Normal distribution

A continuous random variable (X) is said to follow a normal distribution if its probability density function is

[ f(x)=\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2},\qquad -\infty<x<\infty ]

where (\mu) is the mean and (\sigma) is the standard deviation. It is a symmetric bell-shaped distribution. The mean, median and mode are equal, and the total area under the curve is 1. It is denoted by

[ X\sim N(\mu,\sigma^2) ]

For calculations, we use the standard normal variable

[ z=\frac{x-\mu}{\sigma} ]

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Given:

[ \mu=72,\qquad \sigma=7.5 ]

(i) Percentage of students in grade A

Grade A means score (X\ge 80).

[ z=\frac{80-72}{7.5}=\frac{8}{7.5}=1.0667 ]

[ P(X\ge 80)=P(z\ge 1.0667) ]

From the standard normal table,

[ P(z\le 1.0667)\approx 0.8569 ]

Therefore,

[ P(z\ge 1.0667)=1-0.8569=0.1431 ]

So the percentage of students in grade A is

[ 0.1431\times 100=14.31% ]

[ \boxed{\text{Grade A students} \approx 14.31%} ]

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(ii) Lowest score of the top 10% students

Top 10% students means the score above which only 10% of students lie. Hence, we need the 90th percentile.

Let the required score be (x). Then

[ P(X\ge x)=0.10 ]

or,

[ P(X\le x)=0.90 ]

For (P(z\le z_0)=0.90), from the standard normal table,

[ z_0\approx 1.2816 ]

Now,

[ z_0=\frac{x-\mu}{\sigma} ]

[ 1.2816=\frac{x-72}{7.5} ]

[ x=72+1.2816\times 7.5 ]

[ x=72+9.612 ]

[ x=81.612 ]

Therefore, the lowest score of the top 10% students is approximately

[ \boxed{81.61} ]

If marks are taken as whole numbers, then the lowest score is

[ \boxed{82} ]

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Final answers

[ \boxed{\text{(i) Grade A students} \approx 14.31%} ]

[ \boxed{\text{(ii) Lowest score of top 10% students} \approx 81.61 \text{ or } 82} ]

Given:
Population mean, $\mu_0 = 100$
Sample size, $n = 10$
Significance level, $\alpha = 0.05$
Sample observations:
$118,\ 32,\ 96,\ 106,\ 124,\ 91,\ 122,\ 90,\ 102,\ 104$

Calculation of sample mean and standard deviation

[ \sum x = 985 ]

[ \bar{x} = \frac{\sum x}{n} = \frac{985}{10} = 98.5 ]

[ \sum x^2 = 103261 ]

[ s^2 = \frac{\sum x^2 - \frac{(\sum x)^2}{n}}{n-1} ]

[ s^2 = \frac{103261 - \frac{(985)^2}{10}}{9} ]

[ s^2 = \frac{103261 - 97022.5}{9} ]

[ s^2 = \frac{6238.5}{9} = 693.1667 ]

[ s = \sqrt{693.1667} = 26.3281 ]

Standard error:

[ SE = \frac{s}{\sqrt{n}} = \frac{26.3281}{\sqrt{10}} = 8.3257 ]

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i. Test at 5% significance level

Since the population standard deviation is not given and the sample size is small $(n=10)$, we use the t-test.

Hypotheses

[ H_0: \mu = 100 ]

[ H_1: \mu \ne 100 ]

This is a two-tailed test.

Test statistic

[ t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} ]

[ t = \frac{98.5 - 100}{8.3257} ]

[ t = \frac{-1.5}{8.3257} = -0.1802 ]

Degrees of freedom:

[ df = n - 1 = 10 - 1 = 9 ]

At 5% significance level for a two-tailed test,

[ t_{0.025,9} = 2.262 ]

Decision

[ |t| = 0.1802 < 2.262 ]

So we fail to reject $H_0$.

Conclusion

At the 5% significance level, there is no significant evidence that sufferers from the disease differ from the general population in their performance on the test.
The p-value is approximately $0.861 > 0.05$.

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ii. 95% confidence limits of the true population mean

The 95% confidence interval is:

[ \bar{x} \pm t_{0.025,9} \cdot \frac{s}{\sqrt{n}} ]

[ = 98.5 \pm 2.262 \times 8.3257 ]

[ = 98.5 \pm 18.834 ]

Lower limit:

[ 98.5 - 18.834 = 79.666 ]

Upper limit:

[ 98.5 + 18.834 = 117.334 ]

[ \boxed{95%\ \text{confidence limits} = (79.67,\ 117.33)} ]

Since this interval includes $100$, it agrees with the test result that the sufferers’ mean score does not differ significantly from the general population mean.

Let
(x) = dosage, (y) = duration of relief, (n=8).

