Question

Schizophrenia and Dopamine. Previous research has suggested that changes in the activity of dopamine, a neurotransmitter

in the brain, maybe a causative factor for schizophrenia. In the paper "Schizophrenia: Dopamine $\beta$-Hydroxylase Activity and Treatment Response" (Science, Vol. 216, pp. 1423-1425), D. Sternberg et al. published the results of their study in which they examined $25$schizophrenic patients who had been classified as either psychotic or not psychotic by hospital staff. The activity of dopamine was measured in each patient by using the enzyme dopamine $\beta$-hydroxylase to assess differences in dopamine activity between the two groups. The following are the data, in nanomoles per milliliter-hour per milligram ( $\mathrm{nmol}/\mathrm{mL}-\mathrm{hr}/\mathrm{mg})$.

At the $1\%$significance level, do the data suggest that dopamine activity is higher, on average, in psychotic patients? (Note: ${\overline{x}}_1=0.02426$, $s_1=0.00514,x_2=0.01643$, and $s_2=0.00470$.)

Short answer

The interval is$0.235\text{to}7.165$.

Step-by-step solution

Given Information

Given data:

Explanation

Null hypothesis $H_0$:${\mu }_1={\mu }_2$

Alternative hypothesis $H_1$: ${\mu }_1>{\mu }_2$

Test Statistic:

$t=\frac{{\overline{x}}_1-{\overline{x}}_2}{\sqrt{{\displaystyle \frac{n_1{s_1}^2+n_2{s_2}^2}{n_1+n_2-2}}}({\displaystyle \frac{1}{n_1}}+{\displaystyle \frac{1}{n_2}})}$

Since $n_1=10$and $n_2=15$

small samples apply t test

$$\begin{gathered} t=\frac{0.02426-0.01643}{\sqrt{{\displaystyle \frac{10\times 0.{00514}^2+15\times 0.{00470}^2}{10+15-2}}}({\displaystyle \frac{1}{10}}+{\displaystyle \frac{1}{15}})} \\ =\frac{0.00783}{\sqrt{{\displaystyle \frac{0.0002642+0.0003314}{23}}}(0.167)} \\ =\frac{0.00783}{\sqrt{0.000004325}} \\ =\frac{0.00783}{0.002079} \\ =3.7662 \end{gathered}$$

Table value at 1% level of significance

$$\begin{gathered} df=n_1+n_2-2 \\ =10+15-2 \\ =23 \end{gathered}$$

$t_a=1.319$

Since calculated value is greater than table value. We reject the null hypothesis is localid="1654774693748" $H_1:{\mu }_1>{\mu }_2$.

There is significance difference between means.

The endpoint of the interval are,

$\left({\overline{x}}_1-{\overline{x}}_2\right)\pm t_{\frac{a}{2}}·\sqrt{\left(\frac{s_1^2}{n_1}\right)+\left(\frac{s_2^2}{n_2}\right)}=(48.5-33.7)\pm 2.787·\sqrt{\left(\frac{9.2^2}{32}+\frac{5.7^2}{20}\right)}$

$=0.235\text{to}7.165$