Chapter 11.2, Problem 64E

a)

To determine

To find the mean and standard deviation of the ratings for the students given in 1 to 5 scale.

Answer to Problem 64E

Mean = 3.5455 and standard deviation = 1.1843

Explanation of Solution

The mean is,

  $\overline{x}=\frac{1\times 2+2\times 3+3\times 1+4\times 13+5\times 3}{2+3+1+13+3}=3.5455$

The standard deviation,

  $\begin{array}{l}s=\sqrt{\frac{{(1-3.5455)}^2\times 2+{(2-3.5455)}^2\times 3+{(3-3.5455)}^2\times 1+{(4-3.5455)}^2\times 13+{(5-3.5455)}^2\times 3}{2+3+1+13+3-1}}\\ s=1.1843\end{array}$

b)

To determine

To find the mean and standard deviation of the adjusted ratings.

Answer to Problem 64E

The adjusted mean = 4.21 and adjusted sd = 0.568

Explanation of Solution

Given:

Mean and standard deviation for scale 0-4:

  $\begin{array}{l}{\overline{x}}_{0-4}=3.21\\ s_{0-4}=0.568\end{array}$

The adjusted mean and standard deviationare,

  $\begin{array}{l}{\overline{x}}_{(0-4)+1}={\overline{x}}_{0-4}+1=3.21+1=4.21\\ s_{(0-4)+1}=0.568\end{array}$

c)

To determine

To explain whether it is appropriate to test means using a two-sample t test.

Answer to Problem 64E

No.

Explanation of Solution

The given variables having ratings with corresponding to their frequency. Therefore, these are categorical variables. For two sample t test, data should be numerical. Hence, it is not appropriate to use a two-sample t test.