B.Sc. 2nd Year Applied Statistics

Easy

Example 1 — Arithmetic Mean

Moderate

Example 2 — Mean from Frequency Distribution

\[ \Sigma f=4+7+10+6+3=30 \] \[ \Sigma fX=40+140+300+240+150=870 \] \[ \bar X=\frac{\Sigma fX}{\Sigma f} \] \[ \bar X=\frac{870}{30} \] \[ \boxed{\bar X=29} \]

Hard

Example 3 — Combined Mean

\[ n_1=24 \] \[ \bar X_1=68 \] \[ n_2=36 \] \[ \bar X_2=74 \] The combined mean is: \[ \bar X_c= \frac{n_1\bar X_1+n_2\bar X_2} {n_1+n_2} \] \[ = \frac{24(68)+36(74)} {24+36} \] \[ = \frac{1632+2664}{60} \] \[ = \frac{4296}{60} \] \[ \boxed{\bar X_c=71.6} \]

Competitive

Example 4 — Finding an Unknown Observation from Mean

\[ \bar X=24 \] \[ n=5 \] Therefore, \[ \Sigma X=n\bar X \] \[ \Sigma X=5(24) \] \[ \Sigma X=120 \] The sum of the known observations is: \[ 16+21+25+29=91 \] Let the missing observation be \(x\). \[ 91+x=120 \] \[ x=120-91 \] \[ \boxed{x=29} \]

Easy

Example 5 — Median of Ungrouped Data

\[ 7,\ 12,\ 15,\ 19,\ 22 \] Since \(n=5\), \[ \text{Median position} = \frac{n+1}{2} \] \[ = \frac{6}{2}=3 \] Therefore, \[ \boxed{\text{Median}=15} \]

Moderate

Example 6 — Median of Even Number of Observations

\[ 8,\ 12,\ 15,\ 19,\ 24,\ 30 \] \[ n=6 \] The two middle observations are the 3rd and 4th observations. \[ \text{Median} = \frac{15+19}{2} \] \[ \boxed{\text{Median}=17} \]

Hard

Example 7 — Grouped Data Median

\[ N=50 \] \[ \frac N2=25 \] The 25th observation lies in the \(20-30\) class. Therefore, \[ L=20,\quad f=16,\quad CF=14,\quad h=10 \] Formula: \[ Median = L+ \left( \frac{\frac N2-CF}{f} \right)h \] \[ = 20+ \left( \frac{25-14}{16} \right)10 \] \[ = 20+\frac{110}{16} \] \[ = 26.875 \] \[ \boxed{Median=26.875} \]

Easy

Example 8 — Simple Mode

\[ 4,\ 7,\ 7,\ 8,\ 9,\ 7,\ 10,\ 12 \] The value occurring most frequently is 7. \[ \boxed{Mode=7} \]

Hard

Example 9 — Grouped Data Mode

Modal class: \[ 20-30 \] Therefore, \[ L=20 \] \[ f_1=18 \] \[ f_0=11 \] \[ f_2=14 \] \[ h=10 \] Formula: \[ Mode= L+ \frac{f_1-f_0} {2f_1-f_0-f_2} h \] \[ = 20+ \frac{18-11} {2(18)-11-14} (10) \] \[ = 20+ \frac7{11}(10) \] \[ = 26.36 \] \[ \boxed{Mode\approx26.36} \]

Moderate

Example 10 — Range and Coefficient of Range

Given: \[ 12,\ 18,\ 21,\ 25,\ 31,\ 40 \] Largest value: \[ L=40 \] Smallest value: \[ S=12 \] Therefore, \[ Range=L-S \] \[ =40-12 \] \[ \boxed{Range=28} \] Coefficient of range: \[ \frac{L-S}{L+S} \] \[ = \frac{40-12}{40+12} \] \[ = \frac{28}{52} \] \[ \boxed{0.5385} \]

Hard

Example 11 — Standard Deviation

For: \[ 2,\ 4,\ 6,\ 8,\ 10 \] \[ \bar X=\frac{2+4+6+8+10}{5}=6 \] Construct deviations: \[ \Sigma(X-\bar X)^2=40 \] For population SD: \[ \sigma= \sqrt{ \frac{\Sigma(X-\bar X)^2}{n} } \] \[ = \sqrt{\frac{40}{5}} \] \[ =\sqrt8 \] \[ \boxed{\sigma=2.828} \] For sample SD: \[ s= \sqrt{ \frac{40}{5-1} } \] \[ =\sqrt{10} \] \[ \boxed{s=3.162} \]

Competitive

Example 12 — Coefficient of Variation and Comparison

Two laboratory methods have: Method A: \[ \bar X_A=80,\quad SD_A=6 \] Method B: \[ \bar X_B=50,\quad SD_B=5 \] Calculate CV for both. For Method A: \[ CV_A= \frac{SD_A}{\bar X_A}\times100 \] \[ = \frac6{80}\times100 \] \[ \boxed{CV_A=7.5\%} \] For Method B: \[ CV_B= \frac5{50}\times100 \] \[ \boxed{CV_B=10\%} \] Thus Method A has lower relative variability.

Moderate

Example 13 — Karl Pearson's Coefficient of Skewness

Given: \[ Mean=42 \] \[ Mode=36 \] \[ SD=8 \] Formula: \[ Sk= \frac{Mean-Mode}{SD} \] \[ = \frac{42-36}{8} \] \[ =\frac68 \] \[ \boxed{Sk=0.75} \] Since skewness is positive, the distribution is positively skewed.

Hard

Example 14 — Skewness using Mean and Median

Given: \[ Mean=52 \] \[ Median=47 \] \[ SD=10 \] Pearson's second coefficient: \[ Sk= \frac{3(Mean-Median)}{SD} \] \[ = \frac{3(52-47)}{10} \] \[ = \frac{15}{10} \] \[ \boxed{Sk=1.5} \] The distribution is positively skewed.

Competitive

Example 15 — Kurtosis from Central Moments

Given: \[ \mu_2=25 \] \[ \mu_4=1875 \] Calculate the coefficient of kurtosis. \[ \beta_2= \frac{\mu_4}{\mu_2^2} \] \[ = \frac{1875}{25^2} \] \[ = \frac{1875}{625} \] \[ \boxed{\beta_2=3} \] Since: \[ \beta_2=3 \] the distribution is mesokurtic.

Bar Graph — Example 1

Example 16 — Simple Bar Graph

The horizontal axis represents laboratory groups and the vertical axis represents number of students. Equal-width bars are drawn with heights 20, 30, 25 and 15.

Bar Graph — Example 2

Example 17 — Comparative Bar Graph

Section	Male	Female
A	28	35
B	32	30
C	24	38

This is a comparative/grouped bar graph because two categories are compared within each section.

Bar Graph — Example 3

Example 18 — Biological Data

Organism	Positive samples
E. coli	45
Klebsiella	32
Staphylococcus	27
Pseudomonas	18

The tallest bar represents the organism with the greatest observed number of positive samples.

Bar Graph — Example 4

Example 19 — Interpretation Question

\[ Increase=60-35 \] \[ \boxed{Increase=25} \] Percentage increase: \[ \frac{25}{35}\times100 \] \[ \boxed{71.43\%} \]

Histogram — Example 1

Example 20 — Equal Class Intervals

Class	Frequency
0–10	5
10–20	9
20–30	14
30–40	10
40–50	6

Because the class intervals are equal, the height of each rectangle is proportional to its frequency.

Histogram — Example 2

Example 21 — Unequal Class Intervals

Class	Frequency	Class width	Frequency density
0–5	10	5	2
5–15	20	10	2
15–30	30	15	2
30–50	20	20	1

Formula: \[ Frequency\ density= \frac{Frequency}{Class\ width} \] For class \(0-5\): \[ =\frac{10}{5}=2 \] For class \(5-15\): \[ =\frac{20}{10}=2 \] The histogram should use frequency density rather than raw frequency when class widths differ.

Histogram — Example 3

Example 22 — Finding the Modal Class

Class	Frequency
10–20	6
20–30	15
30–40	24
40–50	18
50–60	7

The highest rectangle corresponds to: \[ \boxed{30-40} \] Therefore the modal class is \(30-40\).

Histogram — Example 4

Example 23 — Biological Measurement

Colony count	Frequency
0–20	4
20–40	11
40–60	19
60–80	25
80–100	16
100–120	5

Since all intervals have equal width, the frequency is used directly as the height. \[ \boxed{\text{Modal class}=60-80} \] The histogram will have adjacent rectangles without gaps.

