Business Statistics Lab

The percentages add up to $45+20+15+10+10+5=105$, so each item is taken as a share of 105 when finding the angle of its sector:

$\text{Angle}=\frac{\text{value}}{105}\times 360^\circ$

Construction. Draw a circle, then starting from a radius (say the vertical one) draw the sectors in the order above, clockwise, with the angles shown. Label each sector with its item name and percentage, and shade or colour the sectors differently. The largest sector is Food and the smallest is Misc.

In Excel/SPSS/R, enter the items and values, choose Insert > Pie chart, and add data labels.

Let $X$ = Economics, $Y$ = Statistics, $n=10$.

$\sum X=470,\ \sum Y=420,\ \sum X^2=23110,\ \sum Y^2=17968,\ \sum XY=19685$

$\bar X=47,\quad \bar Y=42$

(a) Correlation coefficient.

$r=\frac{\sum XY-n\bar X\bar Y}{\sqrt{(\sum X^2-n\bar X^2)(\sum Y^2-n\bar Y^2)}}=\frac{19685-19740}{\sqrt{(1020)(328)}}=\frac{-55}{578.4}=-0.095$

There is a very weak negative (almost no) correlation between the marks.

(b) Expected Statistics marks when Economics $=55$.

$b_{YX}=\frac{\sum XY-n\bar X\bar Y}{\sum X^2-n\bar X^2}=\frac{-55}{1020}=-0.0539$

$Y-42=-0.0539(X-47)\ \Rightarrow\ Y=44.53-0.0539X$

At $X=55$:

$Y=44.53-0.0539\times 55=44.53-2.97=41.57$

Expected marks in Statistics $\approx 41.6$.

Here $n=4$ (trials), $N=\sum f=110$.

Mean.

$\bar x=\frac{\sum fx}{N}=\frac{0+20+80+75+60}{110}=\frac{235}{110}=2.136$

Estimate of $p$. From $np=\bar x$:

$p=\frac{2.136}{4}=0.534,\qquad q=1-p=0.466$

Expected frequencies. $f(x)=N\binom{4}{x}p^xq^{4-x}$:

The fitted distribution is $110\,(0.466+0.534)^4$, with expected frequencies $5, 24, 41, 31, 9$.

Given $n=100$, $\bar x=55$, $s=5$, $\mu_0=52$ kg (the question's "height" is taken to mean the weight, since all values are in kg). Large sample, so a Z-test is used.

Hypotheses. $H_0:\mu=52$ against $H_1:\mu\ne 52$ (two-tailed), $\alpha=0.05$.

$Z=\frac{\bar x-\mu_0}{s/\sqrt n}=\frac{55-52}{5/10}=\frac{3}{0.5}=6$

Critical value. $Z_{0.05}=1.96$.

Decision. $|Z|=6>1.96$, so $H_0$ is rejected. The population mean is significantly different from 52 kg.

Given $n=50$, $\bar x=7.5$, $s=1.5$ (large sample).

$S.E.=\frac{s}{\sqrt n}=\frac{1.5}{\sqrt{50}}=0.212$

95% confidence limits: $\bar x\pm 1.96\times S.E.$

$7.5\pm 1.96\times 0.212=7.5\pm 0.416$

Lower limit $=7.08$ kg and upper limit $=7.92$ kg.

So the population mean lies between 7.08 kg and 7.92 kg with 95% confidence.

Mid-values $m$: $11, 13, 15, 17, 19, 21, 23$.

$\bar X=\frac{\sum fm}{N}=\frac{2480}{150}=16.53\ (\text{in 00 hours, i.e. about 1653 hours})$

$\sigma=\sqrt{\frac{\sum fm^2}{N}-\bar X^2}=\sqrt{280.73-273.35}=\sqrt{7.38}=2.72\ (\text{about 272 hours})$

$N=105$, so $N/2=52.5$, which falls in the class $20\text{-}30$ (median class). Here $L=20$, $c.f.=35$, $f=40$, $h=10$.

$\text{Median}=L+\frac{N/2-c.f.}{f}\times h=20+\frac{52.5-35}{40}\times 10=20+4.375=24.375$

Median $\approx 24.38$.

Coefficient of kurtosis: $\beta_2=\dfrac{\mu_4}{\mu_2^2}$, with $\gamma_2=\beta_2-3$.

Mid-values: $5.5, 6.5, 7.5, 8.5, 9.5, 10.5, 11.5$ with $f = 10, 14, 18, 24, 16, 12, 6$ and $N=100$.

Mean.

