Homework Sourse2 You want to rent an unfurnished onebedroom apartment for n
2. You want to rent an unfurnished one-bedroom apartment for next semester. You take a random sample of 10 apartments advertised in the local newspaper and record the monthly rent. Here are the sample data: 500, 650, 600, 505, 450, 550, 515, 495, 650, 395 a. Find the sample mean b. Find the sample median c. Find the standard deviation d. Find the variance e. Find the range f. Construct a 95% confidence interval estimate of the population mean. (You may assume that the original population is normally distributed.) g. Use a 0.05 significance level to test the claim that the mean rent of all advertised apartments is greater than $500 per month.
Solution
a)
Getting the mean, X,
X = Sum(x) / n
Summing the items, Sum(x) = 5310
As n = 10
Thus,
X = 531 [ANSWER]
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b)
Ordering the data,
395
450
495
500
505
515
550
600
650
650
Thus, the median is the average of the 5th and 6th entries,
median = (505+515)/2 = 510 [ANSWER]
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c)
Setting up tables,
x x - X (x - X)^2
500 -31 961
650 119 14161
600 69 4761
505 -26 676
450 -81 6561
550 19 361
515 -16 256
495 -36 1296
650 119 14161
395 -136 18496
Thus, Sum(x - X)^2 = 61690
Thus, as
s^2 = Sum(x - X)^2 / (n - 1)
As n = 10
s^2 = 6854.444444
Thus,
s = 82.7915723 [ANSWER, STANDARD DEVIATION]
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d)
s^2 = 6854.444444 [ANSWER, VARIANCE]
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e)
Also, we have
Maximum = 650
Minimum = 395
Range = max - min = 255 [ANSWER, RANGE]
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