Pie Chart — Example 1

Example 24 — Converting Frequency to Angle

A class contains 40 students:

Subject preference	Students
Microbiology	16
Chemistry	10
Botany	8
Zoology	6

For Microbiology: \[ Angle= \frac{16}{40}\times360^\circ \] \[ =144^\circ \] For Chemistry: \[ Angle= \frac{10}{40}\times360^\circ \] \[ =90^\circ \] For Botany: \[ Angle= \frac8{40}\times360^\circ \] \[ =72^\circ \] For Zoology: \[ Angle= \frac6{40}\times360^\circ \] \[ =54^\circ \] Check: \[ 144+90+72+54=360^\circ \]

Pie Chart — Example 2

Example 25 — Finding Missing Frequency

A pie chart represents 200 students. The sector for Microbiology is \(72^\circ\). Find the number of students represented. \[ Frequency= \frac{Angle}{360^\circ}\times Total \] \[ = \frac{72}{360}\times200 \] \[ =\frac15(200) \] \[ \boxed{40} \]

Pie Chart — Example 3

Example 26 — Health Data

A health survey gives:

Condition	Frequency
Condition A	45
Condition B	30
Condition C	15
Condition D	10

Total: \[ N=100 \] For Condition A: \[ Angle=\frac{45}{100}\times360^\circ \] \[ =\boxed{162^\circ} \] For B: \[ =\frac{30}{100}\times360^\circ \] \[ =\boxed{108^\circ} \] For C: \[ =\frac{15}{100}\times360^\circ \] \[ =\boxed{54^\circ} \] For D: \[ =\frac{10}{100}\times360^\circ \] \[ =\boxed{36^\circ} \]

Pie Chart — Example 4

Example 27 — Competitive Type

The total number of samples is 720. The sector for Group A is \(90^\circ\) and Group B is \(120^\circ\). Group C contains 180 samples. Find the number of samples in Group D. Group A: \[ A=\frac{90}{360}\times720 \] \[ A=180 \] Group B: \[ B=\frac{120}{360}\times720 \] \[ B=240 \] Therefore: \[ D=720-(180+240+180) \] \[ D=720-600 \] \[ \boxed{D=120} \] The angle for D is: \[ \frac{120}{720}\times360^\circ \] \[ \boxed{60^\circ} \]

Moderate

Example 28 — Karl Pearson's Correlation Coefficient

Given: \[ X:1,\ 2,\ 3,\ 4,\ 5 \] \[ Y:2,\ 4,\ 5,\ 4,\ 6 \] Use: \[ r= \frac{ n\Sigma XY-(\Sigma X)(\Sigma Y) }{ \sqrt{ [n\Sigma X^2-(\Sigma X)^2] [n\Sigma Y^2-(\Sigma Y)^2] } } \] Calculate: \[ \Sigma X=15 \] \[ \Sigma Y=21 \] \[ \Sigma X^2=55 \] \[ \Sigma Y^2=97 \] \[ \Sigma XY=67 \] Therefore, \[ r= \frac{ 5(67)-(15)(21) }{ \sqrt{ [5(55)-15^2] [5(97)-21^2] } } \] \[ = \frac{335-315} {\sqrt{(275-225)(485-441)}} \] \[ = \frac{20} {\sqrt{50(44)}} \] \[ = \frac{20}{46.904} \] \[ \boxed{r\approx0.426} \] There is a moderate positive linear relationship in this sample.

Hard

Example 29 — Spearman Rank Correlation

Suppose ranks in two examinations are: \[ X:1,2,3,4,5,6 \] \[ Y:2,1,4,3,6,5 \] Calculate \(d=X-Y\).

X	Y	d	d²
1	2	-1	1
2	1	1	1
3	4	-1	1
4	3	1	1
5	6	-1	1
6	5	1	1

\[ \Sigma d^2=6 \] \[ n=6 \] Formula: \[ r_s= 1- \frac{6\Sigma d^2} {n(n^2-1)} \] \[ = 1- \frac{6(6)} {6(36-1)} \] \[ = 1-\frac{36}{210} \] \[ \boxed{r_s=0.829} \]

Competitive

Example 30 — Spearman Rank with Tied Ranks

When tied observations occur, assign the average rank to the tied values. Suppose: \[ X=10,20,20,40,50 \] The two observations equal to 20 occupy ranks 2 and 3. Therefore their average rank is: \[ \frac{2+3}{2}=2.5 \] The same procedure is applied to all ties before calculating rank differences.

Hard

Example 31 — Kendall's Tau

For five observations, suppose the paired ranking sequence is: \[ 2,\ 1,\ 4,\ 3,\ 5 \] There are: \[ \frac{n(n-1)}2 \] possible pairs. \[ = \frac{5(4)}2 \] \[ =10 \] Count concordant and discordant pairs. For this sequence: \[ C=8 \] \[ D=2 \] Therefore, \[ \tau= \frac{C-D} {n(n-1)/2} \] \[ = \frac{8-2}{10} \] \[ \boxed{\tau=0.6} \] This indicates positive association.

Moderate

Example 32 — Simple Linear Regression

The regression equation is: \[ Y=a+bX \] Suppose: \[ a=5,\quad b=2 \] For \(X=10\): \[ Y=5+2(10) \] \[ Y=5+20 \] \[ \boxed{Y=25} \]

Hard

Example 33 — Regression Coefficients

Given: \[ \bar X=10 \] \[ \bar Y=25 \] \[ b_{YX}=1.5 \] The regression equation of \(Y\) on \(X\) is: \[ Y-\bar Y=b_{YX}(X-\bar X) \] Therefore: \[ Y-25=1.5(X-10) \] \[ Y-25=1.5X-15 \] \[ Y=1.5X+10 \] \[ \boxed{Y=10+1.5X} \] For \(X=14\): \[ Y=10+1.5(14) \] \[ Y=31 \] \[ \boxed{Y=31} \]

Competitive

Example 34 — Coefficient of Determination

Suppose: \[ r=0.82 \] Then: \[ R^2=r^2 \] \[ =(0.82)^2 \] \[ \boxed{R^2=0.6724} \] Percentage explained: \[ 0.6724\times100 \] \[ \boxed{67.24\%} \] Unexplained proportion: \[ 100-67.24 \] \[ \boxed{32.76\%} \]

Hard

Example 35 — Partial Correlation

Given: \[ r_{12}=0.75 \] \[ r_{13}=0.60 \] \[ r_{23}=0.50 \] Find \(r_{12.3}\). Formula: \[ r_{12.3} = \frac{ r_{12}-r_{13}r_{23} }{ \sqrt{ (1-r_{13}^{2})(1-r_{23}^{2}) } } \] Substitution: \[ = \frac{ 0.75-(0.60)(0.50) }{ \sqrt{ (1-0.60^2)(1-0.50^2) } } \] \[ = \frac{0.75-0.30} {\sqrt{(1-0.36)(1-0.25)}} \] \[ = \frac{0.45} {\sqrt{0.64(0.75)}} \] \[ = \frac{0.45}{\sqrt{0.48}} \] \[ \boxed{r_{12.3}\approx0.650} \]

Competitive

Example 36 — Multiple Correlation

Given: \[ r_{12}=0.70 \] \[ r_{13}=0.60 \] \[ r_{23}=0.50 \] Formula: \[ R_{1.23} = \sqrt{ \frac{ r_{12}^{2}+r_{13}^{2} -2r_{12}r_{13}r_{23} }{ 1-r_{23}^{2} } } \] Substitution: \[ R_{1.23} = \sqrt{ \frac{ 0.70^2+0.60^2 -2(0.70)(0.60)(0.50) }{ 1-0.50^2 } } \] \[ = \sqrt{ \frac{ 0.49+0.36-0.42 }{ 0.75 } } \] \[ = \sqrt{\frac{0.43}{0.75}} \] \[ \boxed{R_{1.23}\approx0.757} \]

Competitive

Example 37 — Fisher Decomposition of Total Sum of Squares

Suppose: \[ SST=500 \] and: \[ SSR=320 \] Then: \[ SST=SSR+SSE \] Therefore: \[ SSE=SST-SSR \] \[ =500-320 \] \[ \boxed{SSE=180} \] Coefficient of determination: \[ R^2=\frac{SSR}{SST} \] \[ =\frac{320}{500} \] \[ \boxed{R^2=0.64} \] Thus 64% of the total variation is explained by the fitted model.

Hard

Example 38 — Exponential Curve Fitting

Suppose: \[ Y=ab^X \] Taking logarithms: \[ \log Y=\log a+X\log b \] Suppose the fitted transformed equation is: \[ \log Y=1.20+0.08X \] Then: \[ \log a=1.20 \] and: \[ \log b=0.08 \] Therefore: \[ a=10^{1.20} \] \[ a\approx15.849 \] and: \[ b=10^{0.08} \] \[ b\approx1.202 \] Thus: \[ \boxed{Y\approx15.849(1.202)^X} \] For \(X=5\): \[ Y=15.849(1.202)^5 \] \[ \boxed{Y\approx39.5} \]

Moderate

Example 39 — Expected Frequency

Suppose a contingency table has: \[ Row\ total=80 \] \[ Column\ total=50 \] \[ Grand\ total=200 \] Then: \[ E= \frac{Row\ total\times Column\ total} {Grand\ total} \] \[ = \frac{80(50)}{200} \] \[ \boxed{E=20} \]

Hard

Example 40 — Yule's Q

Given a fourfold table: \[ a=40,\quad b=10,\quad c=5,\quad d=30 \] Formula: \[ Q= \frac{ad-bc} {ad+bc} \] \[ = \frac{(40)(30)-(10)(5)} {(40)(30)+(10)(5)} \] \[ = \frac{1200-50}{1200+50} \] \[ = \frac{1150}{1250} \] \[ \boxed{Q=0.92} \] This indicates strong positive association.

