Respuesta :
The test is to find out if there is a difference in the proportion of the number of autism in two samples differentiated by the use of vitamins.
The test statistic for the test is approximately 2.98
The p-value associated is 0.0028
Reasons:
The given parameters are;
[tex]\begin{array}{|l|c|c|c|}&\mathbf{\underline{Autism}}&\mathbf{\underline{Typical \ Development }}&\mathbf{\underline{Total}}\\No \ vitamin&111&70&181\\Vitamin&143&159&302\\Total&254&229&483\end{array}\right][/tex]
The null hypothesis is H₀; [tex]\hat{p}_1 = \hat{p}_2[/tex]
The alternative hypothesis is Hₐ; [tex]\hat{p}_1 \neq \hat{p}_2[/tex]
Where:
[tex]\hat{p}_1[/tex] = The proportion of the sample that did not use vitamin and had autism
[tex]\hat{p}_2[/tex] = The proportion of the sample that used vitamin and had autism
The test statistics for the difference of two proportions is given as follows;
[tex]Z=\dfrac{\hat{p}_1-\hat{p}_2}{\sqrt{\hat{p}\cdot (1-\hat{p}) \cdot \left (\dfrac{1}{n_{1}}+\dfrac{1}{n_{2}} \right )}}[/tex]
Where;
n₁ = 181, n₂ = 302, [tex]\hat p_1 = \dfrac{111}{181}[/tex], [tex]\hat p_2 = \dfrac{143}{302}[/tex], [tex]\hat p = \dfrac{111 + 143}{181 + 302 } = \dfrac{254}{283}[/tex]
[tex]The\ test\ statistics,\, Z=\dfrac{\dfrac{111}{181}-\dfrac{143}{302}}{\sqrt{ \dfrac{254}{283}\cdot (1-\dfrac{254}{283}) \times \left (\dfrac{1}{181}+\dfrac{1}{302} \right )}} \approx 2.98[/tex]
The test statistic for the test, Z ≈ 2.98
From the Z-table, we have Z > 2.98 = 1 - 0.9986, therefore, given that the
alternative hypothesis [tex]\hat p_1 \neq \hat p_2[/tex], the p-value is 2 × (1 - 0.9986) = 0.0028
The p-value associated with the test is 0.0028
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