Refer to Buena School District bus data. Prepare a report on the maintance cost

Question

Refer to Buena School District bus data. Prepare a report on the maintance cost for last month. Be sure to answer the following questions in your report.

A. Around what values do the data tend to cluster? Specially what was the mean maintenance cost last month? What is the median cost? Is one measure more representative of the typical cost of others?

B. What is the range of the maintenance costs? What is the standard deviation? About 95% of the maintenance costs are between what two values?

http://highered.mheducation.com/sites/0078020522/student_view0/data_sets.html (for all data)

Maintenance cumulative lower upper midpoint width frequency percent      frequency percent 320 < 340 330 20 3   3.8   3   3.8   340 < 360 350 20 3   3.8   6   7.5   360 < 380 370 20 1   1.3   7   8.8   380 < 400 390 20 8   10.0   15   18.8   400 < 420 410 20 5   6.3   20   25.0   420 < 440 430 20 12   15.0   32   40.0   440 < 460 450 20 12   15.0   44   55.0   460 < 480 470 20 15   18.8   59   73.8   480 < 500 490 20 7   8.8   66   82.5   500 < 520 510 20 8   10.0   74   92.5   520 < 540 530 20 1   1.3   75   93.8   540 < 560 550 20 3   3.8   78   97.5   560 < 580 570 20 2   2.5   80   100.0   Descriptive statistics Maintenance count 80 mean 450.29 sample variance 2,882.18 sample standard deviation 53.69 minimum 329 maximum 570 range 241 1st quartile 420.00 median 456.00 3rd quartile 489.75 interquartile range 69.75 mode 329.00 low extremes 0 low outliers 0 high outliers 0 high extremes 0

Explanation / Answer

(A) The data tend to cluster around the values between 410 to 490.
Mean maintenance cost = 450.29
Median maintenance cost = 456
As difference between mean and median is not much we can say one of them is more representative than other.

(B)
Range of maintenance cost = 241
standard deviation = 53.69

In order to calculate 95% CI, z-value = 1.96
n = 80, s = 53.69
Margin of error = 1.96*(53.69/sqrt(80)) = 11.7653
Confidence interval = (450.29-11.7633, 450.29+11.7633)
Confidence interval = (438.5267, 462.0533)

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