Competitive

Example 41 — Yule's Coefficient of Colligation

Given: \[ a=40,\quad b=10,\quad c=5,\quad d=30 \] Formula: \[ Y= \frac{\sqrt{ad}-\sqrt{bc}} {\sqrt{ad}+\sqrt{bc}} \] \[ = \frac{ \sqrt{(40)(30)}-\sqrt{(10)(5)} }{ \sqrt{(40)(30)}+\sqrt{(10)(5)} } \] \[ = \frac{\sqrt{1200}-\sqrt{50}} {\sqrt{1200}+\sqrt{50}} \] \[ = \frac{34.641-7.071} {34.641+7.071} \] \[ \boxed{Y\approx0.661} \]

Hard

Example 42 — Pearson Coefficient of Contingency

Suppose: \[ \chi^2=18 \] and: \[ N=200 \] Formula: \[ C= \sqrt{ \frac{\chi^2} {\chi^2+N} } \] \[ = \sqrt{ \frac{18}{18+200} } \] \[ = \sqrt{\frac{18}{218}} \] \[ \boxed{C\approx0.287} \]

Easy

Example 43 — Simple Probability

A box contains 5 red and 3 blue balls. \[ N=8 \] Probability of red: \[ P(R)=\frac58 \] \[ \boxed{P(R)=0.625} \]

Moderate

Example 44 — Addition Theorem

Suppose: \[ P(A)=0.40 \] \[ P(B)=0.35 \] \[ P(A\cap B)=0.10 \] Then: \[ P(A\cup B) = P(A)+P(B)-P(A\cap B) \] \[ =0.40+0.35-0.10 \] \[ \boxed{P(A\cup B)=0.65} \]

Hard

Example 45 — Conditional Probability

Suppose: \[ P(A\cap B)=0.18 \] and: \[ P(B)=0.30 \] Then: \[ P(A|B) = \frac{P(A\cap B)}{P(B)} \] \[ = \frac{0.18}{0.30} \] \[ \boxed{P(A|B)=0.60} \]

Competitive

Example 46 — Bayes' Theorem

A disease occurs in 1% of the population. \[ P(D)=0.01 \] Sensitivity: \[ P(+|D)=0.90 \] False-positive probability: \[ P(+|\bar D)=0.05 \] Then: \[ P(D|+) = \frac{ P(+|D)P(D) }{ P(+|D)P(D)+P(+|\bar D)P(\bar D) } \] Since: \[ P(\bar D)=1-0.01=0.99 \] Therefore: \[ P(D|+) = \frac{(0.90)(0.01)} {(0.90)(0.01)+(0.05)(0.99)} \] \[ = \frac{0.009} {0.009+0.0495} \] \[ = \frac{0.009}{0.0585} \] \[ \boxed{P(D|+)=0.1538} \] Thus: \[ \boxed{15.38\%} \]

Easy

Example 47 — Binomial Probability

Given: \[ n=5 \] \[ p=0.4 \] \[ q=1-p=0.6 \] Find: \[ P(X=2) \] Formula: \[ P(X=x)= {n\choose x}p^xq^{n-x} \] Therefore: \[ P(X=2) = {5\choose2} (0.4)^2(0.6)^3 \] \[ = 10(0.16)(0.216) \] \[ \boxed{P(X=2)=0.3456} \]

Hard

Example 48 — At Least Two Successes

Suppose: \[ n=10 \] \[ p=0.30 \] Find: \[ P(X\ge2) \] Use complement: \[ P(X\ge2)=1-P(X<2) \] \[ =1-[P(X=0)+P(X=1)] \] Now: \[ P(X=0) = {10\choose0}(0.3)^0(0.7)^{10} \] \[ \approx0.02825 \] And: \[ P(X=1) = {10\choose1}(0.3)(0.7)^9 \] \[ \approx0.12106 \] Therefore: \[ P(X\ge2) = 1-(0.02825+0.12106) \] \[ \boxed{P(X\ge2)\approx0.85069} \]

Competitive

Example 49 — Binomial Mean and Variance

Given: \[ n=80,\quad p=0.25 \] Mean: \[ \mu=np \] \[ =80(0.25) \] \[ \boxed{\mu=20} \] Variance: \[ \sigma^2=npq \] where: \[ q=0.75 \] Therefore: \[ \sigma^2=80(0.25)(0.75) \] \[ \boxed{\sigma^2=15} \] and: \[ \sigma=\sqrt{15} \] \[ \boxed{\sigma\approx3.873} \]

Easy

Example 50 — Poisson Probability

Given: \[ \lambda=3 \] Find: \[ P(X=2) \] Formula: \[ P(X=x)= \frac{e^{-\lambda}\lambda^x}{x!} \] Therefore: \[ P(X=2) = \frac{e^{-3}3^2}{2!} \] \[ = \frac{e^{-3}(9)}2 \] \[ \boxed{P(X=2)\approx0.2240} \]

Hard

Example 51 — No Events

If the average number of mutations observed in a particular interval is: \[ \lambda=2.5 \] Find the probability of observing no mutations. \[ P(X=0) = \frac{e^{-2.5}(2.5)^0}{0!} \] \[ =e^{-2.5} \] \[ \boxed{P(X=0)\approx0.0821} \]

Competitive

Example 52 — At Least One Event

If: \[ \lambda=4 \] Find: \[ P(X\ge1) \] Use complement: \[ P(X\ge1)=1-P(X=0) \] \[ =1-e^{-4} \] \[ \boxed{P(X\ge1)\approx0.9817} \]

Normal Distribution

Example 53 — Pressure Greater Than 125 mmHg

Let \(X\) be the random variable following the Normal distribution with: \[ \mu=115 \] \[ \sigma=10 \] Therefore: \[ X\sim N(115,10^2) \] Let: \[ Z=\frac{X-\mu}{\sigma} \] Therefore: \[ Z=\frac{X-115}{10} \] be the standard Normal variate: \[ Z\sim N(0,1) \] ### (a) Pressure greater than \(125\) mmHg \[ Z= \frac{125-115}{10} \] \[ =1 \] Therefore: \[ P(X>125)=P(Z>1) \] \[ =0.5-P(0125) = 0.5-0.3413 \] \[ \boxed{P(X>125)=0.1587} \] ### (b) Pressure less than \(125\) mmHg \[ P(X<125)=P(Z<1) \] \[ =0.5+0.3413 \] \[ \boxed{P(X<125)=0.8413} \] ### (c) Pressure below which 55% of the population lies Let the required value be \(X_1\). \[ P(X

Normal Distribution

Example 54 — Probability Between Two Values

Let: \[ X\sim N(70,8^2) \] Find: \[ P(62

Competitive Normal Distribution

Example 55 — Central 90% Range

Suppose: \[ X\sim N(100,15^2) \] Find the values that contain the central 90% of observations. The central area is: \[ 0.90 \] Therefore the two tails contain: \[ 1-0.90=0.10 \] Each tail: \[ 0.05 \] From the Normal table: \[ P(0

Competitive Normal Distribution

Example 56 — Finding Cut-off from Percentage

Suppose: \[ X\sim N(115,10^2) \] Find the pressure below which 90% of the population lies. Given: \[ P(X

Moderate

Example 57 — Confidence Interval for Mean with Known SD

Given: \[ \bar X=80 \] \[ \sigma=12 \] \[ n=36 \] At 95% confidence: \[ Z_{\alpha/2}=1.96 \] Standard error: \[ SE= \frac{\sigma}{\sqrt n} \] \[ = \frac{12}{6} \] \[ =2 \] Margin of error: \[ E=1.96(2) \] \[ =3.92 \] Confidence interval: \[ \bar X\pm E \] \[ =80\pm3.92 \] \[ \boxed{76.08<\mu<83.92} \]

Hard

Example 58 — Confidence Interval for Proportion

Given: \[ \hat p=0.60 \] \[ n=100 \] \[ q=1-p=0.40 \] At 95% confidence: \[ Z=1.96 \] Standard error: \[ SE= \sqrt{\frac{pq}{n}} \] \[ = \sqrt{\frac{(0.60)(0.40)}{100}} \] \[ =\sqrt{0.0024} \] \[ \approx0.04899 \] Margin: \[ E=1.96(0.04899) \] \[ \approx0.0960 \] Therefore: \[ CI=0.60\pm0.096 \] \[ \boxed{0.504< p <0.696} \]

Competitive

Example 59 — Sample Size for Mean

The desired margin of error is \(2\). Population SD is estimated as 10. Find the required sample size for 95% confidence. Formula: \[ n= \left( \frac{Z\sigma}{E} \right)^2 \] \[ = \left( \frac{1.96(10)}2 \right)^2 \] \[ =(9.8)^2 \] \[ =96.04 \] Since sample size must be an integer: \[ \boxed{n=97} \]

Competitive

Example 60 — Sample Size for Proportion

For 95% confidence, estimate a population proportion with margin of error 0.05. If no prior estimate of \(p\) is available, use \(p=0.5\). \[ p=0.5 \] \[ q=0.5 \] \[ E=0.05 \] Formula: \[ n= \frac{Z^2pq}{E^2} \] \[ = \frac{(1.96)^2(0.5)(0.5)} {(0.05)^2} \] \[ = \frac{3.8416(0.25)} {0.0025} \] \[ = 384.16 \] Therefore: \[ \boxed{n=385} \]

Hypothesis Testing 1 — Easy

Example 61 — Single Mean Z Test

Given: \[ \bar X=120 \] \[ \mu_0=115 \] \[ \sigma=10 \] \[ n=25 \] ### Test hypothesis Null hypothesis: \[ H_0:\mu=115 \] i.e. there is no significant difference between the population mean and 115. Alternative hypothesis: \[ H_1:\mu\ne115 \] i.e. there is a significant difference. ### Under null hypothesis Test statistic: \[ Z= \frac{\bar X-\mu_0} {\sigma/\sqrt n} \] \[ = \frac{120-115} {10/\sqrt{25}} \] \[ = \frac5{10/5} \] \[ =\frac52 \] \[ \boxed{Z_{\text{cal}}=2.50} \] ### Level of significance \[ \alpha=5\%=0.05 \] Since the alternative hypothesis is two-tailed, use: \[ Z_{\text{tab}}=\pm1.96 \] from the Standard Normal Z table. ### Decision \[ |Z_{\text{cal}}|=2.50 \] \[ 2.50>1.96 \] Therefore, reject \(H_0\). ### Conclusion There is a significant difference between the population mean and 115 at the 5% level of significance.