[ \sum x=56,\quad \sum y=140,\quad \sum x^2=424,\quad \sum y^2=2676,\quad \sum xy=1053 ]

[ \bar x=\frac{56}{8}=7,\qquad \bar y=\frac{140}{8}=17.5 ]

Compute sums of squares:

[ S_{xx}=\sum x^2-\frac{(\sum x)^2}{n}=424-\frac{56^2}{8}=32 ]

[ S_{yy}=\sum y^2-\frac{(\sum y)^2}{n}=2676-\frac{140^2}{8}=226 ]

[ S_{xy}=\sum xy-\frac{(\sum x)(\sum y)}{n}=1053-\frac{56\times140}{8}=73 ]

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i. Relationship between (x) and (y)

Karl Pearson’s correlation coefficient:

[ r=\frac{S_{xy}}{\sqrt{S_{xx}S_{yy}}} ]

[ r=\frac{73}{\sqrt{32\times226}} =\frac{73}{\sqrt{7232}} =\frac{73}{85.041} =0.8584 ]

[ \boxed{r=0.858 \text{ approximately}} ]

Interpretation: There is a strong positive linear relationship between dosage and duration of relief. As dosage increases, the duration of relief also tends to increase. Also,

[ r^2=(0.8584)^2=0.7369 ]

So about 73.69% of the variation in duration of relief is explained by dosage.

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ii. Regression equation of (y) on (x)

Regression coefficient:

[ b_{yx}=\frac{S_{xy}}{S_{xx}}=\frac{73}{32}=2.28125 ]

Intercept:

[ a=\bar y-b_{yx}\bar x ]

[ a=17.5-(2.28125)(7)=17.5-15.96875=1.53125 ]

Hence the regression equation is

[ \boxed{\hat y=1.53125+2.28125x} ]

or approximately,

[ \boxed{\hat y=1.53+2.28x} ]

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iii. Estimate (y) when (x=8)

[ \hat y=1.53125+2.28125(8) ]

[ \hat y=1.53125+18.25=19.78125 ]

[ \boxed{\hat y \approx 19.78} ]

So when dosage is (8), the estimated duration of relief is about 19.78 units.

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iv. Interpretation of the slope

The slope is

[ b_{yx}=2.28125 ]

This means that for every one unit increase in dosage, the duration of relief increases on average by about

[ \boxed{2.28 \text{ units}} ]

The positive slope confirms the positive relationship between dosage and duration of relief.

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v. Standard error of estimate

The standard error of estimate for regression of (y) on (x) is

[ S_e=\sqrt{\frac{S_{yy}-b_{yx}S_{xy}}{n-2}} ]

[ S_e=\sqrt{\frac{226-\left(\frac{73}{32}\right)(73)}{8-2}} ]

[ S_e=\sqrt{\frac{226-166.53125}{6}} ]

[ S_e=\sqrt{\frac{59.46875}{6}} ]

[ S_e=\sqrt{9.91146}=3.148 ]

[ \boxed{S_e \approx 3.15 \text{ units}} ]

Thus the typical prediction error in estimating duration of relief from dosage is about 3.15 units.

i. Conditions for a binomial experiment

Let (X) be the number of defective parts in the two selected parts. The situation is a binomial experiment if:

1. Fixed number of trials: Exactly (n=2) parts are selected.
2. Two possible outcomes: Each part is either defective (success) or non-defective (failure).
3. Constant probability of success: The probability that a part is defective remains (p=0.03) for each part.
4. Independent trials: The result for one part does not affect the result for the other part. This is satisfied if the parts are selected randomly from a very large population or if sampling is done with replacement.
5. Variable of interest: We count the total number of defective parts, (X), in the two trials.

Thus, (X \sim B(n=2,\ p=0.03)).

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ii. Computation of probabilities

Here, [ n=2,\qquad p=0.03,\qquad q=1-p=0.97 ]

The binomial probability formula is: [ P(X=x)={}^nC_x p^x q^{n-x} ]

1. Probability of no defective part: [ P(X=0)={}^2C_0(0.03)^0(0.97)^2 ] [ =1\times 1\times 0.9409 ] [ =0.9409 ] So, [ P(X=0)=0.9409=94.09% ]

2. Probability of one defective part: [ P(X=1)={}^2C_1(0.03)^1(0.97)^1 ] [ =2\times 0.03\times 0.97 ] [ =0.0582 ] So, [ P(X=1)=0.0582=5.82% ]

3. Probability of two defective parts: [ P(X=2)={}^2C_2(0.03)^2(0.97)^0 ] [ =1\times 0.0009\times 1 ] [ =0.0009 ] So, [ P(X=2)=0.0009=0.09% ]

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Probability distribution table

Thus, the probabilities are: [ \boxed{P(X=0)=0.9409,\quad P(X=1)=0.0582,\quad P(X=2)=0.0009} ]

Interval estimation: Interval estimation is the method of estimating an unknown population parameter by constructing a range of values, called a confidence interval, within which the true parameter is expected to lie with a specified degree of confidence. Unlike a point estimate, it also indicates the precision of the estimate.

