Chapter 10.2, Problem 10.49E

To determine

To find: that there is convincing evidence that the drug is effective and carries out a significance test and state the conclusion.

Answer to Problem 10.49E

There is enough evidence to help the claim that the drug is effective.

Explanation of Solution

Given:

  $n=10$

Claim: Mean is greater than 0 (drug is effective).

Formula used:

  $\overline{x}=\frac{{\displaystyle \sum x}}{n}$

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

Calculation:

Assume

  $\alpha =0.05$

The mean is

  $\overline{x}=\frac{{\displaystyle \sum x}}{n}$

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

The variance is

  $\begin{array}{c}s=\sqrt{\frac{{(1.9-2.33)}^2+\mathrm{....}+{\left(3.4-2.33\right)}^2}{10-1}}\\ =2.0022\end{array}$

The claim is either the null hypothesis or the alternative hypothesis

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

The value of the test statistic:

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

The P-value is the probability of obtaining the value of the test statistic and $df=n-1=10-1=9:$

  $0.0025<P<0.005$

If the P-value is lesser than the significance level $\alpha$ then the alternative hypothesis is accepted and null hypothesis is rejected.

  $P<0.05\Rightarrow {\text{Reject H}}_0$

There is enough evidence to help the claim that the drug is effective.