Hypothesis Testing 2 — Hard

Example 62 — One-Sample Z Test, Left-Tailed

A filling machine is designed to fill bottles with an average of 500 mL. Population SD is 20 mL. A sample of 64 bottles has mean 495 mL. Test whether the machine is under-filling at 5% significance. \[ H_0:\mu=500 \] \[ H_1:\mu<500 \] This is a left-tailed test. Test statistic: \[ Z= \frac{495-500} {20/\sqrt{64}} \] \[ = \frac{-5}{20/8} \] \[ = \frac{-5}{2.5} \] \[ \boxed{Z_{\text{cal}}=-2.00} \] At 5% left-tailed level: \[ Z_{\text{tab}}=-1.645 \] Since: \[ -2.00<-1.645 \] reject \(H_0\).

Hypothesis Testing 3 — Competitive

Example 63 — One-Sample Z Test with Decision

A biological assay is expected to have mean response 75 units. The population SD is known to be 12. A sample of 144 observations has mean response 77.1. Test at 1% significance whether the mean response differs from 75. \[ H_0:\mu=75 \] \[ H_1:\mu\ne75 \] \[ Z= \frac{77.1-75} {12/\sqrt{144}} \] \[ = \frac{2.1}{12/12} \] \[ =2.1 \] \[ \boxed{Z_{\text{cal}}=2.10} \] At 1% significance for a two-tailed test: \[ Z_{\text{tab}}=\pm2.576 \] Since: \[ 2.10<2.576 \] do not reject \(H_0\).

Hypothesis Testing 4 — Competitive

Example 64 — Very Large Sample

A population has known SD \(=15\). A sample of \(225\) observations has mean \(103\). Test: \[ H_0:\mu=100 \] against: \[ H_1:\mu>100 \] at 5%. \[ Z= \frac{103-100} {15/\sqrt{225}} \] \[ = \frac3{15/15} \] \[ =3 \] At 5% right-tailed: \[ Z_{\text{tab}}=1.645 \] Since: \[ 3>1.645 \] reject \(H_0\). \[ \boxed{\text{There is significant evidence that }\mu>100.} \]

Hypothesis Testing 5 — Competitive

Example 65 — Biological Measurement

The known population SD of a laboratory measurement is 4. A sample of 49 patients has mean 21.2. Test whether the mean differs from 20 at 5%. \[ H_0:\mu=20 \] \[ H_1:\mu\ne20 \] \[ Z= \frac{21.2-20} {4/\sqrt{49}} \] \[ = \frac{1.2}{4/7} \] \[ = \frac{1.2}{0.5714} \] \[ \boxed{Z_{\text{cal}}\approx2.10} \] Two-tailed 5%: \[ Z_{\text{tab}}=\pm1.96 \] Since: \[ 2.10>1.96 \] reject \(H_0\).

Hypothesis Testing 6

Example 66 — One-Sample t Test

A sample of 10 observations has: \[ \bar X=24 \] \[ s=4 \] Test whether the population mean is 20. \[ H_0:\mu=20 \] \[ H_1:\mu\ne20 \] Under null hypothesis: \[ t= \frac{\bar X-\mu_0} {s/\sqrt n} \] \[ = \frac{24-20} {4/\sqrt{10}} \] \[ = \frac4{1.2649} \] \[ \boxed{|t_{\text{cal}}|\approx3.162} \] Level: \[ \alpha=0.05 \] Degrees of freedom: \[ d.f.=n-1 \] \[ =10-1 \] \[ =9 \] From the t-distribution table, for two-tailed 5% and \(df=9\): \[ t_{\text{tab}}=2.262 \] Since: \[ 3.162>2.262 \] reject \(H_0\).

Hypothesis Testing 7

Example 67 — One-Sample t, One-Tailed

A researcher expects average growth to be at least 30 cm. A sample of 12 plants gives: \[ \bar X=32 \] \[ s=3.6 \] Test whether the mean growth is greater than 30 cm at 5%. \[ H_0:\mu=30 \] \[ H_1:\mu>30 \] \[ t= \frac{32-30} {3.6/\sqrt{12}} \] \[ = \frac2{1.0392} \] \[ \boxed{t_{\text{cal}}\approx1.925} \] \[ d.f.=12-1=11 \] From t table, at 5% right-tailed with \(df=11\): \[ t_{\text{tab}}\approx1.796 \] Since: \[ 1.925>1.796 \] reject \(H_0\). \[ \boxed{\text{Mean growth is significantly greater than 30 cm.}} \]

Hypothesis Testing 8 — Competitive

Example 68 — Small Sample

For \(n=8\): \[ \bar X=17.5 \] \[ s=2.4 \] Test: \[ H_0:\mu=16 \] against: \[ H_1:\mu\ne16 \] at 5%. \[ t= \frac{17.5-16} {2.4/\sqrt8} \] \[ = \frac{1.5}{0.8485} \] \[ \boxed{t_{\text{cal}}\approx1.768} \] \[ d.f.=8-1=7 \] From the t table: \[ t_{\text{tab}}=2.365 \] Since: \[ 1.768<2.365 \] do not reject \(H_0\).

Hypothesis Testing 9

Example 69 — Before and After Reaction Time

Let \(X\) and \(Y\) be before and after reaction times.

Before X	After Y	d = X − Y	d²
7	13	-6	36
23	24	-1	1
12	18	-6	36
12	13	-1	1

\[ \Sigma d=-14 \] \[ \Sigma d^2=74 \] \[ n=4 \] \[ \bar d= \frac{\Sigma d}{n} \] \[ = \frac{-14}{4} \] \[ =-3.5 \] Standard deviation of differences: \[ S_d= \sqrt{ \frac{1}{n-1} \left[ \Sigma d^2- \frac{(\Sigma d)^2}{n} \right] } \] \[ = \sqrt{ \frac13 \left[ 74-\frac{(-14)^2}{4} \right] } \] \[ = \sqrt{ \frac13(74-49) } \] \[ = \sqrt{\frac{25}{3}} \] \[ \boxed{S_d\approx2.887} \] Test hypothesis: \[ H_0:\mu_X=\mu_Y \] i.e. there is no significant difference in reaction time. Alternative: \[ H_1:\mu_X<\mu_Y \] Under null hypothesis: \[ t= \frac{\bar d} {S_d/\sqrt n} \] \[ = \frac{-3.5} {2.887/\sqrt4} \] \[ = \frac{-3.5}{1.4435} \] \[ \boxed{t_{\text{cal}}=-2.425} \] Therefore: \[ |t_{\text{cal}}|=2.425 \] Level: \[ \alpha=0.05 \] \[ d.f.=n-1 \] \[ =4-1=3 \] From the t table, for a one-tailed 5% test and \(df=3\): \[ t_{\text{tab}}=2.353 \] Since: \[ 2.425>2.353 \] reject \(H_0\).

Hypothesis Testing 10

Example 70 — Two-Sample Z Test

Given: \[ \bar X_1=82.5,\quad \bar X_2=76.2 \] \[ \sigma_1=8,\quad \sigma_2=7.5 \] \[ n_1=16,\quad n_2=20 \] Test whether the two means differ at 1%. \[ H_0:\mu_1=\mu_2 \] \[ H_1:\mu_1\ne\mu_2 \] Test statistic: \[ Z= \frac{ \bar X_1-\bar X_2 }{ \sqrt{ \frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2} } } \] \[ = \frac{82.5-76.2} { \sqrt{ \frac{64}{16}+ \frac{56.25}{20} } } \] \[ = \frac{6.3} {\sqrt{4+2.8125}} \] \[ = \frac{6.3}{2.610} \] \[ \boxed{Z_{\text{cal}}\approx2.414} \] At 1% two-tailed: \[ Z_{\text{tab}}=\pm2.576 \] Since: \[ 2.414<2.576 \] do not reject \(H_0\).

Hypothesis Testing 11

Example 71 — Two Independent Means, Competitive

Suppose: \[ \bar X_1=105 \] \[ \bar X_2=98 \] \[ \sigma_1=12 \] \[ \sigma_2=10 \] \[ n_1=36 \] \[ n_2=49 \] Test at 5%. \[ H_0:\mu_1=\mu_2 \] \[ H_1:\mu_1\ne\mu_2 \] \[ Z= \frac{105-98} { \sqrt{ \frac{12^2}{36} + \frac{10^2}{49} } } \] \[ = \frac7 { \sqrt{ 4+\frac{100}{49} } } \] \[ = \frac7{\sqrt{6.0408}} \] \[ \boxed{Z_{\text{cal}}\approx2.848} \] \[ Z_{\text{tab}}=1.96 \] Since: \[ 2.848>1.96 \] reject \(H_0\).