Given:

• Population standard deviation, $\sigma = 50$ hours
• Sample size, $n = 100$
• Sample mean, $\bar{x} = 875$ hours
• Confidence level = 95%, so $z_{0.025} = 1.96$

(i) Standard error of sample mean

$$SE_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{50}{\sqrt{100}} = \frac{50}{10} = 5 \text{ hours}$$

(ii) Margin of error for 95% confidence

$$E = z_{\alpha/2} \cdot SE_{\bar{x}} = 1.96 \times 5 = 9.8 \text{ hours}$$

(iii) 95% confidence interval for the population mean

$$CI = \bar{x} \pm E = 875 \pm 9.8$$ $$= (875 - 9.8,\; 875 + 9.8) = (865.2,\; 884.8) \text{ hours}$$

Conclusion: The 95% confidence interval for the population mean life of all bulbs produced by the firm is 865.2 hours to 884.8 hours.

Given:
Mean, $\mu = 10$ cm
Standard deviation, $\sigma = 0.04$ cm
Let $X$ be the inside diameter of a piston ring.
$X \sim N(10,\ 0.04^2)$

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(i) Proportion of rings with inside diameter exceeding 10.08 cm

We need $P(X > 10.08)$.

Standardize: $z = \frac{X - \mu}{\sigma} = \frac{10.08 - 10}{0.04} = \frac{0.08}{0.04} = 2$

So, $P(X > 10.08) = P(Z > 2) = 1 - P(Z \le 2)$

From the standard normal table: $P(Z \le 2) = 0.9772$

Therefore, $P(Z > 2) = 1 - 0.9772 = 0.0228$

Answer: The proportion of rings exceeding 10.08 cm is 0.0228 (about 2.28%).

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(ii) Value below which 10% of the piston rings fall

Let $x_0$ be the required value such that: $P(X < x_0) = 0.10$

Standardize: $P\left(Z < \frac{x_0 - 10}{0.04}\right) = 0.10$

From the standard normal table, the $z$-value corresponding to a left-tail area of 0.10 is: $z = -1.28$

So, $\frac{x_0 - 10}{0.04} = -1.28$

$x_0 - 10 = -1.28 \times 0.04 = -0.0512$

$x_0 = 10 - 0.0512 = 9.9488 \text{ cm}$

Answer: 10% of the piston rings will fall below 9.9488 cm (approximately 9.95 cm).

(i) Association of Attributes

When two or more attributes (qualitative characteristics such as literacy, gender, or employment) are studied together, we examine whether they are associated — that is, whether the presence or absence of one attribute influences the other. If the attributes are independent, they are said to have no association; if they are related, they have association.

Association is measured by comparing observed frequencies with the frequencies expected under the assumption of independence. The main methods are:

• Yule's coefficient of association, $Q = \frac{(AB)(ab) - (Ab)(aB)}{(AB)(ab) + (Ab)(aB)}$
• Yule's coefficient of colligation, $Y = \frac{\sqrt{(AB)(ab)} - \sqrt{(Ab)(aB)}}{\sqrt{(AB)(ab)} + \sqrt{(Ab)(aB)}}$

where $A$ and $a$ denote the presence and absence of the first attribute, and $B$ and $b$ those of the second. The value of $Q$ lies between $-1$ and $+1$: $Q = 0$ means independence, $Q = +1$ means perfect positive association, and $Q = -1$ means perfect negative association. For a $2 \times 2$ contingency table, the chi-square test can also be used to test the significance of association.

(iii) Type I Error and Type II Error in Testing of Hypothesis

In hypothesis testing we decide either to reject or not reject the null hypothesis $H_0$. Since the decision is based on sample data, two kinds of errors are possible:

Type I error: Rejecting $H_0$ when it is actually true. The probability of this error is denoted by $\alpha$, called the level of significance. For example, concluding that a medicine is effective when it is actually not.

Type II error: Failing to reject $H_0$ when it is actually false. Its probability is denoted by $\beta$. For example, concluding that a medicine is not effective when it actually is.

The quantity $1 - \beta$ is called the power of the test — the probability of correctly rejecting a false $H_0$. In practice, $\alpha$ is fixed in advance (commonly 0.05 or 0.01), and the test is designed to keep $\beta$ as small as possible. For a fixed sample size, reducing one type of error generally increases the other; both can be reduced simultaneously only by increasing the sample size.

Assumptions required for regression analysis

1. Linearity: The relationship between the dependent variable and independent variable(s) is linear.
2. Zero mean of errors: The error term has mean zero, (E(\varepsilon)=0).
3. Homoscedasticity: The variance of the error term is constant for all values of the independent variable.
4. Independence of errors: Error terms are independent of each other; there is no autocorrelation.
5. Normality of errors: For hypothesis testing and confidence intervals, errors should be normally distributed.
6. No multicollinearity: In multiple regression, independent variables should not be highly correlated with each other.
7. No serious outliers/influential points: Extreme observations should not distort the regression results.
8. Correct model specification: The model includes the relevant variables and is correctly specified.
9. Variables measured without error: Independent variables are assumed to be measured accurately.

