## 17.31B

Learning Blissymbols. Blissymbols are pictographs sometimes used to help learning-disabled children.  In a study of computer-assisted learning, 12 normal-ability schoolchildren were assigned at random to each of four computer learning programs.  After they used the program, they attempted to recognize 24 Blissymbols.  Here are the counts correct for one of the programs:

$\begin{array}{rrrrrrrrrrrr} 12 & 22 & 9 & 14 & 20 & 15 & 9 & 10 & 11 & 11 & 15 & 6 \end{array}$

Give a 90% confidence interval for the mean count correct among all children of this age who use the program.

## Solution

Notice first that the use of t statistics is acceptable, by inspection of a stemplot of the data:

$\begin{array}{r|lllllll} 0 & 6 & 9 & 9 & &&& \\ 1 & 0 & 1 & 1 & 2 & 4 & 5 & 5 \\ 2 & 0 & 2 & & & & & \end{array}$

The shape suggests a very thin Normal distribution (small standard deviation).  Assuming the SRS is proper, the interval of confidence can be found by the formula

$\bar{x} \pm t^\ast \displaystyle{ \frac{s}{\sqrt{n}} },$

where except $n =12,$ we need to compute the rest of values involved.  Let us start with the mean $\bar{x}$ and standard deviation $s.$

$\begin{array}{|r|r|r|r|} \hline k & x_k & x_k-\bar{x}&(x_k-\bar{x})^2 \\ \hline 1 & 12 & -0.83&0.69 \\ \hline 2 & 22 & 9.17&84.09 \\ \hline 3&9&-3.83&14.67 \\ \hline 4&14&1.17&1.37 \\ \hline 5&20&7.17&51.41 \\ \hline 6&15&2.17&4.71 \\ \hline 7&9&-3.83&14.67 \\ \hline 8&10&-2.83&8.01 \\ \hline 9&11&-1.83&3.35 \\ \hline 10&11&-1.83&3.35 \\ \hline 11&15&2.17&4.71 \\ \hline 12&6&-6.83&46.65 \\ \hline \end{array}$

The mean is $\bar{x}=12.83$, and the standard deviation is $s=4.648.$  Since we have 11 degrees of freedom, we look up the eleventh row in Table C for the corresponding value under 90% (the confidence) to obtain $t^\ast = 1.796.$

With these values computed, we infere that a 90% confidence interval for the mean count correct among all children of this age who use the program is

$12.83 \pm 2.4098046,$ or the interval from $10.4201954$ to $15.2398046.$

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