Hypothesis Testing 12

Example 72 — Two-Sample t Test with Pooled Variance

Suppose: \[ n_1=10,\quad \bar X_1=72,\quad s_1=5 \] \[ n_2=12,\quad \bar X_2=68,\quad s_2=4 \] First calculate pooled variance: \[ S_p^2= \frac{ (n_1-1)s_1^2+ (n_2-1)s_2^2 } { n_1+n_2-2 } \] \[ = \frac{ 9(25)+11(16) } {20} \] \[ = \frac{225+176}{20} \] \[ =20.05 \] Therefore: \[ S_p=\sqrt{20.05} \] \[ \approx4.478 \] Test statistic: \[ t= \frac{\bar X_1-\bar X_2} { S_p \sqrt{ \frac1{n_1}+\frac1{n_2} } } \] \[ = \frac4 { 4.478 \sqrt{ \frac1{10}+\frac1{12} } } \] \[ \boxed{t_{\text{cal}}\approx2.111} \] Degrees of freedom: \[ d.f.=10+12-2 \] \[ =20 \] From t table at 5% two-tailed: \[ t_{\text{tab}}\approx2.086 \] Since: \[ 2.111>2.086 \] reject \(H_0\).

Hypothesis Testing 13 — Competitive

Example 73 — Unequal Variance / Welch Type

Suppose: \[ n_1=15,\quad \bar X_1=45,\quad s_1=6 \] \[ n_2=12,\quad \bar X_2=40,\quad s_2=3 \] Because the sample variances are substantially different, use an unequal variance approach rather than automatically pooling them. Test statistic: \[ t= \frac{\bar X_1-\bar X_2} { \sqrt{ \frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} } } \] \[ = \frac5 { \sqrt{ \frac{36}{15}+\frac9{12} } } \] \[ = \frac5{\sqrt{2.4+0.75}} \] \[ = \frac5{\sqrt{3.15}} \] \[ \boxed{t_{\text{cal}}\approx2.817} \] The Welch-Satterthwaite degrees of freedom are approximately: \[ d.f.\approx21 \] At 5% two-tailed: \[ t_{\text{tab}}\approx2.080 \] Since: \[ 2.817>2.080 \] reject \(H_0\).

Hypothesis Testing 14

Example 74 — One Proportion Z Test

A laboratory claims that 60% of samples are positive. In a sample of 100, 68 are positive. Test the claim at 5%. \[ H_0:p=0.60 \] \[ H_1:p\ne0.60 \] Sample proportion: \[ \hat p=\frac{68}{100} \] \[ =0.68 \] \[ q=1-p=0.40 \] Test statistic: \[ Z= \frac{\hat p-p} {\sqrt{pq/n}} \] \[ = \frac{0.68-0.60} {\sqrt{(0.60)(0.40)/100}} \] \[ = \frac{0.08}{0.04899} \] \[ \boxed{Z_{\text{cal}}\approx1.633} \] At 5% two-tailed: \[ Z_{\text{tab}}=1.96 \] Since: \[ 1.633<1.96 \] do not reject \(H_0\).

Hypothesis Testing 15

Example 75 — One Proportion, One-Tailed

A manufacturer claims that at least 90% of kits pass quality control. In a sample of 200 kits, 172 pass. \[ \hat p=\frac{172}{200}=0.86 \] Test: \[ H_0:p=0.90 \] \[ H_1:p<0.90 \] \[ q=0.10 \] \[ Z= \frac{0.86-0.90} {\sqrt{(0.90)(0.10)/200}} \] \[ = \frac{-0.04}{\sqrt{0.00045}} \] \[ = \frac{-0.04}{0.02121} \] \[ \boxed{Z_{\text{cal}}\approx-1.886} \] At 5% left-tailed: \[ Z_{\text{tab}}=-1.645 \] Since: \[ -1.886<-1.645 \] reject \(H_0\).

Hypothesis Testing 16 — Competitive

Example 76 — One Proportion

A previous study reports a contamination rate of 8%. A new sample of 500 specimens has 52 contaminated specimens. Test whether the contamination rate has increased. \[ \hat p=\frac{52}{500}=0.104 \] \[ p_0=0.08 \] \[ q_0=0.92 \] Hypotheses: \[ H_0:p=0.08 \] \[ H_1:p>0.08 \] Test statistic: \[ Z= \frac{0.104-0.08} {\sqrt{(0.08)(0.92)/500}} \] \[ = \frac{0.024} {\sqrt{0.0001472}} \] \[ \boxed{Z_{\text{cal}}\approx1.978} \] At 5% right-tailed: \[ Z_{\text{tab}}=1.645 \] Therefore: \[ 1.978>1.645 \] Reject \(H_0\).

Hypothesis Testing 17

Example 77 — Two Proportions

Group A: \[ x_1=72,\quad n_1=100 \] Group B: \[ x_2=55,\quad n_2=100 \] \[ p_1=0.72 \] \[ p_2=0.55 \] Pooled proportion: \[ p= \frac{x_1+x_2}{n_1+n_2} \] \[ = \frac{72+55}{200} \] \[ =0.635 \] \[ q=0.365 \] Test statistic: \[ Z= \frac{p_1-p_2} { \sqrt{ pq \left( \frac1{n_1}+\frac1{n_2} \right) } } \] \[ = \frac{0.17} { \sqrt{ (0.635)(0.365) (0.02) } } \] \[ \boxed{Z_{\text{cal}}\approx2.51} \] At 5% two-tailed: \[ Z_{\text{tab}}=1.96 \] Since: \[ 2.51>1.96 \] reject \(H_0\).

Hypothesis Testing 18

Example 78 — Two Proportions, Competitive

Group 1 has 84 successes out of 120. Group 2 has 68 successes out of 110. \[ p_1=\frac{84}{120}=0.70 \] \[ p_2=\frac{68}{110}\approx0.6182 \] Pooled: \[ p= \frac{84+68}{120+110} \] \[ = \frac{152}{230} \] \[ \approx0.6609 \] \[ q=1-0.6609 \] \[ =0.3391 \] Then: \[ Z= \frac{0.70-0.6182} { \sqrt{ (0.6609)(0.3391) \left( \frac1{120}+\frac1{110} \right) } } \] \[ \boxed{Z_{\text{cal}}\approx1.26} \] At 5% two-tailed: \[ Z_{\text{tab}}=1.96 \] Do not reject \(H_0\).

Hypothesis Testing 19

Example 79 — Paired t Test

Differences are: \[ d=4,5,3,6,2,4 \] Then: \[ \Sigma d=24 \] \[ n=6 \] \[ \bar d=\frac{24}{6}=4 \] Calculate: \[ \Sigma d^2= 16+25+9+36+4+16 \] \[ =106 \] Standard deviation: \[ S_d= \sqrt{ \frac{ \Sigma d^2-(\Sigma d)^2/n } {n-1} } \] \[ = \sqrt{ \frac{ 106-576/6 } 5 } \] \[ = \sqrt{ \frac{106-96}{5} } \] \[ =\sqrt2 \] \[ \boxed{S_d=1.414} \] Test statistic: \[ t= \frac{\bar d} {S_d/\sqrt n} \] \[ = \frac4{1.414/\sqrt6} \] \[ \boxed{t_{\text{cal}}\approx6.928} \] \[ d.f.=6-1=5 \] At 5% two-tailed: \[ t_{\text{tab}}=2.571 \] Since: \[ 6.928>2.571 \] reject \(H_0\).

Hypothesis Testing 20 — Competitive

Example 80 — Paired Medical Measurements

Suppose the differences after minus before are: \[ 3,\ 5,\ 2,\ 6,\ 4,\ 5,\ 3,\ 7 \] \[ n=8 \] \[ \Sigma d=35 \] \[ \bar d=\frac{35}{8}=4.375 \] \[ \Sigma d^2 = 9+25+4+36+16+25+9+49 \] \[ =173 \] Then: \[ S_d= \sqrt{ \frac{ 173-\frac{35^2}{8} } 7 } \] \[ = \sqrt{ \frac{ 173-153.125 } 7 } \] \[ = \sqrt{2.8393} \] \[ \boxed{S_d\approx1.685} \] \[ t= \frac{4.375} {1.685/\sqrt8} \] \[ \boxed{t_{\text{cal}}\approx7.35} \] \[ d.f.=7 \] At 5% two-tailed: \[ t_{\text{tab}}=2.365 \] Therefore: \[ 7.35>2.365 \] Reject \(H_0\).

Hypothesis Testing 21

Example 81 — F Test

Two samples have: \[ s_1^2=25 \] \[ s_2^2=10 \] Use the larger variance in the numerator: \[ F= \frac{25}{10} \] \[ \boxed{F_{\text{cal}}=2.5} \] Suppose: \[ n_1=10,\quad n_2=8 \] Then: \[ v_1=9 \] \[ v_2=7 \] At 5%, use the F table with numerator \(df=9\) and denominator \(df=7\). The exact critical value depends on the tail convention and table used. The decision is made by comparing: \[ F_{\text{cal}} \] with: \[ F_{\text{tab}} \]

Hypothesis Testing 22 — Competitive

Example 82 — Variance Comparison

Sample A: \[ n_1=12,\quad s_1^2=36 \] Sample B: \[ n_2=10,\quad s_2^2=16 \] Since 36 is larger: \[ F= \frac{36}{16} \] \[ \boxed{F_{\text{cal}}=2.25} \] Degrees of freedom: \[ v_1=11 \] \[ v_2=9 \] Use the F-distribution table for the chosen significance level.

