Chapter 10.2, Problem 10.48E

(a)

To determine

To find: the confidence intervals.

Answer to Problem 10.48E

Therefore there is no enough evidence to help the claim that the mean dissolved oxygen content is less than 5 mg per litre

Explanation of Solution

Given:

  $\begin{array}{c}n=15\\ \overline{x}=4.77\\ s=0.94\end{array}$

Assume:

Result example (4.25, 5.29)

It is observed that the confidence interval contains 5 mg per litre, which implies that it is possible that the population mean is 5 and therefore there is not enough evidence to help the claim that the mean dissolved oxygen content is less than 5 mg per litre

(b)

To determine

To Calculate: a significance test.

Answer to Problem 10.48E

There is no enough evidence to help the claim that the mean dissolved oxygen content is less than 5 mg per litre.

Explanation of Solution

Given:

  $\begin{array}{c}n=15\\ \overline{x}=4.77\\ s=0.94\end{array}$

Claim: mean are less than 5

Formula used:

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

Calculation:

The claim is either the null hypothesis or the alternative hypothesis.

  $\begin{array}{c}{H}_0:\mu =5\\ H_1:\mu <5\end{array}$

The value of the test statistic is

  $\begin{array}{c}t=\frac{\overline{x}-{\mu }_0}{s/\sqrt{n}}\\ =\frac{4.77-5}{0.94/\sqrt{15}}\\ =-0.95\end{array}$

The P-value is the probability of getting the value of the test Statistic, or a value more extreme and row $df=n-1=15-1=14:$

  $0.15<P<0.20$

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

  $P>0.05\Rightarrow {\text{Fail to reject H}}_0$

There is no enough evidence to help the claim that the mean dissolved oxygen content is less than 5 mg per litre.