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Let
(x) = mileage in 1000 miles
(y) = sale price in 100 dollars

[ n=5,\qquad \bar x=\frac{340}{5}=68,\qquad \bar y=\frac{340}{5}=68 ]

i. Relationship between mileage and sale price

[ r=\frac{n\Sigma xy-\Sigma x\Sigma y} {\sqrt{[n\Sigma x^2-(\Sigma x)^2][n\Sigma y^2-(\Sigma y)^2]}} ]

[ r=\frac{5(22830)-(340)(340)} {\sqrt{[5(23836)-340^2][5(23250)-340^2]}} ]

[ r=\frac{114150-115600} {\sqrt{(119180-115600)(116250-115600)}} ]

[ r=\frac{-1450}{\sqrt{(3580)(650)}}=-0.9505 ]

There is a strong negative linear relationship between mileage and sale price. As mileage increases, sale price decreases. Also,

[ r^2=(-0.9505)^2=0.9035 ]

So about 90.35% of the variation in sale price is explained by mileage.

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ii. Estimated regression equation by least squares method

[ b=\frac{n\Sigma xy-\Sigma x\Sigma y}{n\Sigma x^2-(\Sigma x)^2} ]

[ b=\frac{-1450}{3580}=-0.4050 ]

[ a=\bar y-b\bar x=68-(-0.4050)(68) ]

[ a=68+27.5419=95.5419 ]

Therefore, the estimated regression equation is:

[ \boxed{\hat y=95.5419-0.4050x} ]

where (x) is mileage in 1000 miles and (\hat y) is sale price in 100 dollars.

In dollar terms:

[ \boxed{\text{Predicted sale price}=9554.19-40.50x\text{ dollars}} ]

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iii. Interpretation of the slope

The slope is:

[ b=-0.4050 ]

It means that for every additional 1,000 miles of mileage, the predicted sale price decreases by 0.4050 hundred dollars, i.e.

[ 0.4050\times 100=40.50\text{ dollars} ]

So, each additional 1,000 miles reduces the predicted sale price by about 40.50 dollars.

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iv. Predict the sale price for 70,000 miles

[ x=\frac{70000}{1000}=70 ]

[ \hat y=95.5419-0.4050(70) ]

[ \hat y=95.5419-28.3519=67.1899 ]

Since (\hat y) is in 100 dollars,

[ \text{Predicted sale price}=67.1899\times 100=6718.99\text{ dollars} ]

[ \boxed{\text{Predicted sale price}\approx 6,719\text{ dollars}} ]

So, a 2015 Honda Accord with 70,000 miles is predicted to sell for about 6,719 dollars.

Note: If the table unit is strictly 1000 km, then 70,000 miles = 112.654 thousand km, and the predicted price would be about 4,991 dollars. The standard version of this problem uses mileage in 1000 miles.

Chi-square test of independence

Hypotheses

$H_0$: The proportion of social media users who use a mobile phone to access social networking is the same across all age groups (no association between age group and use of mobile phone).

$H_1$: The proportions differ among the age groups (there is an association).

Observed frequencies ($f_o$)

Expected frequencies $f_e = \dfrac{(\text{row total})(\text{column total})}{\text{grand total}}$

Calculation of $\chi^2$

$\chi^2 = \sum \frac{(f_o - f_e)^2}{f_e}$

$\chi^2_{cal} = 78.08$

Critical value

Degrees of freedom $= (r-1)(c-1) = (2-1)(3-1) = 2$

At $\alpha = 0.05$, $\chi^2_{0.05,\,2} = 5.991$

Decision

Since $\chi^2_{cal} = 78.08 > \chi^2_{tab} = 5.991$, we reject $H_0$.

Conclusion

At the 0.05 level of significance there is sufficient evidence that the proportions of social media users who use a mobile phone to access social networking differ among the age groups. The 65+ group uses mobile phones far less than the younger groups.

This is a paired t-test (same subjects measured before and after the drug). Using the recorded reaction times (tenths of a second):

Hypotheses: $H_0: \mu_d = 0$ (the drug does not change reaction time); $H_1: \mu_d < 0$ (After > Before, i.e. the drug increases reaction time). One-tailed (left) test.

Level of significance: $\alpha = 0.05$.

Test statistic:

$\bar d = \frac{\sum d}{n} = \frac{-1.6}{5} = -0.32$

$s_d^2 = \frac{\sum d^2 - (\sum d)^2/n}{n-1} = \frac{0.54 - 0.512}{4} = 0.007,\qquad s_d = 0.0837$

$t = \frac{\bar d}{s_d/\sqrt n} = \frac{-0.32}{0.0837/\sqrt5} = \frac{-0.32}{0.0374} = -8.55$

Critical value: $df = n-1 = 4$; for the left-tailed test, $t_{0.05,4} = -2.132$.

Decision: Since $t_{cal} = -8.55 < -2.132$, it lies in the rejection region, so $H_0$ is rejected.

Conclusion: At the 5% level there is sufficient evidence that the drug increases the reaction time.

Conditional probability

The conditional probability of an event $A$ given that an event $B$ has already occurred is the probability that $A$ occurs when it is known that $B$ has occurred. It is defined as

$P(A \mid B) = \frac{P(A \cap B)}{P(B)}, \qquad P(B) > 0.$

Here $P(A \cap B)$ is the probability that both $A$ and $B$ occur, and $P(B)$ is the probability of the given (conditioning) event.