Hypothesis Testing 23

Example 83 — Complete One-Way ANOVA

Three groups: \[ A=(8,9,10) \] \[ B=(12,11,13) \] \[ C=(15,14,16) \] Group means: \[ \bar X_A=9 \] \[ \bar X_B=12 \] \[ \bar X_C=15 \] Overall mean: \[ \bar X=12 \] Between-group sum of squares: \[ SS_B = 3(9-12)^2 + 3(12-12)^2 + 3(15-12)^2 \] \[ = 3(9)+0+3(9) \] \[ \boxed{SS_B=54} \] Within-group sum of squares: For A: \[ (8-9)^2+(9-9)^2+(10-9)^2 \] \[ =1+0+1=2 \] For B: \[ (12-12)^2+(11-12)^2+(13-12)^2 \] \[ =0+1+1=2 \] For C: \[ (15-15)^2+(14-15)^2+(16-15)^2 \] \[ =0+1+1=2 \] Therefore: \[ SS_W=2+2+2 \] \[ \boxed{SS_W=6} \] Degrees of freedom: \[ df_B=k-1 \] \[ =3-1=2 \] \[ df_W=N-k \] \[ =9-3=6 \] Mean squares: \[ MS_B=\frac{54}{2}=27 \] \[ MS_W=\frac6{6}=1 \] Therefore: \[ F= \frac{MS_B}{MS_W} \] \[ =\frac{27}{1} \] \[ \boxed{F_{\text{cal}}=27} \] At 5% with \(df_1=2\), \(df_2=6\): \[ F_{\text{tab}}\approx5.143 \] Since: \[ 27>5.143 \] reject \(H_0\).

Hypothesis Testing 24 — Competitive

Example 84 — ANOVA with Four Groups

Consider: \[ A=(10,12,11,13) \] \[ B=(15,14,16,15) \] \[ C=(20,18,19,21) \] \[ D=(13,12,14,13) \] The procedure is: 1. Calculate each group total. 2. Calculate each group mean. 3. Calculate the grand mean. 4. Calculate \(SS_B\). 5. Calculate \(SS_W\). 6. Calculate degrees of freedom. 7. Calculate \(MS_B\). 8. Calculate \(MS_W\). 9. Calculate \(F\). 10. Compare with the F-table critical value. 11. State the scientific conclusion. This is the format to use in a competitive examination even when the arithmetic is lengthy.

Hypothesis Testing 25

Example 85 — Two-Way ANOVA without Replication

Consider the following observations:

	Method A	Method B	Method C
Lab 1	10	12	14
Lab 2	9	11	13
Lab 3	11	13	15
Lab 4	10	12	14

For two-way ANOVA without replication: \[ SS_T= \sum X^2- \frac{G^2}{N} \] Row sum of squares: \[ SS_R= \sum\frac{R_i^2}{c} - \frac{G^2}{N} \] Column sum of squares: \[ SS_C= \sum\frac{C_j^2}{r} - \frac{G^2}{N} \] Error: \[ SS_E=SS_T-SS_R-SS_C \] Then calculate: \[ F_R=\frac{MS_R}{MS_E} \] and: \[ F_C=\frac{MS_C}{MS_E} \] Compare each with the appropriate F-table value.

Hypothesis Testing 26 — Competitive

Example 86 — Two-Way ANOVA Interpretation

A microbiology experiment compares three culture media across four laboratory conditions. The question asks separately whether: 1. Culture medium has a significant effect. 2. Laboratory condition has a significant effect. Therefore two separate F statistics are required: \[ F_{\text{media}} = \frac{MS_{\text{media}}} {MS_E} \] and: \[ F_{\text{condition}} = \frac{MS_{\text{condition}}} {MS_E} \] The final answer must not simply state that ANOVA is significant. The student must state which factor is significant.

Hypothesis Testing 27

Example 87 — Test of Significance of r

Given: \[ r=0.65 \] \[ n=20 \] Test whether the correlation is significant. \[ H_0:\rho=0 \] \[ H_1:\rho\ne0 \] Test statistic: \[ t= \frac{ r\sqrt{n-2} }{ \sqrt{1-r^2} } \] \[ = \frac{ 0.65\sqrt{18} }{ \sqrt{1-0.65^2} } \] \[ = \frac{0.65(4.243)} {\sqrt{0.5775}} \] \[ = \frac{2.758} {0.760} \] \[ \boxed{t_{\text{cal}}\approx3.63} \] \[ d.f.=20-2=18 \] At 5% two-tailed: \[ t_{\text{tab}}\approx2.101 \] Since: \[ 3.63>2.101 \] reject \(H_0\). There is significant linear correlation.

Hypothesis Testing 28

Example 88 — Significance of Negative Correlation

Suppose: \[ r=-0.52 \] \[ n=25 \] \[ t= \frac{ -0.52\sqrt{23} }{ \sqrt{1-0.52^2} } \] \[ \boxed{t_{\text{cal}}\approx-2.91} \] \[ d.f.=23 \] At 5% two-tailed: \[ |t_{\text{tab}}|\approx2.069 \] Since: \[ 2.91>2.069 \] reject \(H_0\).

Hypothesis Testing 29

Example 89 — Test of Regression Coefficient

Suppose: \[ b=2.4 \] and: \[ SE(b)=0.8 \] Test whether the regression coefficient differs significantly from zero. \[ H_0:b=0 \] \[ H_1:b\ne0 \] Test statistic: \[ t= \frac{b}{SE(b)} \] \[ = \frac{2.4}{0.8} \] \[ \boxed{t_{\text{cal}}=3} \] If: \[ d.f.=18 \] then at 5% two-tailed: \[ t_{\text{tab}}\approx2.101 \] Since: \[ 3>2.101 \] reject \(H_0\). The regression coefficient is statistically significant.

Hypothesis Testing 30

Example 90 — Negative Regression Coefficient

Suppose: \[ b=-1.8 \] \[ SE(b)=0.6 \] Then: \[ t= \frac{-1.8}{0.6} \] \[ \boxed{t_{\text{cal}}=-3} \] For a two-tailed test compare: \[ |t_{\text{cal}}|=3 \] with the positive critical value. The sign indicates the direction of the regression coefficient, while the absolute value is used when comparing with a two-tailed critical value.

Hypothesis Testing 31

Example 91 — Critical Value Method

Suppose a two-tailed Z test gives: \[ Z_{\text{cal}}=2.31 \] At: \[ \alpha=0.05 \] critical values are: \[ Z_{\text{tab}}=\pm1.96 \] Since: \[ 2.31>1.96 \] reject \(H_0\).

Hypothesis Testing 32

Example 92 — p-value Method

Suppose: \[ Z_{\text{cal}}=2.58 \] For a two-tailed test, the p-value is approximately: \[ p\approx0.0099 \] At: \[ \alpha=0.01 \] compare: \[ p<\alpha \] \[ 0.0099<0.01 \] Therefore reject \(H_0\).

Hypothesis Testing 33

Example 93 — Identifying the Error

Suppose: \[ H_0: \] A laboratory test has no harmful effect. If \(H_0\) is actually true but the researcher rejects it, this is: \[ \boxed{\text{Type I error}} \] Probability: \[ P(\text{Type I error})=\alpha \] If \(H_0\) is false but the researcher fails to reject it, this is: \[ \boxed{\text{Type II error}} \] Probability: \[ P(\text{Type II error})=\beta \]

Hypothesis Testing 34

Example 94 — CI and Two-Tailed Test

Suppose a 95% confidence interval for a population mean is: \[ (48.2,\ 53.8) \] Test: \[ H_0:\mu=55 \] against: \[ H_1:\mu\ne55 \] Since 55 does not lie inside the 95% confidence interval, reject \(H_0\) at the corresponding 5% two-sided significance level. \[ \boxed{\text{Reject }H_0} \]

Non-parametric 1

Example 95 — Equal Expected Frequencies

Observed: \[ 42,\ 38,\ 31,\ 29 \] Total: \[ N=140 \] For four equally likely categories: \[ E=\frac{140}{4}=35 \] The calculation table is:

Category	O	E	O−E	(O−E)²/E
A	42	35	7	1.400
B	38	35	3	0.257
C	31	35	-4	0.457
D	29	35	-6	1.029

\[ \chi^2_{\text{cal}} = 1.400+0.257+0.457+1.029 \] \[ \boxed{\chi^2_{\text{cal}}=3.143} \] Hypotheses: \[ H_0: \] The four categories occur in equal proportions. \[ H_1: \] The categories do not occur in equal proportions. Degrees of freedom: \[ d.f.=k-1 \] \[ =4-1=3 \] At 5%: \[ \chi^2_{\text{tab}}=7.815 \] Since: \[ 3.143<7.815 \] do not reject \(H_0\).

Non-parametric 2

Example 96 — Goodness of Fit with Unequal Expected Proportions

Suppose expected proportions are: \[ 0.50,\ 0.30,\ 0.20 \] and total sample size is: \[ N=200 \] Expected frequencies: \[ E_1=200(0.50)=100 \] \[ E_2=200(0.30)=60 \] \[ E_3=200(0.20)=40 \] Suppose observed frequencies are: \[ 110,\ 55,\ 35 \] Then: \[ \chi^2= \frac{(110-100)^2}{100} + \frac{(55-60)^2}{60} + \frac{(35-40)^2}{40} \] \[ = 1+\frac{25}{60}+\frac{25}{40} \] \[ \boxed{\chi^2_{\text{cal}}\approx2.042} \] \[ d.f.=3-1=2 \] At 5%: \[ \chi^2_{\text{tab}}=5.991 \] Since: \[ 2.042<5.991 \] do not reject \(H_0\).

