Question

The level of monoamine oxidase (MAO) activity in blood platelets (nm/mg protein/h) was determined for each individual in a sample of \(43\) chronic schizophrenics, resulting in \(\bar x = 2.69\) and \({s_1} = 2.30,\), as well as for \(45\) normal subjects, resulting in \(\bar y = 6.35\) and \({s_2} = 4.03.\). Does this data strongly suggest that true average MAO activity for normal subjects is more than twice the activity level for schizophrenics? Derive a test procedure and carry out the test using \(\alpha = .01\)

. (Hint: \({H_0}\) and \({H_a}\) here have a different form from the three standard cases. Let \({\mu _1}\) and \({\mu _2}\) refer to true average MAO activity for schizophrenics and normal subjects, respectively, and consider the parameter \(\theta = 2{\mu _1} - {\mu _2}\). Write \({H_0}\) and \({H_a}\) in terms of \(\theta \), estimate \(\theta \), and derive \({\hat \sigma _{\tilde \theta }}\) (“Reduced Monoamine Oxidase Activity in Blood Platelets from Schizophrenic Patients,” Nature, July 28, 1972: 225–226).) \(\alpha = .01\)

Short answer

the solution is

There is insufficient evidence to support the idea that normal people' genuine average MAO activity is more than twice that of schizophrenics.

Step-by-step solution

find mean and variance

given

\(\begin{array}{l}{{\bar x}_1} = 2.69\\{{\bar x}_2} = 6.35\\{s_1} = 2.30\\{s_2} = 4.03\end{array}\)

\(\begin{array}{l}{n_1} = 43\\{n_2} = 45\\\alpha = 0.01\end{array}\)

The sampling distribution of the sample mean \(\mu \) and standard deviation \(\frac{\sigma }{{\sqrt n }}:\)

\(\begin{array}{*{20}{r}}{E\left( {{{\bar X}_1}} \right) = {\mu _1}}\\{E\left( {{{\bar X}_2}} \right) = {\mu _2}}\\{V\left( {{{\bar X}_1}} \right) = \frac{{\sigma _1^2}}{{{n_1}}}}\\{V\left( {{{\bar X}_2}} \right) = \frac{{\sigma _2^2}}{{{n_2}}}}\end{array}\)

For linear combination \(W = a {X_1} + b {X_2}\) the following properties hold for the mean and variance:

\(E(W) = aE\left( {{X_1}} \right) + bE\left( {{X_1}} \right)\)

\(V(W) = {a^2}E\left( {{X_1}} \right) + {b^2}E\left( {{X_1}} \right)\left( { If {X_ - }1 and {X_ - }2} \right.\)are independent)

Let us define the statistic \(c\hat \theta = 2{\bar X_1} - {\bar X_2}\) for the parameter \(\theta = 2{\mu _1} - {\mu _2}\).

Determine its mean and variance

\(\begin{array}{l}E(\hat \theta ) = E\left( {2{{\bar X}_1} - {{\bar X}_2}} \right) = 2E\left( {{{\bar X}_1}} \right) - E\left( {{{\bar X}_2}} \right) = 2{\mu _1} - {\mu _2}\\V(\hat \theta ) = V\left( {2{{\bar X}_1} - {{\bar X}_2}} \right) = {2^2}V\left( {{{\bar X}_1}} \right) + {( - 1)^2}V\left( {{{\bar X}_2}} \right) = 4\frac{{\sigma _1^2}}{{{n_1}}} + \frac{{\sigma _2^2}}{{{n_2}}}\end{array}\)

The standard deviation is the square root of the variance:

\({\sigma _{\hat \theta }} = \sqrt {4\frac{{\sigma _1^2}}{{{n_1}}} + \frac{{\sigma _2^2}}{{{n_2}}}} \)

determine probability

Because the sample sizes are high, the sampling distributions of the sample means are about normal (by the central limit theorem), and so the sampling distribution of \(\hat \theta \) is also approximately normal.

We can utilize the \(z\)-test because the samples are large \(\left( {n > 30} \right)\). (instead of a t-test).

More for the normal subjects, according to the assertion \({X_2}\).

Either the null hypothesis or the alternative hypothesis is the assertion. The null hypothesis and alternative hypothesis are diametrically opposed. An equality must be present in the null hypothesis.

\(\begin{array}{l}{H_0}:2{\mu _1} - {\mu _2} = 0\\{H_a}:2{\mu _1} - {\mu _2} < 0\end{array}\)

The test statistic's value is the statistic's value multiplied by its standard deviation divided by its mean:

\(z = \frac{{2{{\bar x}_1} - {{\bar x}_2} - \left( {2{\mu _1} - {\mu _2}} \right)}}{{\sqrt {\frac{{4s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} }} = \frac{{2(2.69) - 6.35 - 0}}{{\sqrt {\frac{{4 \times 2.3{0^2}}}{{43}} + \frac{{4.0{3^2}}}{{45}}} }} \approx - 1.05\)

If the null hypothesis is true, the P-value is the chance of getting a value that is more extreme or equal to the standardized test statistic \(z\). Using the usual probability table, calculate the probability.

\({\rm{P = P(Z < - 1}}{\rm{.05) = 0}}{\rm{.1469}}\)

If the p-value is smaller than the significance level \(\alpha \), then the null hypothesis is rejected.

\(P > 0.01 \Rightarrow Fail to reject {H_0}\)

There is insufficient evidence to support the idea that normal people' genuine average MAO activity is more than twice that of schizophrenics.

conclusion

There is insufficient evidence to support the idea that normal people' genuine average MAO activity is more than twice that of schizophrenics.