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Numerical part

Let the events be:

• $M$: the selected student is male
• $F$: the selected student is female
• $A$: the selected student got an "A" grade

Given data:

$P(M) = 0.60, \quad P(F) = 0.40$ $P(A \mid M) = 0.25, \quad P(A \mid F) = 0.20$

Step 1: Probability that a randomly selected student got an "A" grade

By the theorem of total probability,

$P(A) = P(A \mid M)\,P(M) + P(A \mid F)\,P(F)$

$P(A) = (0.25)(0.60) + (0.20)(0.40)$

$P(A) = 0.15 + 0.08 = 0.23$

Step 2: Required conditional probability

We need $P(F \mid A)$, the probability that the student is female given that the student got an "A" grade. By Bayes' theorem,

$P(F \mid A) = \frac{P(A \mid F)\,P(F)}{P(A)}$

$P(F \mid A) = \frac{0.20 \times 0.40}{0.23}$

$P(F \mid A) = \frac{0.08}{0.23} = \frac{8}{23} \approx 0.3478$

Result

$P(F \mid A) = \frac{8}{23} \approx 0.348$

So, if a student is observed to have an "A" grade, the probability that the student is female is approximately 0.348 (about 34.8%).

Given: $n = 500$, $x = 340$, confidence level $90\%$, so $z_{\alpha/2} = 1.645$.

i. Point estimate

$\hat{p} = \frac{x}{n} = \frac{340}{500} = 0.68$

ii. Margin of error at 90% confidence

$SE = \sqrt{\frac{\hat{p}\hat{q}}{n}} = \sqrt{\frac{0.68\times 0.32}{500}} = \sqrt{0.0004352} \approx 0.02086$

$E = z_{\alpha/2}\,SE = 1.645\times 0.02086 \approx 0.0343$

iii. 90% confidence interval

$\hat{p} \pm E = 0.68 \pm 0.0343 = (0.6457,\ 0.7143)$

We are 90% confident that between about 64.6% and 71.4% of all adults consider the future health of social security a major economic concern.

Step 1: Hypotheses

• $H_0$: The theoretical model fits the observed frequencies.
• $H_1$: The theoretical model does not fit the observed frequencies.

Step 2: Test statistic

$\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i}$

Step 3: Computation

Step 4: Degrees of freedom

No parameter is estimated from the data, so $df = 9 - 1 = 8$.

Step 5: Critical value

At the 5% level, $\chi^2_{0.05,\,8} = 15.507$.

Step 6: Decision

Since $\chi^2_{cal} = 10.14 < 15.507$, we fail to reject $H_0$.

Conclusion: At the 5% level of significance, we cannot conclude that the theoretical model fails to explain the observed values. The model fits the data adequately.

Short Notes

(i) Nominal Scale and Ordinal Scale

Nominal Scale: It is the simplest and lowest level of measurement in which data are classified into categories or classes that have no order or ranking. Numbers are used only as labels or names to identify the categories. The categories are mutually exclusive and exhaustive, but no mathematical operation (addition, subtraction, etc.) can be performed on them. For example, classifying people by religion (Hindu, Muslim, Christian), by gender (male, female), or by blood group (A, B, AB, O). The only statistical measure applicable is the mode or frequency count.

Ordinal Scale: It is a scale of measurement in which the categories possess a meaningful order or rank, but the difference between two successive ranks is not known or equal. The data can be arranged from lowest to highest or vice versa. For example, ranking students as first, second, third; measuring satisfaction as strongly agree, agree, neutral, disagree, strongly disagree; or socio-economic status as high, middle, low. Besides the mode and median, rank correlation and other non-parametric tests can be applied. However, arithmetic operations like addition and subtraction are not meaningful.

Key difference: In a nominal scale the categories are merely named with no order, whereas in an ordinal scale the categories are arranged in a definite order or rank.

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(ii) Poisson Distribution

Definition: The Poisson distribution is a discrete probability distribution that gives the probability of a specified number of events occurring in a fixed interval of time or space, when these events occur independently and at a constant average rate. It is named after the French mathematician S. D. Poisson.

Probability mass function: If $X$ is a Poisson variate with parameter $m$ (the mean number of occurrences), then

$P(X = x) = \frac{e^{-m} m^x}{x!}, \quad x = 0, 1, 2, \dots, \ m > 0$

where $e = 2.71828$ is the base of natural logarithms.

Properties:

1. It is a discrete distribution, since $X$ takes only non-negative integer values.
2. The mean and variance are equal, both being equal to $m$: $E(X) = \text{Var}(X) = m$.
3. It is the limiting form of the binomial distribution when $n \to \infty$, $p \to 0$ and $np = m$ remains finite (the law of rare events).
4. The sum of independent Poisson variates is also a Poisson variate.
5. It is positively skewed for small $m$ and becomes symmetrical as $m$ increases.

Applications: It is used to model the number of telephone calls received at a switchboard per hour, the number of accidents on a road per day, the number of defective items in a large lot, the number of radioactive particles emitted per second, and the number of customers arriving at a service counter.

