768.8843193733
Sample Size: 10 Sample Size: 10
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 0%
4 P(39,089
.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 Type/Class Year Make Model Price MPG (City) MPG (Highway) Drive Type
Qualitative Quantitative Qualitative Qualitative Quantitative Quantitative Quantitative Qualitative
SUV 2021 Hyundai Genesis GV80 $ 48,900.00 20 21 All Wheel Drive
Hybrid/SUV 2021 Lexus GX $ 58,665.00 16 20 All Wheel Drive
Coupe 2017 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 2 Wheel Drive-rear
SUV 2021 Volkswagen Atlas $ 43,320.00 16 22 2 Wheel Drive-Front
Sedan 2020 BMW M340i $ 50,998.00 22 30 All Wheel Drive
Coupe 2020 Chevrolet Camaro LT $ 27,996.00 19 29 2 Wheel Drive-rear
Coupe 2020 BMW M8 $ 119,994.00 15 21 All Wheel Drive
Coupe 2020 Nissan GT-R $ 109,706.00 16 22 All Wheel Drive
Sports Car 2006 Maserati BirdCage 75th $ 3,000,000.00 7 12 All Wheel Drive
Before Adding Outlier DATA
Mean: $ 56,813.70 19 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