91Ó°ÊÓ

$$\begin{gathered} \mathrm{Î\pm }=0.01 \\ n=10 \end{gathered}$$

We know, the test statistic formula is:

$t=\frac{\stackrel{¯}{x}-{\mathrm{μ}}_0}{sl\sqrt{n}}$

The three conditions are as follows: random, independent (10% condition), and normal/large sample.

Random: Satisfied because the sample was chosen at random.

Independent: satisfied, because the sample of ten patients suffering from insomnia represents less than $10\%$of the total population of patients suffering from insomnia.

Normal/large sample: satisfied because the pattern in the normal quantile plot is roughly linear, indicating that the distribution is approximately Normal.

Because all conditions are met, a hypothesis test for the population mean is appropriate.

The average is:

$$\begin{gathered} 1.9+0.8+1.1+0.1-0.1 \\ \stackrel{¯}{x}=\frac{+4.4+5.5+1.6+4.6+3.4}{10} \\ =\frac{23.3}{10} \\ =2.33 \end{gathered}$$

The variance is:

$$\begin{gathered} s=\sqrt{\frac{{(1.9-2.33)}^2+{(0.8-2.33)}^2+{(1.1-2.33)}^2+{(0.1-2.33)}^2+{(-0.1-2.33)}^2+{(4.4-2.33)}^2+{(5.5-2.33)}^2+{(1.6-2.33)}^2+{(4.6-2.33)}^2+{(3.4-2.33)}^2}{10-1}} \\ =2.0022 \end{gathered}$$

Hypothesis test

The null hypothesis or the alternative hypothesis is the claim. The null hypothesis statement states that the population means equals the value specified in the claim. If the claim is the null hypothesis, then the alternative hypothesis statement is the inverse of the null hypothesis.

$$\begin{gathered} H_0:\mathrm{μ}=0 \\ {H}_a:\mathrm{μ}>0 \end{gathered}$$

The statistic is:

$$\begin{gathered} t=\frac{\stackrel{¯}{x}-{\mathrm{μ}}_0}{s/\sqrt{n}} \\ =\frac{2.33-0}{2.0022\sqrt{10}} \\ =3.680 \end{gathered}$$

The P-value is the probability of receiving the test static value, or a more extreme value, assuming that the null hypothesis is true.

$df=n-1=10-1=9$

$0.0025<P<0.005$

Command Ti83/84- calculator: tcdf $(3.051,1\mathrm{E}99,9)$which will return a P-value of $0.00689.$Note: it could replace$1E99$ by any other very large positive number.
If the P-value is smaller than the significance level $\mathrm{Î\pm },$then the null hypothesis is rejected.

$P<0.01⇒\text{Reject}{H}_0$

There is sufficient evidence to suggest that the drug's average sleep increase is beneficial to insomnia patients.

In part (a), the null hypothesis $H_0$is rejected.

Type I error: reject the null hypothesis $H_0$once the null hypothesis is true.

Type II error: Failure to reject the null hypothesis $H_0$when it is false.

We can only have made a type I error if we reject the null hypothesis $H_0$

This would imply that there is sufficiently convincing evidence that the mean sleep increase for insomnia patients taking this drug is positive, whereas the mean sleep increase is actually $0$ This would imply that we would give insomnia patients a drug that does not work.