Develop a 95% confidence interval for the mean number of miles driven until transmission failure for the population of automobiles with transmission failure. Provide a managerial interpretation of the interval estimate. 3. Discuss the implication of your statistical findings in terms of the belief that some owners of the automobiles experienced early transmission failures. 4. How many repair records should be sampled if the research firm wants the population mean number of miles driven until transmission failure to

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.

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

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