TMA03 - Question 3

This document provides the solution to Question 3 of TMA03, which involves calculating the mean, median, and standard deviation of spam email messages. It also includes hypothesis testing for the difference in median and the confidence interval for the number of spam email messages.

TMA03 - Question 3
3.
a. i. Mean number of spam emails =
x
¿
=12+15+202+2+18+20+8
7 =277
7 =39 . 57
Rounding off to nearest tenth, the mean number of spam emails = 40
ii. We arrange the spam emails in ascending order: 2, 8, 12, 15, 18, 20, and 202.
Number of observations = 7 => odd number of observations.
Hence, Median = the middle most observation = 15.
Using the formula for Median = ( n+1
2 )
th
observation
=4th observation, Median = 15.
iii. Median is less than mean and the distribution is highly left or negatively skewed.
This happens due to presence of outlier values. Here, the observation 202 of spam
emails is an unusual or outlier observation, which affects the mean of the distribution.
Therefore, Mean and Median are very different in this case (Doane, & Seward, 2011,
p.3-15).
iv. We considered that xi' s denote the number of spam emails. Standard deviation or
Root Mean Square deviation is calculated as
SD= √ ∑ ( xi−x
¿
)
2
n = √ ∑ ( xi
2 )
n − ( x
¿
)
2
= √ ( 122+152+ 2022+22+182+202+ 82 )
7 −39 . 572
 SD = 71.88 ¿ 72
 Hence, a standard deviation of 72 spam emails was evaluated.
b. According to the problem, Median = 15 phishing attacks/month
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Members whose email accounts were monitored = N = 20.
Recorded phishing attacks are recorded in ascending order as:
4 8 8 10 12 12 12 12 13 13 14 16
16 16 17 18 19 20 20 21
Approach A: Hypothesis testing for difference in Median
Let, M d be the difference between sample median and hypothesized median.
Null hypothesis: H0 : ( M d =0 )
Alternate hypothesis: H A : ( M d≠0 )
The difference between population median and the sample medians was noted to
follow a symmetric binomial distribution (according to the problem). Now, using the
normal approximation to the binomially distributed deviations, the Z-test is used for
assessing the hypotheses.
The population distribution for deviation from the median has been used to calculate
the mean and the standard deviation (population). Mean for deviation from median
M d=∑ xf
∑ f =0 and the standard deviation has been calculated as
σ = √ ∑ x2 f
∑ f −Md
2= √ 4 . 94
0 . 996 =2 . 23
Arranging the observations in ascending order, the median of the sample is evaluated
as 13.5.
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Table 1: Excel output for calculation of Mean and SD
X f c.f xf x^2f
-7 0.001 0.001 -0.007 0.049
-6 0.005 0.006 -0.03 0.18
-5 0.015 0.021 -0.075 0.375
-4 0.035 0.056 -0.14 0.56
-3 0.074 0.13 -0.222 0.666
-2 0.12 0.25 -0.24 0.48
-1 0.16 0.41 -0.16 0.16
0 0.176 0.586 0 0
1 0.16 0.746 0.16 0.16
2 0.12 0.866 0.24 0.48
3 0.074 0.94 0.222 0.666
4 0.035 0.975 0.14 0.56
5 0.015 0.99 0.075 0.375
6 0.005 0.995 0.03 0.18
7 0.001 0.996 0.007 0.049
0 0.996 7.968 0.0000 4.94
Therefore the z-statistic is calculated as z=(13 .5−15)−0
2. 23 =−0 . 673
We noted that at 20% confidence level or 80% significance level the p-value is
evaluated using NORM.S.DIST function in MS Excel as 0.25 < 0.8 (significance
level). Calculated p-value is less than the level of significance. Hence, we conclude
that Median = 13.5 for the samples was an enough evidence to say that the 20 sample
observations had median significantly different from 15.
Approach B: Hypothesis testing for Median
Null hypothesis: H0 : ( M =15 )
Alternate hypothesis: H A : ( M ≠15 )
Let, s = the LARGER count of
1. Observations less than the median value (11)
2. Observation greater than the median value (9)
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The difference between population median and the sample medians was noted to
follow a Binomial distribution (according to the probability distribution of the
problem).
Now, using the normal approximation to the binomially distributed deviations, the
Sign test is used for assessing the null and the alternate hypotheses (Pagano,
Gauvreau, & Gauvreau, 2018, p.300-330).
The mean for the count of deviations is calculated as n∗p=20∗0 .5=10 .
The standard deviation is calculated as √ p ( 1− p ) n=0 .5∗√ n=0 .5∗√ 20=2 . 236
The test statistic is calculated as
zcal= s−0 .5∗n
0 .5 √ n =11−10
2. 236 =0 . 447
The critical value is calculated for 20% confidence limit as zcrit = 0.2533. Hence,
zcal> zcrit implied that z cal falls in the critical region.
Hence, at 20% confidence level the null hypothesis is rejected, concluding that 11
observations greater than median was enough evidence to say that the 20 sample
observations had median significantly different from 15.
Taking the significance level at 20%, the confidence limit is calculated as
zcrit =1. 281 . In that case the null hypothesis cannot be rejected, concluding that 11
observations greater than median will not provide enough evidence to say that the 20
sample observations had median significantly different from 15.
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c. 95.46% of all email accounts have a daily number of spam email messages that lie
between the values of _____4.2______ and ________12.6________.
Explanation:
For 95.46% probability of all emails, the distribution is assumed to be normally
distributed with mean = 8.4 and S.D = 2.1.
Using standard normal table, P ( −2≤Z≤2 ) =0 . 9546
Hence, the confidence limits will be within 2 standard deviations of the mean
(Chatfield 2018, p.106-126). Hence, the confidence interval will be
x
¿
±2 s=8 . 4±2∗2. 1=[ 4 . 2, 12. 6 ] .
With 95.46% confidence we can say that email accounts have a daily number of
spam email messages approximately between 4 and 13.
References
Chatfield, C 2018, Statistics for Technology : A Course in Applied Statistics, Third Edition,
Routledge, New York, pp. 106-126, retrieved February 21, 2019, from
<https://www.taylorfrancis.com/books/9781351414081>.
Doane, DP & Seward, LE 2011, ‘Measuring Skewness: A Forgotten Statistic?’, Journal of
Statistics Education, vol. 19, no. 2, p. 3-15, retrieved from
<https://doi.org/10.1080/10691898.2011.11889611>.
Pagano, M, Gauvreau, K & Gauvreau, K 2018, Principles of Biostatistics, Chapman and
Hall/CRC, New York, pp.300-330, retrieved February 21, 2019, from
<https://www.taylorfrancis.com/books/9780429952463>.
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