Question 924119: An online used car company sells second-hand cars. For 30 randomly selected transactions, the mean price is 2100 dollars.
Part a) Assuming a population standard deviation transaction prices of 100 dollars, obtain a 99% confidence interval for the mean price of all transactions.
So my thought:
Is this the right set-up? mean(x)-/+ z*(standard deviation/sqrt(n))
where x=2100, s.d. =100 n=30 and z* being equal to 2.57?
Because I tried my set-up, and got an incorrect solution. I would be thankful for any help!
Answer by ewatrrr(24785)   (Show Source):
You can put this solution on YOUR website!
Hi
This is the right set-up: mean(x)-/+ z*(standard deviation/sqrt(n))
mean = 2100, sd = 100
I. try z = 2.576 0r (2.58)
2100 ± 47 rounded (Do know what rounding is preferred)
CI: ( 2053, 2147) (Use Format desired for CI)
0r
II. Possibly they are looking for the t-value to be used(DF = 29): t = 2.76
2100 ± 50
....Small Sample t-values
DF Probability, p 2-tailed
0.1 0.05 0.01 0.001
1 6.31 12.71 63.66 636.62
2 2.92 4.3 9.93 31.6
3 2.35 3.18 5.84 12.92
4 2.13 2.78 4.6 8.61
5 2.02 2.57 4.03 6.87
6 1.94 2.45 3.71 5.96
7 1.89 2.37 3.5 5.41
8 1.86 2.31 3.36 5.04
9 1.83 2.26 3.25 4.78
10 1.81 2.23 3.17 4.59
11 1.8 2.2 3.11 4.44
12 1.78 2.18 3.06 4.32
13 1.77 2.16 3.01 4.22
14 1.76 2.14 2.98 4.14
15 1.75 2.13 2.95 4.07
16 1.75 2.12 2.92 4.02
17 1.74 2.11 2.9 3.97
18 1.73 2.1 2.88 3.92
19 1.73 2.09 2.86 3.88
20 1.72 2.09 2.85 3.85
21 1.72 2.08 2.83 3.82
22 1.72 2.07 2.82 3.79
23 1.71 2.07 2.82 3.77
24 1.71 2.06 2.8 3.75
25 1.71 2.06 2.79 3.73
26 1.71 2.06 2.78 3.71
27 1.7 2.05 2.77 3.69
28 1.7 2.05 2.76 3.67
29 1.7 2.05 2.76 3.66
30 1.7 2.04 2.75 3.65
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