Respuesta :
The table for the sales data is missing, so i have attached it.
Also, the last part of the question is incomplete.
The complete part 3 is;
3) How many repair records should be sampled if the research firm wants the population men's number of miles drive until, the transmission failure to be estimated with a margin of error of 5,000 miles? Use 95% confidence.
Answer:
A) (66438.35, 80241.65)
B) there's very likely to be a genuine case of early transmission failures, that needs be further investigated.
C) N = 96
Step-by-step explanation:
From the table, the values of the miles driven are;
85092
39323
64342
74276
74425
37831
77539
32609
89641
61978
66998
67202
89341
88798
59465
94219
67998
40001
118444
73341
77437
116803
59817
72069
53500
85288
32534
92857
101769
25066
79294
138114
64090
63436
95774
77098
64544
53402
32464
65605
121352
69922
86813
85586
59902
85861
69568
35662
116269
82256
There are 50 number (n) of miles driven.
Thus;
Mean = sum of all miles driven/50
Sum of all miles driven = 85092 + 39323 + 64342 + 74276 + 74425 + 37831 + 77539 + 32609 + 89641 + 61978 + 66998 + 67202 + 89341 + 88798 + 59465 + 94219 + 67998 + 40001 + 118444 + 73341 + 77437 + 116803 + 59817 + 72069 + 53500 + 85288 + 32534 + 92857 + 101769 + 25066 + 79294 + 138114 + 64090 + 63436 + 95774 + 77098 + 64544 + 53402 + 32464 + 65605 + 121352 + 69922 + 86813 + 85586 + 59902 + 85861 + 69568 + 35662 + 116269 + 82256 = 3667015
Mean;x' = 3667015/50 ≈ 73340
From online calculations, standard deviation; s ≈ 24899
Now, margin of error is given by;
E = zs/√n
We are told confidence interval is 95%. z - value at that is 1.96 from online sources.
Thus;
E = (1.96 × 24899)/√50
E = 6901.65
Now the 95% confidence interval is given by;
(x' - E), (x' + E)
This gives;
(73340 - 6901.65), (73340 + 6901.65) = (66438.35, 80241.65)
B) From the miles driven, the median is;
(53500 + 85288)/2 = 69394
This value is less than the mean of 73340. Thus, half of the transmission failures already happened before the average running miles. Therefore, there's very likely to be a genuine case of early transmission failures, that needs be further investigated.
C) we are told to use 5000 miles and 95% confidence interval.
Thus; E = 5000 and s remains 24899
The required sample size is given by;
N = (z•s/E)²
N = (1.96 × 24899/5000)²
N ≈ 96
