Statistics for Public Administration: Descriptive and Inferential

1. Steps in testing a statistical hypothesis

1. Set up the null and alternative hypotheses
   (H_0): no difference or no effect.
   (H_1): the research claim to be tested.

2. Choose the level of significance
   Usually (\alpha=0.05) or (0.01). This fixes the probability of Type I error.

3. Select the appropriate test statistic
   Depending on the data: (z), (t), (\chi^2), (F), etc.
   For two independent small samples, use the t-test.

4. Determine the critical region or p-value
   Find the critical value from the relevant table using (\alpha) and degrees of freedom.
   For a one-tailed test, use one tail; for a two-tailed test, use both tails.

5. Compute the test statistic from sample data
   Calculate the sample mean, variance, standard error, and then the test statistic.

6. Compare computed value with critical value
   If computed value falls in the critical region, reject (H_0).
   Or compare p-value with (\alpha).

7. Make the decision
   Reject (H_0) or fail to reject (H_0).

8. State the conclusion in context
   Say whether the sample provides sufficient evidence for the claim.

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2. Given data

Let sailors be sample 1 and soldiers be sample 2.

Sailors:
[ 63,\ 65,\ 68,\ 69,\ 71,\ 72 ] [ n_1=6 ]

Soldiers:
[ 61,\ 62,\ 65,\ 66,\ 69,\ 70,\ 71,\ 72,\ 73 ] [ n_2=9 ]

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(a) Mean height and variance of sailors

[ \sum x_1=63+65+68+69+71+72=408 ]

[ \bar x_1=\frac{408}{6}=68\text{ inches} ]

[ \sum x_1^2=63^2+65^2+68^2+69^2+71^2+72^2=27804 ]

Sample variance:

[ s_1^2=\frac{\sum x_1^2-\frac{(\sum x_1)^2}{n_1}}{n_1-1} ]

[ s_1^2=\frac{27804-\frac{408^2}{6}}{5} ]

[ s_1^2=\frac{27804-27744}{5}=\frac{60}{5}=12 ]

[ s_1=\sqrt{12}=3.464\text{ inches} ]

Answer:
Mean height of sailors (=68) inches.
Variance (=12\text{ in}^2).
Standard deviation (=3.464) inches.

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(b) Mean height and variation of soldiers

[ \sum x_2=61+62+65+66+69+70+71+72+73=609 ]

[ \bar x_2=\frac{609}{9}=67.667\text{ inches} ]

[ \sum x_2^2=61^2+62^2+65^2+66^2+69^2+70^2+71^2+72^2+73^2=41361 ]

Sample variance:

[ s_2^2=\frac{\sum x_2^2-\frac{(\sum x_2)^2}{n_2}}{n_2-1} ]

[ s_2^2=\frac{41361-\frac{609^2}{9}}{8} ]

[ s_2^2=\frac{41361-41209}{8}=\frac{152}{8}=19 ]

[ s_2=\sqrt{19}=4.359\text{ inches} ]

Coefficient of variation:

[ CV_2=\frac{s_2}{\bar x_2}\times 100 ]

[ CV_2=\frac{4.359}{67.667}\times 100=6.44% ]

Answer:
Mean height of soldiers (=67.667) inches.
Variance (=19\text{ in}^2).
Standard deviation (=4.359) inches.
Coefficient of variation (=6.44%).

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(c) Are sailors on average taller than soldiers? Test the hypothesis.

Hypotheses

[ H_0:\mu_1=\mu_2 ] [ H_1:\mu_1>\mu_2 ]

This is a right-tailed test.

Take (\alpha=0.05).

Test statistic

Since the samples are independent and small, use the pooled (t)-test.

