Question

Mmabatho randomly selected a sample of 100 children with ASD and found that only 70 of them
are in special need schools.


Botlhale and Mmabatho are at it once again. This time they want to determine whether the true
proportion of ASD children in special need schools in the population is 0:75. Consider information
provided for questions 7 and 8 and assume a 5% level of significance.
Which of the following statements is incorrect?
(1) H0 : n= 0.75 against H1 : != 0.75
(2) The critical value is 1.96:
(3) The value of the test statistic is -1.09:
(4) The p-value is 0:2502:
(5) We do not reject H0

Answer 1

The following information is provided: The sample size is $N=100,$ the number of favorable cases is $X=70,$ and the sample proportion is $\bar{p}=\dfrac{X}{N}=\dfrac{70}{100}=0.7,$ and the significance level is $\alpha=0.05$

The following null and alternative hypotheses need to be tested:

$H_0:p=0.75$

$H_1:p\not=0.75$

This corresponds to a two-tailed test, for which a z-test for one population proportion needs to be used.

Based on the information provided, the significance level is $\alpha=0.05,$ and the critical value for a two-tailed test is $z_c=1.96.$

The rejection region for this two-tailed test is $R=\{z:|z|>1.96\}$

The $z-$ statistic is computed as follows:

$$z={\bar{p}-p_0 \over \sqrt{p_0(1-p_0)/n}}={0.7-0.75 \over \sqrt{0.75(1-0.75)/100}}=-1.1547$$

Since it is observed that $|z|=1.1547<1.96=z_c,$ it is then concluded that the null hypothesis is not rejected. Therefore, there is not enough evidence to claim that the population proportion $p$ is different than $p_0,$ at the $\alpha=0.05$ significance level.

Using the P-value approach: The p-value for $z=-1.1547$ is $p=0.2485,$ and since $p=0.2485\geq0.05,$ it is concluded that the null hypothesis is not rejected. Therefore, there is not enough evidence to claim that the population proportion $p$ is different than $p_0,$ at the $\alpha=0.05$ significance level.

The following statements are wrong:

(3) The value of the test statistic is -1.09.

(4) The p-value is 0.2502.