Chapter 9.3, Problem 41E

Antipsychotic drugs are widely prescribed for conditions such as schizophrenia and bipolar disease. The article “Cardiometabolic Risk of Second- Generation Antipsychotic Medications During First-Time Use in Children and Adolescents” (J. of the Amer. Med. Assoc., 2009) reported on body composition and metabolic changes for individuals who had taken various antipsychotic drugs for short periods of time.

a. The sample of 41 individuals who had taken aripiprazole had a mean change in total cholesterol (mg/dL) of 3.75, and the estimated standard error $s_D/\sqrt{n}$ was 3.878. Calculate a confidence interval with confidence level approximately 95% for the true average increase in total cholesterol under these circumstances (the cited article included this CI).

b. The article also reported that for a sample of 36 individuals who had taken quetiapine, the sample mean cholesterol level change and estimated standard error were 9.05 and 4.256, respectively. Making any necessary assumptions about the distribution of change in cholesterol level, does the choice of significance level impact your conclusion as to whether true average cholesterol level increases? Explain. [Note: The article included a P-value.] c. For the sample of 45 individuals who had taken olanzapine, the article reported (7.38, 9.69) as a 95% CI for true average weight gain (kg). What is a 99% CI?

To determine

Calculate the 95% confidence interval for the true average increase in total cholesterol.

Answer to Problem 41E

The 95% confidence interval for the true average increase in total cholesterol lies between –3.85 and 11.35.

Explanation of Solution

Given info:

The information is based on selecting a sample of 41 individuals has been taken antipsychotic drugs for the short period of time has the mean of 3.75 and the estimated standard error of 3.878.

Calculation:

Let ${\mu }_D$ denotes the true mean change in the total cholesterol.

Confidence interval:

The confidence interval using large sample method is,

$\overline{d}\pm z_{\frac{\alpha }{2}}\frac{s_D}{\sqrt{n}}$ ,

Where $\overline{d}$ the sample mean, $s_D$ is the standard deviation of the mean and n be the sample size.

Confidence interval:

Step-by-step procedure to obtain the confidence interval using the MINITAB software:

• Choose Stat > Basic Statistics > One sample Z.
• Choose Summarized columns. Enter the Sample size as 41, Mean as 3.78 and Standard deviation as 24.8.
• In Perform hypothesis test, enter the 0.
• Choose Options.
• In Confidence level, enter 95.
• In Alternative, select Not equal.
• Click OK in all the dialog boxes.

Output using the MINITAB software is given below:

From the MINITAB output, the confidence interval is (–3.85,11.35).

Thus, the 95% confidence interval for the true average increase in total cholesterol lies between –3.85 and 11.35.

Interpretation:

There is 95% confident that the true average increase in total cholesterol lies between –3.85 and 11.35.

To determine

Explain whether the choice of significance level impact the conclusion that the true average cholesterol level under the quetiapine is positive.

Answer to Problem 41E

The data provides the sufficient evidence that the true average cholesterol level under the quetiapine is positive at the level of significance of 5%.

Explanation of Solution

Given info:

The information is based on selecting a sample of 36 individuals has been taken quetiapine drugs for the short period of time has the mean of 9.05 and the estimated standard error of 4.256.

Calculation:

Let ${\mu }_D$ denotes the true mean change in the total cholesterol.

Assume that the level of significance is 0.05.

The test hypotheses are,

Null hypothesis:

$H_0:{\mu }_D=0$

That is, the true mean change in the total cholesterol under quetiapine is same.

Alternative hypothesis:

$H_a:{\mu }_D>0$

That is, the true mean change in the total cholesterol quetiapine is positive.

Requirements for a paired t-test:

• The sample X and Y taken from the population is selected at random.
• The samples X and Y are dependent of each other.
• Samples must be distributed to normal.

Here, the samples selected at random and dependent. Moreover, the sample size is assumed to be normally distributed. Hence, the assumptions are satisfied.

Test statistic:

The test statistic is,

$t=\frac{\overline{d}-{\Delta }_0}{\frac{s_D}{\sqrt{n}}}$ ,

Where $\overline{d}$ the sample mean, $s_D$ is the standard deviation of the mean and n be the sample size.

The test statistic is obtained given below:

Substitute $\overline{d}=9.05$,${\Delta }_0=0$ and $\frac{s_D}{\sqrt{n}}=4.256$

$\begin{array}{c}t=\frac{9.05-0}{4.256}\\ =\frac{9.05}{4.256}\\ =2.13\end{array}$

The test statistic is 2.13

The P- value is obtained by $P-\text{value =}P\left(T_{n-1}\ge t\right)$.

As there are 36 individuals , thus $n=36$.

Thus,

$\begin{array}{c}P-\text{value =}P\left(T_{36-1}\ge 2.13\right)\\ =P\left(T_{35}\ge 2.13\right)\\ \simeq 0.02\end{array}$

Rejection region:

If $P\text{-value}>\alpha$, then fail to reject the null hypothesis $\left(H_0\right)$.

If $P\text{-value}<\alpha$, then reject the null hypothesis $\left(H_0\right)$.

Conclusion:

Here, the P-value is less than the level of significance.

That is, $P\text{-value}\left(=0.02\right)<\alpha \left(=0.05\right)$.

Therefore, the decision is “reject the null hypothesis”.

Thus, there is enough evidence to infer that the true average cholesterol level under the quetiapine is positive at the level of significance of 5%.

To determine

Find the 99% confidence interval for the true average weight gain.

Answer to Problem 41E

The 99% confidence interval for the true average weight gain is (7.02,10.06)

Explanation of Solution

Given info:

The information is based on selecting a sample of 45 individuals has been taken olanzapine has the 95% confidence interval of (7.38,9.69).

Calculation:

Confidence interval:

The confidence interval using large sample method is,

$\overline{d}\pm z_{\frac{\alpha }{2}}\frac{s_D}{\sqrt{n}}$ ,

Where $\overline{d}$ the sample mean, $s_D$ is the standard deviation of the mean and n be the sample size.

The sample mean $\overline{d}$ is obtained as:

$\begin{array}{c}\overline{d}=\frac{7.38+9.69}{2}\\ =\frac{17.07}{2}\\ =8.535\end{array}$

The standard error of $\overline{d}$ is obtained as:

$\begin{array}{c}\text{Width}=2\left(1.96\left(SE\left(\overline{d}\right)\right)\right)\\ 9.69-7.38=2\left(1.96\right)\left(SE\left(\overline{d}\right)\right)\\ \left(SE\left(\overline{d}\right)\right)=\frac{2.31}{3.92}\\ \simeq 0.59\end{array}$

Level of significance:

The level of significance for both-tailed test,

$\begin{array}{c}\frac{\alpha }{2}=\frac{0.01}{2}\\ =0.005\end{array}$

The 99% confidence interval for the true average weight gain is obtained as:

Substitute $\overline{d}=8.535$, $Z_{0.005}=2.576$ and $\frac{s_D}{\sqrt{n}}=4.256$

$\begin{array}{c}\overline{d}\pm z_{\frac{\alpha }{2}}\frac{s_D}{\sqrt{n}}=8.535\pm \left(2.576\right)\left(0.59\right)\\ =8.535\pm 1.52\\ =\left(7.02,10.06\right)\end{array}$

Thus, there is 99% confidence that the true average weight gain lies between the interval 7.02 and 10.06.