Chapter 10.3, Problem 78E

(a)

To determine

To explain why these are paired data.

Explanation of Solution

Data re paired if a data value in one sample corresponds with a unique data value in the other sample.

In this case, we note that the Male literacy data values belongs to the same countries as the Female literacy rates which implies that we have two data values per day and thus each of these data values per day forms a pair.

(b)

To determine

To describe what the graph reveals about whether males have higher literacy rates than females than these countries on average.

Explanation of Solution

The dots in the dot plot represents the difference between males literacy rates and females literacy rates.

We note that all dots in the dot plot lie to the right of the zero which means that most of the differences are positive and thus the male literacy rates appears to be higher than the female literacy rates in general.

(c)

To determine

To calculate the mean difference and the standard deviation of the differences and interpret the mean difference.

Answer to Problem 78E

The mean is $9.8208$ and the standard deviation is $9.6569$ .

Explanation of Solution

It is given that the table of the difference which is the male literacy rates less by the female literacy rates.

The mean is:

  $\begin{array}{l}\overline{x}=\frac{{\displaystyle \sum _{i=1}^n{x}_i}}{n}\\ =\frac{\begin{array}{l}30.5+0+0.2+8.6+16.8+7.4+15.4+8.7+6.4+0.5+\\ 7.4+7.9+4.7+18.5+25+9.1+12.4+0.2+14.8+4+\\ 1+0.6+34.4+1.2\end{array}}{24}\\ =\frac{235.7}{24}\\ =9.8208\end{array}$

The sample variance is then as:

  $\begin{array}{l}{s}^2=\frac{{\displaystyle \sum {(x-\overline{x})}^2}}{n-1}\\ =\frac{\begin{array}{}$ {(30.5-9.8208)}^2+{(0-9.8208)}^2+{(0.2-9.8208)}^2+{(8.6-9.8208)}^2+{(16.8-9.8208)}^2+$$ {(7.4-9.8208)}^2+{(15.4-9.8208)}^2+{(8.7-9.8208)}^2+{(6.4-9.8208)}^2+{(0.5-9.8208)}^2+$$ {(7.4-9.8208)}^2+{(7.9-9.8208)}^2+{(4.7-9.8208)}^2+{(18.5-9.8208)}^2+{(25-9.8208)}^2+$$ {(9.1-9.8208)}^2+{(12.4-9.8208)}^2+{(0.2-9.8208)}^2+{(14.8-9.8208)}^2+{(4-9.8208)}^2+$$ {(1-9.8208)}^2+{(0.6-9.8208)}^2+{(34.4-9.8208)}^2+{(1.2-9.8208)}^2$\end{array}}{24-1}\\ =93.2548\end{array}$

The sample standard deviation is then,

  $\begin{array}{l}s=\sqrt{s^2}\\ =\sqrt{93.2548}\\ =9.6569\end{array}$

The difference are on average $9.8208$ which tends to vary on average by $9.6569$ from the average of $9.8208$ .