$\bar X=\frac{55+91+135+204+152+126+69}{100}=\frac{832}{100}=8.32$

Central moments (deviations $d=m-8.32$):

$\mu_2=\frac{\sum fd^2}{N}=2.788,\qquad \mu_4=\frac{\sum fd^4}{N}=17.098$

$\beta_2=\frac{17.098}{(2.788)^2}=\frac{17.098}{7.771}=2.20,\qquad \gamma_2=2.20-3=-0.80$

Interpretation. Since $\beta_2=2.20<3$ (equivalently $\gamma_2<0$), the distribution is platykurtic: flatter than the normal curve, with wages spread fairly evenly around the mean and fewer extreme values.

Total days $N = 110+60+15+10+5 = 200$.

$\bar x = \lambda = \frac{0\cdot110 + 1\cdot60 + 2\cdot15 + 3\cdot10 + 4\cdot5}{200} = \frac{140}{200} = 0.7$

Poisson probabilities with $\lambda = 0.7$ and $e^{-0.7} = 0.4966$:

$P(x) = \frac{e^{-0.7}(0.7)^x}{x!}$

Economics: $n_1 = 60$, $\bar x_1 = 68$. MBS: $n_2 = 50$, $\bar x_2 = 69$. $\sigma = 2.24$ for both groups.

$H_0: \mu_1 = \mu_2$ (equal heights). $H_1: \mu_2 > \mu_1$ (MBS students are taller), a one-tailed test at $\alpha = 0.05$.

$z = \frac{\bar x_2 - \bar x_1}{\sigma\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}} = \frac{1}{2.24\sqrt{\frac{1}{60}+\frac{1}{50}}} = \frac{1}{2.24\times0.1915} = 2.33$

The critical value is $z_{0.05} = 1.645$. Since $2.33 > 1.645$, $H_0$ is rejected.

We conclude that management (MBS) students are taller than economics students.

There are 30 observations, ranging from 66 to 192. The class 68-80 cannot hold 66, so the first class is taken as 60-80. The classes are 60-80, 80-100, 100-120, and so on (a value equal to a limit goes into the higher class).

Central tendency.

$\bar x = \frac{\sum fm}{N} = \frac{3680}{30} = 122.7$

Median: the class 120-140 holds the 15th item, so $M_d = 120 + \dfrac{15-13}{6}\times20 = 126.7$.

Mode: the modal class is 80-100, so $M_o = 80 + \dfrac{10-2}{2\cdot10-2-1}\times20 = 89.4$.

Dispersion.

$\sigma = \sqrt{\frac{\sum f(m-\bar x)^2}{N}} = \sqrt{1132.9} = 33.7,\qquad C.V. = \frac{33.7}{122.7}\times100 = 27.4\%$

Skewness. Karl Pearson's coefficient is $\dfrac{\bar x - M_o}{\sigma} = \dfrac{122.7-89.4}{33.7} = 0.99$. The moment measure is $\beta_1 = \mu_3/\sigma^3 = 0.21$. Both are positive, so the distribution is positively skewed. Mean > median > mode.

Kurtosis. $\beta_2 = \mu_4/\sigma^4 = 2.01 < 3$, so the distribution is platykurtic (flatter than normal).

Conclusion. The average value is about 123, with moderate variation (C.V. about 27%). The distribution is slightly positively skewed, flat-topped and bimodal-looking, with clusters at 80-100 and 140-160.

Total expenditure $= 400+120+250+100+50+180 = \text{Rs } 1100$.

Angle of each sector $= \dfrac{\text{item}}{1100}\times360^\circ$.

Construction. Draw a circle of suitable radius. Starting from a radius, mark the sectors one after another with a protractor, in the order above, using the angles in the table. Shade or colour each sector differently and label it with the item name, or give a legend (key).

The classes are inclusive, so they are converted to continuous classes. The frequencies add up to $N = 150$ (not 100), so $N = 150$ is used.

i. Lowest income of the richest 30% is $P_{70}$ (equivalently $D_7$).

Position $= \frac{70\times150}{100} = 105$, which lies in 9.5-14.5.

$P_{70} = 9.5 + \frac{105-45}{70}\times5 = 9.5 + 4.29 = 13.79$

The lowest income of the richest 30% is about Rs 13,790.

ii. Highest income of the poorest 25% is $Q_1 = P_{25}$.

Position $= \frac{150}{4} = 37.5$, which lies in 4.5-9.5.

$Q_1 = 4.5 + \frac{37.5-10}{35}\times5 = 4.5 + 3.93 = 8.43$

The highest income of the poorest 25% is about Rs 8,430.