Descriptive Statistics and Data Presentation

Steps in testing a statistical hypothesis

1. Set up the hypotheses: State the null hypothesis $H_0$ and the alternative hypothesis $H_1$.
2. Choose the level of significance: Usually $\alpha=0.05$ or $0.01$.
3. Select the test statistic: $z$, $t$, $\chi^2$ or $F$, according to sample size and population conditions.
4. Determine the critical region: Find the critical value from the table for $\alpha$ and the degrees of freedom.
5. Compute the test statistic from the sample data.
6. Decide: If the computed value falls in the critical region, reject $H_0$; otherwise do not reject it.
7. State the conclusion in the context of the problem.

Numerical part

Let sailors be sample 1 and soldiers sample 2.

(a) Sailors: $63,65,68,69,71,72$; $n_1=6$, $\sum x_1=408$

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

$\sum(x_1-\bar x_1)^2=25+9+0+1+9+16=60$, so

$s_1^2=\frac{60}{5}=12 \text{ (inches)}^2,\quad s_1=3.46$

(b) Soldiers: $61,62,65,66,69,70,71,72,73$; $n_2=9$, $\sum x_2=609$

$\bar x_2=\frac{609}{9}=67.67 \text{ inches}$

$\sum(x_2-\bar x_2)^2=152$, so

$s_2^2=\frac{152}{8}=19 \text{ (inches)}^2,\quad s_2=4.36$

(c) Test whether sailors are taller on average

Assume independent samples from normal populations with equal variances (small samples, so a pooled $t$-test).

• $H_0:\mu_1=\mu_2$ against $H_1:\mu_1>\mu_2$ (one-tailed)
• $\alpha=0.05$

Pooled variance:

$s_p^2=\frac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}=\frac{5(12)+8(19)}{13}=\frac{212}{13}=16.31,\quad s_p=4.04$

Test statistic:

$t=\frac{\bar x_1-\bar x_2}{s_p\sqrt{\frac1{n_1}+\frac1{n_2}}}=\frac{68-67.67}{4.04\sqrt{\frac16+\frac19}}=\frac{0.33}{4.04\times0.527}=0.16$

Critical value: $t_{0.05}(13\ \text{d.f.})=1.771$ (one-tailed).

Since $0.16<1.771$, $H_0$ is not rejected. There is no significant evidence at the 5% level that sailors are on average taller than soldiers; the 0.33-inch difference is due to sampling variation.

Parametric and non-parametric tests

Parametric tests assume that the sample comes from a population with a known distribution (usually normal) and test hypotheses about population parameters such as the mean, variance or proportion. Data should be on an interval or ratio scale. Examples: $z$-test and $t$-test for means (e.g. testing whether the mean marks of two colleges differ), $F$-test for equality of variances, ANOVA.

Non-parametric (distribution-free) tests make no assumption about the population distribution and can be used with nominal or ordinal data, small samples or skewed data. Examples: chi-square test, sign test, Wilcoxon signed-rank test, Mann-Whitney $U$ test, Kruskal-Wallis $H$ test, run test.

Applications of the chi-square test

1. Goodness of fit: whether observed frequencies fit a theoretical or claimed distribution (e.g. binomial, Poisson, or a given ratio).
2. Test of independence: whether two attributes in a contingency table are independent (e.g. education level and use of a public service).
3. Test of homogeneity: whether several populations have the same distribution of a characteristic.
4. Test for a population variance: whether a normal population has a specified variance.

Numerical part

$H_0$: Renewals in the four quarters are in the ratio $2:3:4:6$. $H_1$: They are not in this ratio. $\alpha=0.05$.

Observed: $115,230,340,370$; $N=1055$. (The stated total of 903 does not match the quarterly figures, so the sum of the quarters, 1055, is used.)