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(iii) Association of Attributes

Meaning: Association of attributes means the relationship or dependence between two or more attributes, i.e., whether the presence or absence of one attribute is accompanied by the presence or absence of another attribute. Attributes are qualitative characteristics such as literacy, gender, employment, or marital status, which cannot be measured numerically but can only be classified.

Notation: For two attributes $A$ and $B$, the data are arranged in a $2 \times 2$ contingency table using the symbols:

• $(A)$ = number of persons possessing attribute $A$
• $(\alpha)$ = number of persons not possessing attribute $A$
• $(B)$ = number of persons possessing attribute $B$
• $(\beta)$ = number of persons not possessing attribute $B$
• $(AB)$ = number possessing both $A$ and $B$
• $(\alpha\beta)$ = number possessing neither $A$ nor $B$
• $(A\beta)$ = number possessing $A$ but not $B$
• $(\alpha B)$ = number possessing $B$ but not $A$
• $N$ = total number of observations

Types of association:

1. Positive association: When the attributes tend to occur together, i.e., the presence of $A$ favours the presence of $B$.
2. Negative association: When the presence of one attribute is accompanied by the absence of the other.
3. No association (independence): When the two attributes are completely unrelated.

Condition of independence: Two attributes $A$ and $B$ are independent if

$(AB) = \frac{(A) \times (B)}{N}$

Measures of association: The degree of association is measured by

1. Yule's coefficient of association: $Q = \frac{(AB)(\alpha\beta) - (A\beta)(\alpha B)}{(AB)(\alpha\beta) + (A\beta)(\alpha B)}$ $Q$ lies between $-1$ and $+1$; $Q = 0$ means no association, $Q = +1$ means perfect positive association and $Q = -1$ means perfect negative association.

2. Yule's coefficient of colligation: $Y = \frac{\sqrt{(AB)(\alpha\beta)} - \sqrt{(A\beta)(\alpha B)}}{\sqrt{(AB)(\alpha\beta)} + \sqrt{(A\beta)(\alpha B)}}$

3. Coefficient of contingency and the chi-square test are also used to test the significance of association.

Uses: Association of attributes is useful in sociology, education, medicine and economics for studying relationships between qualitative characteristics, such as the association between literacy and employment, or between vaccination and immunity.

Normal Distribution

A continuous random variable (X) is said to follow a normal distribution if its probability density function is

[ f(x)=\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}},\qquad -\infty<x<\infty ]

where (\mu) is the mean and (\sigma) is the standard deviation. It is denoted by

[ X\sim N(\mu,\sigma^2) ]

Main properties:

1. The curve is bell-shaped and symmetric about the mean (\mu).
2. Mean, median and mode are equal.
3. Total area under the curve is 1.
4. It is completely determined by (\mu) and (\sigma).
5. The standard normal variable is

[ Z=\frac{X-\mu}{\sigma}\sim N(0,1) ]

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Given

[ \mu=164.2\text{ cm},\qquad \sigma=2.7\text{ cm} ]

Let (X) be the height of an eighteen-year-old girl.

(i) Probability that height is more than 168 cm

[ P(X>168)=P\left(Z>\frac{168-164.2}{2.7}\right) ]

[ =P\left(Z>\frac{3.8}{2.7}\right) ]

[ =P(Z>1.4074) ]

From the standard normal table,

[ P(Z<1.4074)=0.9203 ]

Therefore,

[ P(Z>1.4074)=1-0.9203=0.0797 ]

[ \boxed{P(X>168)=0.0797\approx 7.97%} ]

So the probability that a girl has height more than 168 cm is about 7.97%.

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(ii) Lowest height of the tallest 10% girls

The tallest 10% girls lie above the 90th percentile. Hence,

[ P(X>x)=0.10 ]

[ P(Z>z)=0.10 ]

[ P(Z<z)=0.90 ]

From the standard normal table,

[ z=1.2816 ]

Now,

[ x=\mu+z\sigma ]

[ x=164.2+1.2816(2.7) ]

[ x=164.2+3.4603 ]

[ x=167.6603\text{ cm} ]

[ \boxed{x\approx 167.66\text{ cm}} ]

Therefore, the lowest height of the tallest 10% girls is approximately 167.66 cm.

Given: $n = 8$, $\mu_0 = 16$ ounce, $\alpha = 0.05$. The population is normal and $\sigma$ is unknown, so a one-sample $t$-test is used.

Hypotheses

$H_0: \mu = 16 \qquad H_1: \mu < 16 \quad \text{(left-tailed)}$

Sample mean

$\bar{x} = \frac{15.9+16.1+15.8+15.7+15.9+16.1+15.8+16.0}{8} = \frac{127.3}{8} = 15.9125$

Sample standard deviation

$s = \sqrt{\frac{0.14875}{7}} = \sqrt{0.02125} = 0.1458$

Test statistic

$t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} = \frac{15.9125 - 16}{0.1458/\sqrt{8}} = \frac{-0.0875}{0.05155} = -1.698$

Critical value: for a left-tailed test with $df = 7$, $t_{0.05,\,7} = -1.895$.

