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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.

It is assumed that the sample values are normally distributed.

Assume that the samples are randomly selected.

Therefore, the Student鈥檚 t-distribution would be used here.

The distribution of data of clinical trial for the drug zopiclone follows Student鈥檚 t-distribution.

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

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$渭$

Mathematically, it can be expressed as,

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

The hypothesis is left tailed.

Formula for test statistic is given by,

$t=\frac{\stackrel{炉}{x}-渭}{\frac{s}{\sqrt{n}}}$

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

The statistics is given as,

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

Substituting these values,

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

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

For Student鈥檚 t-distribution, critical value $t_{伪}$is a value corresponding to the area in one tail of $伪$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 $\left(t:t_{0.05}<-1.753\right)$.

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.

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.