Question

2
Question # 1 (15%)
An experiment is conducted to observe the room temperature. Assume the temperature can vary
from 0 oF to 100 oF. A collection of 6 measurements of temperature in a classroom during May is
given below.
Meas. # 1 2 3 4 5 6
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Temperature 64 59 61 58 66 60 oF
(a) Define the sample space for the measurements.
(b) Find at least two statistical measures for location. Indicate which one is preferred.
(c) Find the variance and standard deviation.
(d) Construct the boxplot and indicate 1) the normal range of data and 2) if there is any outlier.

Answer 1

Answer a:
The Sample Space for the measurements is any value in the interval [0, 100] (inclusive)

Answer b:

Measures for Location are –

· Median - Is the centre most value of any data set

· Quartiles – Quartiles divide a data set into 4 or more equal parts. Some Quartiles are – Lower Quartile (median of the first half of the data set), Upper Quartile (median of the second half of the data set) and Median, also called the second quartile

· Percentiles and Deciles are also used as measures to indicate a value below which a given percentage of observations fall in a data set

Median is the most commonly used as a measure of location because it is a measure which is not affected by outliers in the data set

Answer c:

Let the values of temperatures be represented by x

The following table shows all the calculations –

Measurements	Temperatures (x)	x^2
1	64	4096
2	59	3481
3	61	3721
4	58	3364
5	66	4356
6	60	3600
Total	368	22618

Total number of observations, n = 6

Mean of the temperatures, = 368/6 = 61.333 oF

Variance of the temperatures = {(x^2) - (n.^2)} / (n - 1) = (22618 – (6 x 61.333 x 61.333)) / 5 = 9.5157 oF

Standard Deviation = Variance^0.5 = 3.0847 oF

Answer d:
Arranging all the measurements in Ascending Order –

58, 59, 60, 61, 64, 66

Median of the measurements = (3^rd + 4^th ) Measurements / 2 = 60.5

Lower Quartile / First Quartile is the mean of the first half of the data set, that is, the first 3 measurements

Lower Quartile = 2^nd Measurement = 59

Upper Quartile / Third Quartile is the mean of the next half of the data set, that is, the last 3 measurements

Upper Quartile = 2^nd Measurement

Taking 61 as the first measurement, Upper Quartile = 64

The Minimum Value of the Measurement Set is 58 and Maximum Value is 66

The Box – Plot is given below –

1. The Normal Range = Maximum Value – Minimum Value = 66 – 58 = 8

2. To check for outliers, we need to find the range / fences of the measurement set.

The Inter Quartile Range, IQR = Upper Quartile – Lower Quartile = 64 – 59 = 5

To find the fences, we need to multiply 1.5 to the Interquartile Range. Let p = 1.5 x IQR. Then we add this value to the Upper Quartile and also, we subtract p from the Lower Quartile. In this way we get the Upper Fence and Lower Fence of the measurement set respectively

Therefore, p = 1.5 x IQR = 1.5 x 5 = 7.5

Lower Fence = Lower Quartile – p = 59 – 7.5 = 51.5

Upper Fence = Upper Quartile + p = 64 + 7.5 = 71.5

If all the values of the measurement set lie inside the interval [51.5, 71.5] then there are no outliers in the set

Since, all the measurements lie inside the specified interval, there are no outliers