Chapter 11.2, Problem 63E

(a)

To determine

To find: The mean and the standard deviation of the ratings $1to5$.

Answer to Problem 63E

  $\begin{array}{l}\overline{x}=3.545\\ s=1.1857\end{array}$

Explanation of Solution

Given information:

Total number of people $=90$

The table of the frequency count of the ratings $1to5$ is as follows

	$1$ to $5$
Rating	$1$	$2$	$3$	$4$	$5$
Frequency	$2$	$3$	$1$	$13$	$3$

Formula used:

  $\begin{array}{l}\overline{x}=\frac{{\displaystyle \sum \left(x\times f\right)}}{{\displaystyle \sum f}}\\ Sd\left(x\right)=\sqrt{\frac{1}{{\displaystyle \sum f}-1}\times \left({\displaystyle \sum \left(x^2\times f\right)}\right)-{\displaystyle \sum f}\times {\overline{x}}^2}\end{array}$

Calculation:

$x$	$f$	$xf$	$x^{2}f$
$1$	$2$	$2$	$2$
$2$	$3$	$6$	$12$
$3$	$1$	$3$	$9$
$4$	$13$	$52$	$208$
$5$	$3$	$15$	$75$
	${\displaystyle \sum f}=22$	${\displaystyle \sum \left(x\times f\right)=}78$	${\displaystyle \sum \left(x^2\times f\right)=}306$

The mean will be

  $\overline{x}=\frac{78}{22}=3.545$

The standard deviation will be

  $s=\sqrt{\frac{1}{22-1}\times \left(306-22\times {3.545}^2\right)}=1.1857$

(b)

To determine

To find: The adjusted mean for ratings $0to4$

Answer to Problem 63E

Adjusted Mean $=4.21$

Adjusted standard deviation $=0.568$

Explanation of Solution

Given information:

Mean of the ratings $0to4=3.21$

Standard deviation of the ratings $0to4=0.568$

Concept used:

  $\begin{array}{l}E\left[aX+b\right]=a\times E\left[X\right]+b\\ Sd\left(aX+b\right)=a\times Sd\left(x\right)\end{array}$

Calculation:

Using the above concept as an addition or subtraction in all the terms leads to the simultaneous changes in the mean while the standard deviation is affected by only multiplication or division of all the terms.

As $1$ is added to each of the ratings, so

The mean of the adjusted ratings will be

  $\begin{array}{l}\overline{x}\text{'}=\overline{x}+1\\ =3.21+1=4.21\end{array}$

The adjusted standard deviation will remain the same as it remains unaffected by addition or subtraction of the value.

(c)

To determine

To find: The appropriateness to compare means by two sample t test.

Answer to Problem 63E

Two sample test for mean isn’t appropriate.

Explanation of Solution

Two sample test for means is used when the data is nominal in nature.

In this case, the ratings are to be tested which an ordinal data (ordered data)

Since, the data is categorical in nature, so two sample t test is not appropriate.