Expected frequencies (ratio total $=15$):

$E_1=\tfrac{2}{15}(1055)=140.67,\; E_2=\tfrac{3}{15}(1055)=211,\; E_3=\tfrac{4}{15}(1055)=281.33,\; E_4=\tfrac{6}{15}(1055)=422$

$\chi^2=\sum\frac{(O-E)^2}{E}=25.04$

Degrees of freedom $=4-1=3$; $\chi^2_{0.05}(3)=7.815$.

Since $25.04>7.815$, $H_0$ is rejected. The renewals are not in the ratio $2:3:4:6$.

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 selections are independent:

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

i) Probability that none will be selected

$$P(M'\cap F')=P(M')\times P(F')$$ $$=\frac{3}{5}\times \frac{4}{5} =\frac{12}{25} =0.48$$

ii) Probability that at least one will be selected

$$P(M\cup F)=1-P(M'\cap F')$$ $$=1-\frac{12}{25} =\frac{13}{25} =0.52$$

iii) Probability that both will be selected

$$P(M\cap F)=P(M)\times P(F)$$ $$=\frac{2}{5}\times \frac{1}{5} =\frac{2}{25} =0.08$$

iv) Probability that only one will be selected

$$P(\text{only one})=P(M\cap F')+P(M'\cap F)$$ $$=\frac{2}{5}\times \frac{4}{5}+\frac{3}{5}\times \frac{1}{5}$$ $$=\frac{8}{25}+\frac{3}{25} =\frac{11}{25} =0.44$$

Answers:

$$\text{i) } \frac{12}{25}=0.48,\quad \text{ii) } \frac{13}{25}=0.52,\quad \text{iii) } \frac{2}{25}=0.08,\quad \text{iv) } \frac{11}{25}=0.44$$

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b) Normal Distribution

Given:

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

Standard normal variable:

$$Z=\frac{X-\mu}{\sigma}$$

i) Number of employees having expenditure 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,

$$P(150<X<200)=P(-2.5<Z<0)$$ $$=P(0<Z<2.5)=0.4938$$

Number of employees:

$$=4000\times 0.4938=1975.2\approx 1975$$

Approximately 1975 employees.

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ii) Number of employees having expenditure more than Rs 280

For $X=280$:

$$Z=\frac{280-200}{20}=4$$

So,

$$P(X>280)=P(Z>4)$$ $$=0.5-P(0<Z<4)$$ $$=0.5-0.49997=0.00003$$

Number of employees:

$$=4000\times 0.00003=0.12\approx 0$$

Practically no employee has expenditure more than Rs 280.

a) Test whether the two samples have the same variance

Given:

Hypotheses:

• $H_0: \sigma_1^2 = \sigma_2^2$ (the two populations have equal variance)
• $H_1: \sigma_1^2 \neq \sigma_2^2$ (the two populations do not have equal variance)

Test statistic: Since the samples are independent and small, we use the F-test for equality of variances.

Sample variances: $s_1^2 = \frac{\sum(x-\bar x)^2}{n_1-1} = \frac{190}{12-1} = \frac{190}{11} = 17.27$ $s_2^2 = \frac{\sum(x-\bar x)^2}{n_2-1} = \frac{230}{14-1} = \frac{230}{13} = 17.69$

Test statistic: $F = \frac{s_1^2}{s_2^2} = \frac{17.27}{17.69} = 0.976$

Degrees of freedom: $v_1 = n_1 - 1 = 11$, $v_2 = n_2 - 1 = 13$

Critical value: At 5% level of significance for a two-tailed test, the table value is $F_{0.025,\,11,\,13} \approx 3.12$ (upper tail) and the corresponding lower-tail value is about $0.34$.

Decision: Since the calculated $F = 0.976$ lies between the lower and upper critical values ($0.34 < 0.976 < 3.12$), we fail to reject $H_0$.

Conclusion: There is no sufficient evidence to conclude that the two samples have different variances. The two populations may be regarded as having the same variance.

