Answer:
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Metropolitan Research, Inc., a consumer research organization, conducts surveys designed to evaluate a wide variety of products and services available to consumers. In one particular study, Metropolitan looked at consumer satisfaction with the performance of automobiles produced by a major Detroit manufacturer. A questionnaire sent to owners of one of the manufacturer's full-sized cars reviewed several complaints about early transmission problems. To learn more about the transmission failures, Metropolitan used a sample of actual transmission repairs provided by a transmission repair firm in the Detroit area. The following data show the actual number of miles driven for 50 vehicles at the time of transmission failure:
Miles
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
4.) How many repair records should be sampled if the research firm wants the population mean number of miles driven until transmission failure to be estimated with a margin of error of 5000 miles? Use 95% confidence.
answer : 1) Min value = 25070 1st quarter = 60420,Median = 72700,Mean = 73340,3rd quarter = 86580,Max value = 138100
2) ( 66438.73, 80241.87 )
3) The statistical results shows that some owners of the automobiles experienced early transmission failures
4) 95
Step-by-step explanation:
1) The transmission failure ( using descriptive statistics )
Min = 25070
1st quarter = 60420
Median = 72700
Mean = 73340
3rd quarter = 86580
Max = 138100
2) managerial interpretation of interval estimate
given 95% confidence interval
a = 0.05
z(0.025) = 1.96
Hence the 95% confidence interval for the mean number of miles driven until transmission failure
= xbar +/- Z*s/vn
⇒ 73340.3 +/- 1.96*24898.72 * [tex]\sqrt{50}[/tex]
= ( 66438.73, 80241.87 )
3) The findings shows that some owners of the automobiles experienced early transmission failures
4 ) number of repair records
n = ( Z * S/E) ^2 ------- 1
Where : Z = 1.96, S = 24898.72, E = 5000
= 1.96 * 24898.72 / 5000 ) ^2
= 95
