Question

Testing Hypotheses. In Exercises 13–24, assume that a simple random sample has been selected and test the given claim. Unless specified by your instructor, use either the P-value method or the critical value method for testing hypotheses. Identify the null and alternative hypotheses, test statistic, P-value (or range of P-values), or critical value(s), and state the final conclusion that addresses the original claim.

Insomnia Treatment A clinical trial was conducted to test the effectiveness of the drug zopiclone for treating insomnia in older subjects. Before treatment with zopiclone, 16 subjects had a mean wake time of 102.8 min. After treatment with zopiclone, the 16 subjects had a mean wake time of 98.9 min and a standard deviation of 42.3 min (based on data from “Cognitive Behavioral Therapy vs Zopiclone for Treatment of Chronic Primary Insomnia in Older Adults,” by Sivertsenet al.,Journal of the American Medical Association, Vol. 295, No. 24). Assume that the 16 sample values appear to be from a normally distributed population, and test the claim that after treatment with zopiclone, subjects have a mean wake time of less than 102.8 min. Does zopiclone appear to be effective?

Short answer

We reject the claim that the mean wake time for subjects after the treatment with zopiclone is less than 102.8 min.

Zopiclone drug does not appear to be effective.

Step-by-step solution

Given information

To test the effectiveness of the drug zopiclone for treating insomnia, critical trial was conducted for 16 subjects. The mean wake time of subjects before the treatment is 102.8 min; and after the treatment, the mean wake time is 98.9 min with a standard deviation of 42.3 min.

Check the requirements

It is assumed that the sample values are normally distributed.

Assume that the samples are randomly selected.

Therefore, the Student’s t-distribution would be used here.

The distribution of data of clinical trial for the drug zopiclone follows Student’s t-distribution.

Thus, the sample size of clinical trial (n)=16

Describe the hypotheses

Null hypothesis$H_0$is a statement of claim that the mean wake time for subjects after treatment with zopiclone is 102.8 min.

Alternate hypothesis H₁ is a statement of claim that the mean wake time for subjects after treatment with zopiclone is less than 102.8 min.

Assume the population mean wake time after treatment with zopiclone as$\mu$

Mathematically, it can be expressed as,

$$\begin{gathered} H_0:\mu =102.8 \\ {H}_1:\mu <102.8 \end{gathered}$$

The hypothesis is left tailed.

Calculate the test statistic

Formula for test statistic is given by,

$t=\frac{\overline{x}-\mu }{\frac{s}{\sqrt{n}}}$

Where , $\overline{x}$is the sample mean and s is the standard deviation of sample.

The statistics is given as,

$$\begin{gathered} \overline{x}=98.9 \\ s=42.3 \end{gathered}$$

Substituting these values,

$$\begin{gathered} t=\frac{98.9-102.8}{\frac{42.3}{\sqrt{16}}} \\ t=-0.3688 \end{gathered}$$

Calculate the critical value. 

Assume that the level of significance is 0.05 $\left(\alpha \right)$.

For Student’s t-distribution, critical value $t_{\alpha }$is a value corresponding to the area in one tail of $\alpha$the t-distribution.

The sample size is 16 (n).

The degree of freedom is,

$$\begin{gathered} \text{df}=n-1 \\ =15 \end{gathered}$$

The critical value is obtained on left tail of the distribution.

In the t-distribution table, find the value corresponding to the row value of degree of freedom 15 and column value of area in one tail 0.05 is 1.753 which is critical value ${\text{t}}_{0.05}$; but the given test is left tailed therefore use -1.753 as a critical value.

Thus, the critical value ${\text{t}}_{0.05}$ is -1.753.

The rejection region is hence defined as $\left(t:t_{0.05}<-1.753\right)$.

Compare test statistics and critical value

Test statistics is -0.3688 and the critical value $t_{0.05}$is -1.753.

According to this, we can conclude that the test statistic -0.3688 will not fall in the critical region bounded by the critical value -1.753.

Therefore, we failed to reject the null hypothesis.

As the null hypothesis is failed to be rejected, there is enough evidence to support the claim that the mean wake time after treatment is not different from 102.8 min.

Conclusion

There is no significant difference between the mean wake time before the treatment and after the treatment with zopiclone. Hence, the treatment with the drug zopiclone does not have significant effect.

Therefore, zopiclone drug does not appear to be effective.