Question

A study published in Psychiatry Research states that 62% of autistic children are left-handed.[12]

Answer 1

Answer:

The pvalue of the test is 0.7188 > 0.05, which means that there is not enough statistical evidence based on the sample to dispute the results of the study published in Psychiatry Research.

Step-by-step explanation:

Test if there is enough statistical evidence based on the sample to dispute the results of the study published in Psychiatry Research:

This means that at the null hypothesis we test if the proportion is 62%, that is:

$H_0: p = 0.62$

At the alternate hypothesis, we test if the proportion is different from 0.62, that is:

$H_a: p \neq 0.62$

The test statistic is:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

In which X is the sample mean, $\mu$ is the value tested at the null hypothesis, $\sigma$ is the standard deviation and n is the size of the sample

62% is tested at the null hypothesis:

This means that $\mu = 0.62, \sigma = \sqrt{0.62*0.38}$

When a random sample of 26 autistic children were observed, 17 were found to be left-handed.

This means that $n = 26, X = \frac{17}{26} = 0.6538$

Value of the test statistic:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

$z = \frac{0.6538 - 0.62}{\frac{\sqrt{0.62*0.38}}{\sqrt{26}}}$

$z = 0.36$

Pvalue of test and decision:

The pvalue of the test is the probability of a proportion that differs from the mean by at least 0.6538 - 0.62 = 0.0338, which is P(|z| > 0.36), which is two multiplied by the pvalue of z = -0.36

Looking at the z-table, z = -0.36 has a pvalue of 0.3594

2*0.3594 = 0.7188

The pvalue of the test is 0.7188 > 0.05, which means that there is not enough statistical evidence based on the sample to dispute the results of the study published in Psychiatry Research.