Chapter 18, Problem 14P

To determine

a. If 25 out of 36 autistic children select the correct definition, is this significantly more than would be expected if they were simply guessing? Use a two-tailed test with $\alpha =0.05$

b. If only 16 out of 36 children with SLI select the correct definition, is this significantly more than would be expected if they were simply guessing? Use a two-tailed test with $\alpha =0.05$ .

Answer to Problem 14P

Solution:

1. Using the Binomial test and Z scores with $\mu =n\times p,\sigma =\sqrt{n\times p\times q},q=1-p$

   $H_0:p=0.5,H_1:p\ne 0.5,n=36,\widehat{p}=25/36=.6944$

   $\begin{array}{l}{Z}_{calculated}=\frac{\widehat{p}-p}{\sqrt{p\times q/n}}=2.3328>1.96\\ Z_{tabulated,0.05,2tailed}=(-1.96,+1.96)\end{array}$

   $H_0$ of p=0.5 is rejected.

b

$\begin{array}{l}n=36,\widehat{p}=16/36=.4444\\ Z_{calculated}=\frac{\widehat{p}-p}{\sqrt{p\times q/n}}=-0.667,Z_{tabulated}=-1.96\\ \end{array}$

$H_0$ of p=0.5 is accepted.

Explanation of Solution

Given:

The researchers used relatively small samples and found that the children with TLD and those with autism were able to learn novel words using the syntactical cues in sentences. The children with SLI, on the other hand, experienced significantly more difficulty.

Suppose that a similar study is conducted in which each child listens to a set of sentences containing a novel word and then is given a choice of three definitions for the word.

Calculations:

a. The null hypothesis states that for the autistic children, the selection of correct words is done through guessing and cues in sentenes have no effect, hence $H_0:p=0.5,H_1:p\ne 0.5$. With a sample of n= 36, $np=18,qn=18$. Both values are greater than 5, so the distribution of z-scores is approximately normal. With $\alpha =0.05$, the critical region is $(-1.96,+1.96)$ .

For this X = 25, n=36, compute the sample proportion $\widehat{p}=25/36=0.6944$, and the hypothesized proportion is p=0.5(null, chance)

$\begin{array}{l}{Z}_{calculated}=\frac{\widehat{p}-p}{\sqrt{p\times q/n}}=2.3328\\ Z_{tabulated,0.05,2tailed}=(-1.96,+1.96)\\ Z_{calculated}>1.96\end{array}$

Because the data are in the critical region, our decision is to reject $H_0$ . These data

Provide sufficient evidence that the correct guessing is significantly more than would be expected if they were simply guessing for autistic children.

b. The null hypothesis states that for the SLI children, the selection of correct words is done through guessing and cues in sentenes have no effect, hence $H_0:p=0.5,H_1:p\ne 0.5$. With a sample of n= 36, $np=18,qn=18$. Both values are lesser than 5, so the distribution of z-scores is approximately normal. With $\alpha =0.05$, the critical region is $(-1.96,+1.96)$ .

For this X = 16, n=36, compute the sample proportion $\widehat{p}=16/36=0.4444$, p=0.5(null)

$\begin{array}{l}{Z}_{calculated}=\frac{\widehat{p}-p}{\sqrt{p\times q/n}}=(0.4444-.5)/\sqrt{.5\times (1-.5)/36}\\ =-0.6714\\ Z_{tabulated,0.5,2tailed}=(-1.96,+1.96),\end{array}$

$H_{0}accepted$ $Z_{calculated}$ is in the acceptance region.

Because the data are in the acceptance region, our decision is to accept $H_0$ . These data do not provide sufficient evidence that the correct guessing is significantly more than would be expected if they were simply guessing for the SLI children.

Conclusion:

a. For the first sample data provide sufficient evidence that the correct guessing is significantly more than would be expected if they were simply guessing for autistic children.

b. For the second sample data provide sufficient evidence that the cues in sentences have no effect on correct guessing .They were not significantly more than would be expected if they were simply guessing for SLI children.

Justification:

a. $H_0$ rejected hence sample (which has proportion of 69.44%) has enough evidence to conclude that the correct guessing is significantly more than would be expected if they were simply guessing for autistic children(that would have given p=50%).

b. $H_0$ accepted hence sample(proportion 44.44%) has enough evidence to suffice that cues in sentences have no effect on correct guessing. They were not significantly more than would be expected if they were simply guessing for SLI children( which is p=50%).