Pooled variance:

[ s_p^2=\frac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2} ]

[ s_p^2=\frac{(6-1)(12)+(9-1)(19)}{6+9-2} ]

[ s_p^2=\frac{60+152}{13}=\frac{212}{13}=16.308 ]

Standard error:

[ SE=\sqrt{s_p^2\left(\frac{1}{n_1}+\frac{1}{n_2}\right)} ]

[ SE=\sqrt{16.308\left(\frac{1}{6}+\frac{1}{9}\right)} ]

[ SE=\sqrt{16.308\times 0.2778} ]

[ SE=\sqrt{4.530}=2.128 ]

Test statistic:

[ t=\frac{\bar x_1-\bar x_2}{SE} ]

[ t=\frac{68-67.667}{2.128} ]

[ t=\frac{0.333}{2.128}=0.157 ]

Degrees of freedom:

[ df=n_1+n_2-2=6+9-2=13 ]

Critical value at (\alpha=0.05), right-tailed, (df=13):

[ t_{0.05,13}=1.771 ]

Decision

[ t_{\text{cal}}=0.157 < t_{\text{crit}}=1.771 ]

So we fail to reject (H_0).

Also, the p-value is approximately (0.44), which is greater than (0.05).

Conclusion

At the 5% level of significance, there is not sufficient evidence to conclude that sailors are on average taller than soldiers. The observed difference of (0.333) inch is statistically insignificant.

Parametric tests assume the data come from a population with a known distribution (usually normal) and test hypotheses about parameters such as the mean or variance. They need interval or ratio data. Examples: the z-test, t-test, F-test and ANOVA.

Non-parametric tests are distribution-free. They make no assumption about the population shape and can be used with nominal or ordinal data, small samples or data with outliers. They are less powerful when the parametric assumptions hold. Examples: chi-square test, sign test, Mann-Whitney U test, Wilcoxon test, Kruskal-Wallis H-test.

Applications of the chi-square test:

• Goodness of fit: whether observed frequencies fit a theoretical or expected distribution.
• Test of independence: whether two attributes are associated (for example, gender and voting preference) in a contingency table.
• Test of homogeneity of proportions across populations.
• Test of a single population variance.

Numerical problem

$H_0$: the renewals are in the ratio 2:3:4:6. $H_1$: they are not.

Total = 903, ratio total = 15, so $E = \frac{903}{15}\times(2,3,4,6) = 120.4,\ 180.6,\ 240.8,\ 361.2$.

$\chi^2_{cal} = \sum\frac{(O-E)^2}{E} = 10.81,\quad df = 4-1 = 3,\quad \chi^2_{0.05,3} = 7.815$

Since $10.81 > 7.815$, reject $H_0$. The renewals are not in the ratio 2:3:4:6 at the 5% level.

a) Probability

Let
$M$ = event that the male is selected,
$F$ = event that the female is selected.

Given: $P(M)=\frac{2}{5},\qquad P(F)=\frac{1}{5}$

Assuming the two selections are independent:

$P(M')=\frac{3}{5},\qquad P(F')=\frac{4}{5}$

i) None of them will be selected

$P(M'\cap F')=P(M')P(F')=\frac{3}{5}\times \frac{4}{5}=\frac{12}{25}$

ii) At least one of them will be selected

$P(M\cup F)=1-P(M'\cap F')=1-\frac{12}{25}=\frac{13}{25}$

iii) Both of them will be selected

$P(M\cap F)=P(M)P(F)=\frac{2}{5}\times \frac{1}{5}=\frac{2}{25}$

iv) Only one of them will be selected

$P(\text{only one})=P(M\cap F')+P(M'\cap F)$

$$=\frac{2}{5}\times \frac{4}{5}+\frac{3}{5}\times \frac{1}{5} =\frac{8}{25}+\frac{3}{25} =\frac{11}{25}$$

Answers:

• None selected: $\frac{12}{25}$
• At least one selected: $\frac{13}{25}$
• Both selected: $\frac{2}{25}$
• Only one selected: $\frac{11}{25}$

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b) Normal Distribution

Given: $N=4000,\qquad \mu=200,\qquad \sigma=20$

Standard normal variable: $z=\frac{x-\mu}{\sigma}=\frac{x-200}{20}$

i) Between Rs 150 and Rs 200

For $x=150$: $z=\frac{150-200}{20}=-2.5$

For $x=200$: $z=\frac{200-200}{20}=0$

So, $P(150<X<200)=P(-2.5<Z<0)$

Using the standard normal table: $P(0<Z<2.5)=0.4938$

Therefore, $P(150<X<200)=0.4938$

Number of employees: $=4000\times 0.4938=1975.2\approx 1975$

So, about 1975 employees have expenditure between Rs 150 and Rs 200.