Non-parametric 3

Example 97 — Competitive Goodness-of-Fit

A genetic model predicts a ratio: \[ 9:3:3:1 \] Suppose \(N=160\). Expected frequencies: \[ E_1=160\left(\frac9{16}\right)=90 \] \[ E_2=160\left(\frac3{16}\right)=30 \] \[ E_3=160\left(\frac3{16}\right)=30 \] \[ E_4=160\left(\frac1{16}\right)=10 \] Observed: \[ 86,\ 34,\ 28,\ 12 \] Then: \[ \chi^2= \frac{(86-90)^2}{90} + \frac{(34-30)^2}{30} + \frac{(28-30)^2}{30} + \frac{(12-10)^2}{10} \] \[ = \frac{16}{90} + \frac{16}{30} + \frac4{30} + \frac4{10} \] \[ \boxed{\chi^2_{\text{cal}}\approx1.511} \] \[ d.f.=4-1=3 \] \[ \chi^2_{\text{tab}}=7.815 \] Therefore: \[ 1.511<7.815 \] Do not reject \(H_0\).

Non-parametric 4

Example 98 — Goodness-of-Fit Decision

Suppose: \[ \chi^2_{\text{cal}}=12.4 \] and: \[ d.f.=4 \] At 5%: \[ \chi^2_{\text{tab}}=9.488 \] Since: \[ 12.4>9.488 \] reject \(H_0\).

Non-parametric 5

Example 99 — \(2\times2\) Independence

Observed:

	Disease +	Disease −	Total
Exposed	40	60	100
Unexposed	20	80	100
Total	60	140	200

Expected exposed/disease+: \[ E= \frac{100(60)}{200} \] \[ =30 \] Expected exposed/disease−: \[ E= \frac{100(140)}{200} \] \[ =70 \] The remaining expected frequencies are 30 and 70. Therefore: \[ \chi^2= \frac{(40-30)^2}{30} + \frac{(60-70)^2}{70} + \frac{(20-30)^2}{30} + \frac{(80-70)^2}{70} \] \[ = 3.333+1.429+3.333+1.429 \] \[ \boxed{\chi^2_{\text{cal}}\approx9.524} \] Hypotheses: \[ H_0: \] The two attributes are independent. \[ H_1: \] The two attributes are associated. Degrees of freedom: \[ d.f.=(r-1)(c-1) \] \[ =(2-1)(2-1) \] \[ =1 \] At 5%: \[ \chi^2_{\text{tab}}=3.841 \] Since: \[ 9.524>3.841 \] reject \(H_0\).

Non-parametric 6

Example 100 — \(2\times3\) Independence

Observed:

	A	B	C	Total
Group 1	30	20	10	60
Group 2	20	25	15	60
Total	50	45	25	120

Expected frequency is calculated separately for every cell: \[ E_{ij}= \frac{Row\ total\times Column\ total} {Grand\ total} \] For Group 1, category A: \[ E= \frac{60(50)}{120} \] \[ =25 \] For Group 1, category B: \[ E= \frac{60(45)}{120} \] \[ =22.5 \] For Group 1, category C: \[ E= \frac{60(25)}{120} \] \[ =12.5 \] The same method is used for Group 2. Then: \[ \chi^2= \sum\frac{(O-E)^2}{E} \] Degrees of freedom: \[ d.f.=(2-1)(3-1)=2 \] At 5%: \[ \chi^2_{\text{tab}}=5.991 \]

Non-parametric 7

Example 101 — \(3\times3\) Contingency Table

Use: \[ d.f.=(3-1)(3-1) \] \[ =4 \] For every cell: \[ E= \frac{Row\ total\times Column\ total} {Grand\ total} \] Then: \[ \chi^2= \sum \frac{(O-E)^2}{E} \] At 5%: \[ \chi^2_{\text{tab}}=9.488 \] The final decision is made by comparing calculated and tabulated values.

Non-parametric 8

Example 102 — Biological Association

A study compares three age groups with three levels of antibody response. The appropriate analysis is a chi-square test of independence because both variables are categorical. The student must: 1. Construct the observed-frequency table. 2. Calculate row totals. 3. Calculate column totals. 4. Calculate grand total. 5. Calculate every expected frequency. 6. Calculate every \((O-E)^2/E\). 7. Sum the contributions. 8. Calculate degrees of freedom. 9. Obtain the chi-square critical value. 10. Make the decision. 11. State whether antibody response is associated with age group.

Non-parametric 9

Example 103 — Expected Frequency Problem

Suppose: \[ Row\ total=75 \] \[ Column\ total=40 \] \[ N=150 \] Then: \[ E= \frac{75(40)}{150} \] \[ \boxed{E=20} \] If the observed value is 25: \[ \frac{(O-E)^2}{E} = \frac{(25-20)^2}{20} \] \[ = \frac{25}{20} \] \[ \boxed{1.25} \] This contribution is then added to the contributions of all other cells.

Non-parametric 10

Example 104 — Sign Test

Suppose 10 paired differences give: \[ +,+,-,+,+,-,+,+,+,- \] Number of positive differences: \[ x=7 \] Number of negative differences: \[ n-x=3 \] Under \(H_0\): \[ p=0.5 \] The number of positive signs follows: \[ X\sim Binomial(n=10,p=0.5) \] For a two-tailed test, calculate the appropriate binomial tail probability. This is an exact non-parametric alternative to a paired t test when the assumptions for the paired t test are not appropriate.

Non-parametric 11

Example 105 — Sign Test with Strong Direction

Suppose: \[ n=12 \] and: \[ x=11 \] positive differences are observed. Under: \[ H_0:p=0.5 \] the probability of 11 or more positive signs is: \[ P(X\ge11) = {12\choose11} (0.5)^{12} + {12\choose12} (0.5)^{12} \] \[ = \frac{12+1}{4096} \] \[ =\frac{13}{4096} \] \[ \approx0.00317 \] For a two-tailed test: \[ p\approx2(0.00317) \] \[ \boxed{p\approx0.00635} \] Since: \[ p<0.05 \] reject \(H_0\).

Non-parametric 12

Example 106 — Wilcoxon Signed-Rank

Suppose paired differences are: \[ 2,\ -3,\ 4,\ -1,\ 5 \] Ignore zero differences if present. Rank absolute differences: \[ 1,\ 2,\ 3,\ 4,\ 5 \] Assign ranks and restore signs. Then calculate: \[ T^+=\sum\text{positive ranks} \] and: \[ T^-=\sum\text{negative ranks} \] The test statistic is commonly: \[ T=\min(T^+,T^-) \] Compare \(T\) with the Wilcoxon signed-rank critical value table for the appropriate sample size and significance level.

Non-parametric 13

Example 107 — Wilcoxon with Tied Absolute Differences

Suppose differences are: \[ 2,\ -2,\ 4,\ -5,\ 5,\ 6 \] The absolute differences are: \[ 2,\ 2,\ 4,\ 5,\ 5,\ 6 \] The two observations with absolute difference 2 occupy ranks 1 and 2. Therefore each receives: \[ \frac{1+2}{2}=1.5 \] Similarly, the two observations with absolute difference 5 receive: \[ \frac{4+5}{2}=4.5 \] This average-rank treatment is essential in competitive examinations.

Non-parametric 14

Example 108 — Wilcoxon Decision

Suppose after ranking: \[ T^+=28 \] \[ T^-=7 \] Therefore: \[ T=\min(28,7) \] \[ \boxed{T=7} \] If the tabulated critical value is 8: \[ T_{\text{cal}}=7

E. Mann–Whitney U Test

Non-parametric 15

Example 109 — Mann–Whitney U

Group A: \[ 8,\ 10,\ 12 \] Group B: \[ 15,\ 18,\ 20 \] Rank all observations: \[ 8(1),\ 10(2),\ 12(3),\ 15(4),\ 18(5),\ 20(6) \] For Group A: \[ R_1=1+2+3=6 \] For Group B: \[ R_2=4+5+6=15 \] Calculate: \[ U_1=n_1n_2+ \frac{n_1(n_1+1)}2-R_1 \] \[ = (3)(3)+ \frac{3(4)}2 -6 \] \[ =9+6-6 \] \[ \boxed{U_1=9} \] Similarly: \[ U_2=n_1n_2-U_1 \] \[ =9-9 \] \[ \boxed{U_2=0} \] Therefore: \[ U=\min(9,0) \] \[ \boxed{U=0} \]

Non-parametric 16

Example 110 — Mann–Whitney with Larger Samples

For two independent groups, rank all observations together. Calculate: \[ R_1 \] and: \[ R_2 \] Then: \[ U_1= n_1n_2+ \frac{n_1(n_1+1)}2-R_1 \] \[ U_2= n_1n_2-U_1 \] Use: \[ U=\min(U_1,U_2) \] For sufficiently large samples, a normal approximation can be used: \[ \mu_U= \frac{n_1n_2}{2} \] \[ \sigma_U= \sqrt{ \frac{n_1n_2(n_1+n_2+1)}{12} } \] and: \[ Z= \frac{U-\mu_U}{\sigma_U} \] with continuity correction where appropriate.