Decision: Since $-1.698 > -1.895$, $t$ does not fall in the rejection region, so we fail to reject $H_0$.

Conclusion: At the 5% level there is not enough evidence to conclude that the mean fill weight is less than 16 ounce.

Assumptions of ANOVA

1. The populations from which the samples are drawn are normally distributed.
2. The populations have equal variances (homogeneity of variance).
3. The samples are random and independent of each other, and the observations within each sample are independent.

Hypotheses

$H_0: \mu_{60} = \mu_{70} = \mu_{80}$ against $H_1$: at least one mean differs.

Calculations ($k=3$ temperatures, $n=3$ each, $N=9$)

Totals: $T_1 = 242,\; T_2 = 275,\; T_3 = 264$, so $G = 781$.

$CF = \frac{G^2}{N} = \frac{781^2}{9} = 67773.44$

$\sum x^2 = 19522 + 25213 + 23234 = 67969$

$TSS = 67969 - 67773.44 = 195.56$

$SST = \frac{242^2 + 275^2 + 264^2}{3} - CF = \frac{203885}{3} - 67773.44 = 188.22$

$SSE = TSS - SST = 195.56 - 188.22 = 7.33$

ANOVA table

$F = \frac{94.11}{1.222} = 77.0$

Decision: The tabulated value is $F_{0.05}(2,6) = 5.14$. Since $77.0 > 5.14$, we reject $H_0$.

Conclusion: At the 5% level, the mean yields differ significantly with temperature.

Positive association: Two attributes (A) and (B) are said to be positively associated when the presence of one attribute favours the presence of the other. In such a case, (A) and (B) occur together more often than expected under independence. Yule’s coefficient of association (Q>0).

Negative association: They are negatively associated when the presence of one attribute favours the absence of the other. Then (Q<0). If (Q=0), the attributes are independent.

Given: [ (AB)=75,\quad (A)=80,\quad (\alpha\beta)=42,\quad N=145 ]

First find the remaining cell frequencies: [ (A\beta)=(A)-(AB)=80-75=5 ] [ (\alpha B)=N-{(AB)+(A\beta)+(\alpha\beta)} =145-(75+5+42)=23 ]

Yule’s coefficient of association: [ Q=\frac{(AB)(\alpha\beta)-(A\beta)(\alpha B)} {(AB)(\alpha\beta)+(A\beta)(\alpha B)} ]

[ Q=\frac{75\times 42-5\times 23}{75\times 42+5\times 23} ]

[ Q=\frac{3150-115}{3150+115} =\frac{3035}{3265} =\frac{607}{653} ]

[ Q\approx 0.9296 ]

Interpretation: Since (Q) is positive and very close to (+1), attributes (A) and (B) are highly positively associated. The presence of (A) strongly favours the presence of (B), and the absence of (A) strongly favours the absence of (B).

Data. Total frequency $N=120$. Treat the classes as continuous with boundaries $0.2,0.3,0.4,0.5,0.6,0.7$ (class width $h=0.1$).

(i) Histogram and frequency polygon

Plot weight (oz) on the x-axis and frequency on the y-axis. The histogram has adjacent rectangles of equal width with heights 30, 46, 24, 14, 6. The frequency polygon joins the points (mid-point, frequency): $(0.25,30),(0.35,46),(0.45,24),(0.55,14),(0.65,6)$, closed to the axis at the empty classes' mid-points $(0.15,0)$ and $(0.75,0)$.

 f
46 |        ###
   |        ###
30 |  ###   ###
   |  ###   ###
24 |  ###   ### ###
   |  ###   ### ###
14 |  ###   ### ### ###
 6 |  ###   ### ### ### ###
   +--------------------------- weight (oz)
     .25    .35  .45  .55 .65

The peak is at the class 0.31-0.40 and the right tail is longer.

(ii) Skewness

Moments about the assumed mean (in units of $h$):

$m_1'=\frac{\sum fd}{N}=-0.6667,\quad m_2'=\frac{\sum fd^2}{N}=1.7,\quad m_3'=\frac{\sum fd^3}{N}=-1.8667.$

Mean $=0.45+0.1(-0.6667)=0.3833$ oz.

$\mu_2=m_2'-m_1'^2=1.7-0.4444=1.2556,\qquad \sigma=0.1\sqrt{1.2556}=0.1121\text{ oz}.$

$\mu_3=m_3'-3m_1'm_2'+2m_1'^3=-1.8667+3.4-0.5926=0.9407$

$\text{Skewness }\gamma_1=\frac{\mu_3}{\mu_2^{3/2}}=\frac{0.9407}{1.4068}\approx 0.67.$

Check with Karl Pearson's coefficient: modal class is 0.31-0.40, mode $=0.3+\dfrac{46-30}{2(46)-30-24}(0.1)=0.342$, so $Sk=\dfrac{0.3833-0.342}{0.1121}\approx 0.37$.