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b) Estimate the loss in production in a week at Kathmandu

Given:

• Number of workers on strike, $X = 1800$
• Mean number of workers on strike, $\bar X = 800$
• Mean loss of daily production, $\bar Y =$ Rs 15 thousand
• SD of workers on strike, $\sigma_x = 100$
• SD of loss of daily production, $\sigma_y =$ Rs 2 thousand
• Coefficient of correlation, $r = 0.80$

Regression line of $Y$ on $X$: $b_{yx} = r \cdot \frac{\sigma_y}{\sigma_x} = 0.80 \times \frac{2}{100} = 0.016$

$Y - \bar Y = b_{yx}(X - \bar X)$ $Y - 15 = 0.016(1800 - 800)$ $Y - 15 = 0.016 \times 1000 = 16$ $Y = 15 + 16 = 31$

Loss in daily production = Rs 31 thousand

Loss in production in a week (7 days): $= 31 \times 7 = 217$

Estimated loss in production in a week at Kathmandu = Rs 217 thousand (Rs 2,17,000).

(a) Variability of paddy yield in the two districts

The question asks whether the standard deviation in district 1 is smaller (more consistent yield) than in district 2.

Given: $n_1=n_2=100$, $\bar x_1=225$, $s_1=11$; $\bar x_2=180$, $s_2=13$. $\alpha=0.01$.

$H_0:\sigma_1=\sigma_2$ against $H_1:\sigma_1<\sigma_2$ (one-tailed).

With large samples, use the $z$-test for the difference of standard deviations:

$z=\frac{s_1-s_2}{\sqrt{\dfrac{s_1^2}{2n_1}+\dfrac{s_2^2}{2n_2}}}=\frac{11-13}{\sqrt{\dfrac{121}{200}+\dfrac{169}{200}}}=\frac{-2}{\sqrt{1.45}}=-1.66$

Critical value: $z_{0.01}=-2.33$. Since $-1.66>-2.33$, $H_0$ is not rejected.

(Check with the $F$-test: $F=\dfrac{13^2}{11^2}=1.40$, which is below the one-tailed 1% critical value of about $1.7$ for $(99,99)$ d.f., giving the same decision.)

Conclusion: At the 1% level there is not enough evidence that yield is less variable in district 1; the difference in standard deviations (11 vs 13) can arise from sampling variation.

(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 observations together (ties get average ranks; e.g. 68 is rank 1, and the four 110s share rank 14.5, the two 140s rank 19.5, the two 160s rank 22.5, the two 95s rank 9.5). Rank sums:

(Total $=300=\frac{24\times25}{2}$, as required.)

$H=\frac{12}{N(N+1)}\sum\frac{R_i^2}{n_i}-3(N+1)=\frac{12}{24\times25}\left(\frac{129.5^2+120.5^2+50^2}{8}\right)-3(25)$

$H=0.02\times\frac{16770.25+14520.25+2500}{8}-75=0.02\times4223.8-75=9.48$

(Correcting for ties gives $H\approx9.53$, with the same conclusion.)

Degrees of freedom $=k-1=2$; $\chi^2_{0.05}(2)=5.991$.

Since $9.48>5.991$, $H_0$ is rejected.

Conclusion: At the 5% level, the dollar amounts differ significantly by payment method. Cash purchases have clearly the lowest ranks, so customers spend less per purchase when paying cash than with credit card or cheque.

Given:

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

Formula:

[ r=\pm \sqrt{b_{yx}\times b_{xy}} ]

Substituting:

[ r=\sqrt{1.25\times 0.8} ]

[ r=\sqrt{1.00}=1 ]

Since both regression coefficients are positive, the correlation coefficient is positive.

[ \boxed{r=+1} ]

Interpretation:
The coefficient of correlation is (+1), which indicates a perfect positive correlation between (X) and (Y). It means the two variables are perfectly linearly related in the same direction; as (X) increases, (Y) also increases in exact proportion. Also, (r^2=1), so 100% of the variation in one variable is explained by the other.