Question

1. Ten students compare their monthly rent. The rents for each of the students is presented below. Y = {384,472,496,503, 527,548,549,554,683,750) a) What are the mean and the median of the rent prices? b) What are the sample variance, standard deviation, range and inter quartile range of the rent prices? c) Assume that rent prices follow a normal distribution. Given the observed sample of values and the mean and standard deviation you calculated above construct a 95% CI for the true mean of the rent prices.

Answer 1

a)

Mean:

$Mean=\frac{\sum x}{n}=\frac{5466}{10}=546.6$

Mean of the rent prices = 546.6

Median of the rent prices:

First arrange the data in increasing order.

384, 472, 496, 503, 527, 548, 549, 554, 683, 750

n = 10 which is even.

Median = [(n/2)th observation + (n/2)+1th observation] / 2

= [5th observation + 6th observation] / 2

= [527 + 548] / 2

= 1075 / 2

= 537.5

Median of the rent prices is 537.5

b)

Let x be the rent prices.

x	$(x-\bar{x})^2$ $=(x-546.6)^2$
384	26438.76
472	5565.16
496	2560.36
503	1900.96
527	384.16
548	1.96
549	5.76
554	54.76
683	18604.96
750	41371.56
	$\sum (x-\bar{x})^2=96888.4$

Sample variance:

52 Σε – 1)2 - 1 η -

$s^2=\sqrt{\frac{96888.4}{10-1}}$

$s^2=\sqrt{\frac{96888.4}{9}}$

$s^2=\sqrt{10765.3778}$

$s^2=103.76$ (Round to 2 decimal)

Sample variance of the rent prices = 103.76

Sample standard deviation:

$s=\sqrt{s^2}=\sqrt{103.76}=10.19$ (Round to 2 decimal)

Sample standard deviation of the rent prices = 10.19

Range = Maximum value - Minimum value

= 683 - 384

Range of the rent prices = 299

Interquartile range (IQR) :

IQR = Q3 - Q1

where Q3 is third quartile and Q1 is first quartile

When number of observations n is even, Q1 is simply median of lower half of data set and Q3 is median of upper half of data set.

Lower half of data set = {384, 472, 496, 503, 527}

Upper half of data set = {548, 549, 554, 683, 750}

Q1 = Median of {384, 472, 496, 503, 527} n = 5 which is odd

= (n+1/2)th observation

= (5+1)/2th observation

= 3rd observation

Q1 = 496

Q3 = Median of {548, 549, 554, 683, 750} n = 5 which is odd

= (n+1/2)th observation

= (5+1)/2th observation

= 3rd observation

Q3 = 554

Interquartile range (IQR) = Q3 - Q1

= 554 - 496

= 58

Interquartile range (IQR) of the rent prices = 58

c)

Confidence level = c = 0.95

95% confidence interval for the true mean of the rent prices is

$\bar{x}-t_c*\frac{s}{\sqrt{n}}<\mu<\bar{x}+t_c*\frac{s}{\sqrt{n}}$

where tc is t critical value for c = 0.95 and degrees of freedom = n - 1 = 10 - 1 = 9

tc = 2.262 (From statistical table of t values)

$546.6-2.262*\frac{10.19}{\sqrt{10}}<\mu<546.6+2.262*\frac{10.19}{\sqrt{10}}$

$546.6-2.262*3.222361<\mu<546.6+2.262*3.222361$

$546.6-7.28898<\mu<546.6+7.28898$

$539.311<\mu<553.889$ (Round to 3 decimal)

95% confidence interval for the true mean of the rent prices is (539.311, 553.889)