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ii) More than Rs 280

For $x=280$: $z=\frac{280-200}{20}=4$

So, $P(X>280)=P(Z>4)$

Using the standard normal table: $P(Z>4)=0.00003 \quad \text{(approximately)}$

Number of employees: $=4000\times 0.00003=0.12\approx 0$

So, practically no employee has expenditure more than Rs 280.

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Final Answers:

a)
i) $\frac{12}{25}$
ii) $\frac{13}{25}$
iii) $\frac{2}{25}$
iv) $\frac{11}{25}$

b)
i) Approximately 1975 employees
ii) Approximately 0 employees

(a) Test of equality of variances (F-test)

$H_0:\sigma_1^2=\sigma_2^2$ against $H_1:\sigma_1^2\neq\sigma_2^2$ (two-tailed, $\alpha=0.05$).

$s_1^2=\frac{190}{11}=17.27,\qquad s_2^2=\frac{230}{13}=17.69$

Taking the larger variance in the numerator:

$F_{cal}=\frac{17.69}{17.27}=1.024,\quad (v_1,v_2)=(13,11)$

The critical value is $F_{0.025}(13,11)\approx 3.4$ (the table gives 3.43 for 12 and 3.33 for 15 numerator d.f.). Since $1.024 < 3.4$, we fail to reject $H_0$. The two samples can be regarded as having the same variance.

(b) Estimated weekly loss

Regression of daily loss $Y$ (Rs '000) on strikers $X$:

$b_{YX}=r\frac{\sigma_Y}{\sigma_X}=0.80\times\frac{2}{100}=0.016$

$Y-15=0.016\,(1800-800)=16\ \Rightarrow\ Y=31$

Estimated daily loss = Rs 31,000.

$\text{Weekly loss}=31\times7=217\ \text{(in '000)}$

Estimated loss in a week = Rs 217,000.

(a) Test of the claim that the first district's mean yield is higher

Take 100 plots in each district. $n_1=n_2=100$, $\bar x_1=225$, $\bar x_2=180$, common $\sigma=12$ (the standard deviation for the entire area is used).

$H_0:\mu_1=\mu_2$; $H_1:\mu_1>\mu_2$ (right-tailed, $\alpha=0.01$).

$z=\frac{\bar x_1-\bar x_2}{\sigma\sqrt{\frac1{n_1}+\frac1{n_2}}}=\frac{45}{12\sqrt{0.02}}=\frac{45}{1.697}=26.5$

Critical value $z_{0.01}=2.33$. Since $26.5>2.33$, reject $H_0$. The mean yield in the first district is significantly better than in the second.

(b) Kruskal-Wallis H-test

$H_0$: there is no difference in the dollar amounts paid by credit card, cheque and cash. $H_1$: at least one differs. $\alpha=0.05$.

Rank all 24 values together (ties get average ranks):

(Check: 122.5 + 135.5 + 42 = 300 = 24·25/2.)

$H=\frac{12}{N(N+1)}\sum\frac{R_i^2}{n_i}-3(N+1)=\frac{12}{24\cdot25}\left(\frac{122.5^2+135.5^2+42^2}{8}\right)-75=12.83$

With the tie correction, $H\approx12.9$. Critical value $\chi^2_{0.05,\,2}=5.991$. Since $12.83>5.991$, reject $H_0$. The amounts differ by payment method: credit card and cheque purchases are larger, and cash purchases are smaller.

Given:

• Regression coefficient of (Y) on (X): (b_{yx}=1.25)
• Regression coefficient of (X) on (Y): (b_{xy}=0.8)

Formula: [ r=\sqrt{b_{yx}\times b_{xy}} ] The sign of (r) is the same as the sign of the regression coefficients. Here both are positive, so (r) is positive.

Calculation: [ r=\sqrt{1.25\times 0.8} ] [ r=\sqrt{1.00} ] [ r=+1 ]

Result: [ \boxed{r=+1} ]

Interpretation: The coefficient of correlation is (+1), which indicates a perfect positive correlation between (X) and (Y). This means the two variables are perfectly linearly related in the same direction: as (X) increases, (Y) also increases proportionally, and all the data points lie exactly on a straight line with positive slope.