Non-parametric 17

Example 111 — Mann–Whitney Competitive

Suppose: \[ n_1=5 \] \[ n_2=6 \] and after ranking: \[ R_1=13 \] Then: \[ U_1= (5)(6)+ \frac{5(6)}2 -13 \] \[ = 30+15-13 \] \[ \boxed{U_1=32} \] Then: \[ U_2=n_1n_2-U_1 \] \[ =30-32 \] This signals that the rank-sum supplied must be checked because \(U\) cannot be negative. In an actual examination, verify the rank sum before proceeding.

This is an important competitive-exam lesson: always check whether your intermediate result is mathematically possible before making a decision.

F. Kruskal–Wallis Test

Non-parametric 18

Example 112 — Kruskal–Wallis Test

Suppose there are three groups: \[ A=(2,4,6) \] \[ B=(8,10,12) \] \[ C=(14,16,18) \] Rank all observations from 1 to 9. Group rank sums: \[ R_1=1+2+3=6 \] \[ R_2=4+5+6=15 \] \[ R_3=7+8+9=24 \] Total: \[ N=9 \] Each group has: \[ n_i=3 \] Formula: \[ H= \frac{12}{N(N+1)} \sum \frac{R_i^2}{n_i} -3(N+1) \] Therefore: \[ H= \frac{12}{9(10)} \left[ \frac{6^2}{3} + \frac{15^2}{3} + \frac{24^2}{3} \right] -30 \] \[ = \frac{12}{90} (12+75+192)-30 \] \[ = 0.1333(279)-30 \] \[ =37.2-30 \] \[ \boxed{H=7.2} \] Degrees of freedom: \[ d.f.=k-1 \] \[ =3-1=2 \] At 5%: \[ \chi^2_{\text{tab}}=5.991 \] Since: \[ 7.2>5.991 \] reject \(H_0\).

Non-parametric 19

Example 113 — Kruskal–Wallis Interpretation

If: \[ H_{\text{cal}}=9.20 \] and: \[ d.f.=3 \] then at 5%: \[ \chi^2_{\text{tab}}=7.815 \] Since: \[ 9.20>7.815 \] reject \(H_0\). There is evidence that at least one group's distribution differs from the others.

G. Additional Chi-Square Competitive Problems

Non-parametric 20

Example 114 — Fourfold Table

For: \[ a=35,\quad b=15,\quad c=20,\quad d=30 \] Yule's Q: \[ Q= \frac{ad-bc}{ad+bc} \] \[ = \frac{35(30)-15(20)} {35(30)+15(20)} \] \[ = \frac{1050-300}{1050+300} \] \[ \boxed{Q=0.556} \]

Non-parametric 21

Example 115 — Pearson Contingency Coefficient

Given: \[ \chi^2=25 \] \[ N=250 \] \[ C= \sqrt{ \frac{25}{25+250} } \] \[ = \sqrt{\frac{25}{275}} \] \[ \boxed{C\approx0.302} \]

Non-parametric 22

Example 116 — Chi-Square Decision

Suppose: \[ \chi^2_{\text{cal}}=11.8 \] \[ d.f.=5 \] At 5%: \[ \chi^2_{\text{tab}}=11.070 \] Since: \[ 11.8>11.070 \] reject \(H_0\).

Non-parametric 23

Example 117 — Chi-Square with Small Contribution

Suppose: \[ O=48 \] \[ E=50 \] Then: \[ \frac{(O-E)^2}{E} \] \[ = \frac{(48-50)^2}{50} \] \[ = \frac4{50} \] \[ \boxed{0.08} \] Repeat for every cell and sum the contributions.

Non-parametric 24

Example 118 — Checking Expected Frequencies

If an expected frequency is less than 5, the validity of the usual chi-square approximation should be considered carefully. Possible examination actions include: - combining adjacent categories where scientifically appropriate, - using an exact method for a suitable \(2\times2\) situation, - or stating that the chi-square approximation may be unreliable. The student should not blindly calculate and report a chi-square statistic without checking expected frequencies.

Non-parametric 25

Example 119 — Choosing the Correct Non-parametric Test

Choose the appropriate test: Situation A: Compare observed categorical frequencies with theoretical frequencies. \[ \boxed{\text{Chi-square goodness-of-fit}} \] Situation B: Determine whether two categorical variables are associated. \[ \boxed{\text{Chi-square test of independence}} \] Situation C: Compare paired ordinal observations. \[ \boxed{\text{Wilcoxon signed-rank or Sign test}} \] Situation D: Compare two independent groups with ordinal or non-normal observations. \[ \boxed{\text{Mann–Whitney U}} \] Situation E: Compare three or more independent groups using ranks. \[ \boxed{\text{Kruskal–Wallis}} \]

Non-parametric 26

Example 120 — Complete Non-parametric Decision Problem

A researcher has three independent groups. The measurements are strongly non-normal and sample sizes are small. The correct reasoning is: 1. Groups are independent. 2. There are more than two groups. 3. The data do not satisfy the assumptions of one-way ANOVA. 4. Therefore use Kruskal–Wallis. Then: \[ H= \frac{12}{N(N+1)} \sum\frac{R_i^2}{n_i} -3(N+1) \] Compare \(H\) with the chi-square critical value with: \[ d.f.=k-1 \] This type of question tests **test selection**, not merely arithmetic.

Non-parametric 27

Example 121 — Comparing Parametric and Non-parametric Alternatives

A paired measurement is strongly skewed. Parametric option: \[ \boxed{\text{Paired t test}} \] Non-parametric alternative: \[ \boxed{\text{Wilcoxon signed-rank test}} \] If only direction is considered: \[ \boxed{\text{Sign test}} \] A two-independent-group comparison can use: \[ \boxed{\text{Independent t test}} \] or its non-parametric alternative: \[ \boxed{\text{Mann–Whitney U}} \] Three or more independent groups: \[ \boxed{\text{One-way ANOVA}} \] or: \[ \boxed{\text{Kruskal–Wallis}} \]

Non-parametric 28

Example 122 — Critical Value Interpretation

Suppose a Kruskal–Wallis test gives: \[ H_{\text{cal}}=6.90 \] with: \[ d.f.=2 \] At 5%: \[ \chi^2_{\text{tab}}=5.991 \] Therefore: \[ 6.90>5.991 \] Reject \(H_0\).

Non-parametric 29

Example 123 — Mann–Whitney Decision

Suppose: \[ U_{\text{cal}}=4 \] and the table gives: \[ U_{\text{tab}}=5 \] For a lower-tail Mann–Whitney critical-value approach: \[ U_{\text{cal}}

Non-parametric 30

Example 124 — Wilcoxon Decision

Suppose: \[ T_{\text{cal}}=6 \] and: \[ T_{\text{tab}}=8 \] For the Wilcoxon signed-rank test: \[ T_{\text{cal}}

Non-parametric 31

Example 125 — Competitive Test Selection

A question gives: - two related measurements, - ordinal/non-normal data, - small sample, - no assumption of normality. The appropriate test is: \[ \boxed{\text{Wilcoxon signed-rank test}} \] If the question says only the signs of differences are to be considered: \[ \boxed{\text{Sign test}} \] If the observations are independent: \[ \boxed{\text{Mann–Whitney U}} \] This distinction is commonly tested in competitive examinations.

Non-parametric 32

Example 126 — Comprehensive Chi-Square Procedure

For any complete chi-square independence problem, write the following order in the examination: Test hypothesis \[ H_0: \] The attributes are independent. \[ H_1: \] The attributes are associated. Expected frequency \[ E_{ij} = \frac{ Row\ total\times Column\ total }{ Grand\ total } \] Test statistic \[ \chi^2= \sum \frac{(O-E)^2}{E} \] Level of significance \[ \alpha=0.05 \] Degrees of freedom \[ d.f.=(r-1)(c-1) \] Critical value From the chi-square table at 5% level and the calculated degrees of freedom: \[ \chi^2_{\text{tab}}=... \] Decision Compare: \[ \chi^2_{\text{cal}} \] with: \[ \chi^2_{\text{tab}} \] Conclusion State whether the attributes are independent or significantly associated.

Before solving a numerical, ask:

1. What type of data are given?
2. Is the variable quantitative or categorical?
3. Is the sample independent or paired?
4. Is the population SD known or unknown?
5. How many groups are being compared?
6. Is the question one-tailed or two-tailed?
7. What is the significance level?
8. What are \(H_0\) and \(H_1\)?
9. Which probability/statistical table is required?
10. What are the degrees of freedom?
11. What is the calculated statistic?
12. What is the critical value?
13. Does the calculated value fall inside the rejection region?
14. What is the statistical decision?
15. What is the biological/scientific interpretation?

Hypothesis Testing Answer Pattern to Memorize

Test hypothesis

\[ H_0:... \] \[ H_1:... \]

Under null hypothesis

Test Statistic

\[ Test\ statistic \] \[ = Formula \] \[ = Substitution \] \[ = Calculation \] \[ =\boxed{Answer} \]

Level of significance

\[ \alpha=5\%=0.05 \]

Degrees of freedom

\[ d.f.=... \]

Critical value

From the appropriate statistical table at the specified level of significance and degrees of freedom:

\[ Critical\ value=... \]

Decision

\[ Calculated\ value \quad \text{vs.}\quad Tabulated\ value \]

Conclusion

Write the conclusion in the context of the actual question. Do not simply write “significant” or “not significant.”