Interpretation. Both measures are positive, so the distribution is positively skewed (skewed to the right): most pears are light, with a tail of a few heavy ones, and mean ($0.383$) $>$ mode ($0.342$).

(iii) Percentage of pears heavier than 0.45 oz

The value 0.45 lies at the middle of the class 0.41-0.50 (frequency 24), so about half of this class, $12$, lies above it. Adding the higher classes:

$12+14+6=32$

$\frac{32}{120}\times100\approx \mathbf{26.7\%}$

About 26.7% of the pears weigh more than 0.45 oz.

Definition

Positive correlation: Two variables are positively correlated when they change in the same direction. If one variable increases, the other also tends to increase. The correlation coefficient is positive, i.e. (r>0).

Negative correlation: Two variables are negatively correlated when they change in opposite directions. If one variable increases, the other tends to decrease. The correlation coefficient is negative, i.e. (r<0).

Let
(X) = family size
(Y) = number of telephone calls per week.

Here, (n=10).

[ \sum X=60,\quad \sum Y=380,\quad \sum X^2=376,\quad \sum Y^2=18646,\quad \sum XY=2488 ]

[ \bar X=\frac{60}{10}=6,\qquad \bar Y=\frac{380}{10}=38 ]

Also,

[ S_{xx}=\sum X^2-\frac{(\sum X)^2}{n}=376-\frac{60^2}{10}=16 ]

[ S_{yy}=\sum Y^2-\frac{(\sum Y)^2}{n}=18646-\frac{380^2}{10}=4206 ]

[ S_{xy}=\sum XY-\frac{(\sum X)(\sum Y)}{n}=2488-\frac{60\cdot 380}{10}=208 ]

i. Correlation coefficient

Karl Pearson’s correlation coefficient is

[ r=\frac{S_{xy}}{\sqrt{S_{xx}S_{yy}}} ]

[ r=\frac{208}{\sqrt{16\cdot 4206}} =\frac{208}{\sqrt{67296}} \approx 0.8018 ]

[ \boxed{r\approx 0.802} ]

Interpretation: Since (r\approx 0.802), there is a strong positive correlation between family size and the number of telephone calls per week. Larger families tend to make more telephone calls. Also,

[ r^2\approx 0.6429 ]

so about (64.3%) of the variation in weekly calls is explained by family size.

ii. Linear regression equation of calls on family size

Regression of (Y) on (X):

[ Y=a+bX ]

[ b=\frac{S_{xy}}{S_{xx}}=\frac{208}{16}=13 ]

[ a=\bar Y-b\bar X=38-13(6)=38-78=-40 ]

Therefore, the regression equation is

[ \boxed{\hat Y=-40+13X} ]

or

[ \boxed{\text{Calls}=-40+13(\text{Family size})} ]

iii. Estimate the number of calls for a family of size 9

Put (X=9):

[ \hat Y=-40+13(9) ]

[ \hat Y=-40+117=77 ]

[ \boxed{\text{Estimated number of calls}=77} ]

iv. Standard error of estimate

The standard error of estimate of (Y) on (X) is

[ S_{yx}=\sqrt{\frac{S_{yy}-bS_{xy}}{n-2}} ]

[ S_{yx}=\sqrt{\frac{4206-13(208)}{10-2}} ]

[ S_{yx}=\sqrt{\frac{4206-2704}{8}} ]

[ S_{yx}=\sqrt{\frac{1502}{8}} ]

[ S_{yx}=\sqrt{187.75}\approx 13.70 ]

[ \boxed{S_{yx}\approx 13.70\text{ calls per week}} ]

So the standard error of estimate is approximately 13.70 calls per week.

Paired $t$-test (because the same 8 patients are measured before and after the drug).

Let
[ d=\text{Temperature before drug}-\text{Temperature after drug} ]

[ n=8,\qquad \sum d=4.8,\qquad \sum d^2=7.44 ]

Sample mean difference: [ \bar d=\frac{\sum d}{n}=\frac{4.8}{8}=0.6 ]

Sample standard deviation of differences: [ s_d=\sqrt{\frac{\sum d^2-\frac{(\sum d)^2}{n}}{n-1}} ] [ s_d=\sqrt{\frac{7.44-\frac{(4.8)^2}{8}}{7}} =\sqrt{\frac{7.44-2.88}{7}} =\sqrt{\frac{4.56}{7}} =\sqrt{0.6514}=0.8071 ]

Standard error: [ SE_{\bar d}=\frac{s_d}{\sqrt n}=\frac{0.8071}{\sqrt 8}=0.2854 ]

Test statistic: [ t=\frac{\bar d}{SE_{\bar d}} =\frac{0.6}{0.2854} =2.103 ]

Degrees of freedom: [ df=n-1=8-1=7 ]

At 5% level of significance for a two-tailed test: [ t_{0.025,,7}=2.365 ]

Decision: [ |t|=2.103 < 2.365 ]

So we fail to reject (H_0). Also, the approximate (p)-value is about (0.073), which is greater than (0.05).

Conclusion:
At the 5% level of significance, there is not enough evidence to conclude that the new drug changes body temperature.