Chapter 9.3, Problem 74E

(a)

To determine

To Explain: that there is convincing at the $\alpha =0.01$ significance level that the average sleep increase is positive for insomnia patients when taking this drug.

Answer to Problem 74E

There is enough convincing proof that the mean sleep increase is positive for insomnia patients when taking the drug.

Explanation of Solution

Given:

  $\begin{array}{l}\alpha =0.01\\ n=10\end{array}$

Formula used:

  $t=\frac{\overline{x}-{\mu }_0}{s/\sqrt{n}}$

Calculation:

Conditions

The three conditions are: Random, independent.(10% condition), Normal/Large sample.

Random: Satisfied, because the sample is a random sample.

Independent: satisfied, because the sample of 10 patients who suffer from insomnia is less than 10% of the population of all patients who suffer from insomnia

Normal/ Large sample: satisfied, because the pattern in the normal quintile plot is roughly linear, this indicates that the distribution is about Normal.

Since all conditions are satisfied, it is suitable to perform a hypothesis test for the population mean

The mean is

  $\begin{array}{c}\overline{x}=\frac{\begin{array}{l}1.9+0.8+1.1+0.1-0.1\\ +4.4+5.5+1.6+4.6+3.4\end{array}}{10}\\ =\frac{23.3}{10}\\ =2.33\end{array}$

The variance is

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

Hypothesis test

The claim is either the null hypothesis or the alternative hypothesis. The null hypothesis statement is the population mean is equal to the value given in the claim. If the null hypothesis is the claim, then the alternative hypothesis statement is the opposite of the null hypothesis.

  $\begin{array}{c}{H}_0:\mu =0\\ H_1:\mu >0\end{array}$

The statistic is

  $\begin{array}{c}t=\frac{\overline{x}-{\mu }_0}{s/\sqrt{n}}\\ =\frac{2.33-0}{2.0022/\sqrt{10}}\\ =3.680\end{array}$

The P-value is the probability of getting the value of the test static, or a value more extreme, 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, 1E99, 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 $\alpha$ , then the null hypothesis is rejected.

  $P<0.01\Rightarrow \text{Reject }{H}_0$

There is enough convincing proof that the mean sleep increase is positive for insomnia patients when taking the drug.

(b)

To determine

To Explain: the conclusion in part (a), which kind of mistakes a Type I error or a Type II error could have made, explain this mistake would mean in context.

Answer to Problem 74E

There is enough convincing proof that the mean sleep increase is positive for insomnia patients when taking this drug.

Explanation of Solution

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

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

Type II error: Fail to reject the null hypothesis $H_0$ , once the null hypothesis is false

Since we reject the null hypothesis $H_0$ , it is only possible to have made a type I error.

This would mean that there is enough convincing proof that the mean sleep increase is positive for insomnia patients when taking this drug, whereas the mean sleep increase is actually 0. This would then also imply that we would give insomnia patients a drug that doesn’t actually work.