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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 $\mathrm{β}$-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 $\mathrm{β}$-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: ${\stackrel{¯}{x}}_1=0.02426$, $s_1=0.00514,x_2=0.01643$, and $s_2=0.00470$.)

Step by step solution

01

Given Information

Given data:

02

Explanation

Null hypothesis $H_0$:${\mathrm{μ}}_1={\mathrm{μ}}_2$

Alternative hypothesis $H_1$: ${\mathrm{μ}}_1>{\mathrm{μ}}_2$

Test Statistic:

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

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

small samples apply t test

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

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" $H_1:{\mathrm{μ}}_1>{\mathrm{μ}}_2$.

There is significance difference between means.

The endpoint of the interval are,

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

$=0.235\text{to}7.165$

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