Two peer replies, must be 100 words each
I’m working on a Mathematics exercise and need support.
Instructions: Responses should be a minimum of 100 words and may include direct questions. In the peer responses, pick a confidence level other than 95% , 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 my results to that of the initial 95% of the two peers, 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?
Peer 1 (Stacia ):
T-Critical Value
To find the 95% T-Critical Value, I first found 1-.95, which was .05. Since we have to both add and subtract from .05, I divided .05/2 = .025. I then subtracted .025 from 1, to get a probability of .975. Then I needed to find the degree of freedom. The degrees of freedom, which is our sample size minus 1. (10-1=9). Now that I have found those two, I can insert them into my formula in Excel.
=T.INV(.975,9)= 2.26215716
I then plug this number into the equation: ̅x± T∗ ( SD/√ N)
37,149.40 + 2.26215716 * (10038.59662/3.16227766) = $44,330
37,149.40 – 2.26215716 * (10038.59662/3.16227766) = $29,968
This concludes that there is a 95% confidence that the sample price of cars will be between $29,968 and $44,330.
Z- Critical Value
For the Z Critical Value, I need to use my p and q from week 3.
p = 0.60
q = 0.40
To find the 95% confidence interval, I will subtract .95 from 1, to give me .05. Since we will be adding and subtracting, I will divide .05/2, to give me .025. In order to use Excel, I need to subtract 1-.025, which gives me .975. The formula I will use in Excel is =NORM.S.INV(.975). This gives me a Z-Critical Value of 1.96 (rounded).
Now, I will plug everything into the equation: ̂p± Z∗ (√ ̂p∗ ̂q /n )
.60 + 1.96*(√.6*.4/10) = 90.36418943, or 90.4%
.60 – 1.96 *(√.6*.4/10) = 29.63581057, or 29.6%
This means that there is a 95% confidence that proportion of cars sampled that falls below the average goes from 29.6% to 90.3%.
I was not surprised by the intervals. After looking through all of my data from the car prices, it all made sense.
Peer 2 ( Martine):
1. T-Critical Value: In order to find the 95% T-Critical Value, I subtracted .95 from 1 to get .05. then divided that by 2 to receive .025, and after subtracting that from 1, I end with a final result of .975. In order to find the degree of freedom, I subtracted 1 from the sample size of 10, resulting in a total of 9.
Using this data within Excel, the formula then looks like this =T.INV(0.975,9) = 2.2621571628
I then take the number and insert it, along with other data from week 1 into the following formula: ̅x± T∗ ( SD/√ N) which ends up looking like so:
a. 41,117+2.2621571628*(16,398/√10) = 52847.42254411
and
b. 41,117-2.2621571628*(16,398/√10) = 29386.57745588
Therefore, there is a 95% confidence that the sample price of vehicles will be between $29,386.57 and $52,847.42.
2. Z-Critical Value: To find this, we’ll look back to week 3 when we found these values:
Success (p) = 0.6
Failure (q) = 0.4
I will use the same intro as I did for the T-Critical Value but I will switch the Excel formula to Norm.S.Inv and end with =NORM.S.INV(0.975), giving us a result of 1.9599639845.
Now I will plug all of my values into the following formula ̂p± Z∗ (√ ̂p∗ ̂q /n ) which gives us:
a. 0.6+1.9599639845*(√0.6*0.4/10) = 0.9036
and
b. 0.6-1.9599639845*(√0.6*0.4/10) = 0.2964
Therefore, there is a 95% confidence that the proportion of cars sampled which fall below the average ranges from 90.36% and 29.64%.
I wouldn’t say I felt surprised by any of these results, as I worked the data and equations it kind of all seemed to fall in line.
