Chapter 9.3, Problem 69E

a.

To determine

To find: The P -value using technology and table. Also, the conclusions that could be drawn using the significance levels 1% and 5%.

Answer to Problem 69E

The P -value is 0.0431.

Explanation of Solution

Given:

The null and alternative hypotheses are:

  $\begin{array}{l}{H}_0:\mu =5\\ H_a:\mu >5\end{array}$

Sample size ( n )= 20

Test statistic ( t ) = 1.81

Calculation:

From alternative hypothesis, it is known that the test is one tailed.

The degree of freedom is computed as:

  $\begin{array}{c}df=n-1\\ =20-1\\ =19\end{array}$

The P -value using Ti-83 calculator can be computed as:

  

Now, P - value using table can be calculated as:

  $\begin{array}{c}P-\text{value}=P\left(t>\left|t\right|\right)\\ =P\left(t>\left|1.81\right|\right)\\ =0.0431\end{array}$

Thus, the P -value is 0.0431.

Here, P -value (0.0431) >$\alpha \left(0.01\right)$ and < $\alpha \left(0.05\right)$ . Thus, the null hypothesis is rejected at $\alpha =\left(0.01\right)$ .

Conclusion:

Thus, at 1% significance there are insufficient evidence to conclude that $\mu =5$ .

b.

To determine

To find: The P -value using technology and table. Also, the conclusions that could be drawn using the significance levels 1% and 5%.

Answer to Problem 69E

The P -value is 0.086.

Explanation of Solution

Given:

The null and alternative hypotheses are:

  $\begin{array}{l}{H}_0:\mu =5\\ H_a:\mu \ne 5\end{array}$

Sample size ( n )= 20

Test statistic ( t ) = 1.81

Calculation:

From alternative hypothesis, it is known that the test is one tailed.

The degree of freedom is computed as:

  $\begin{array}{c}df=n-1\\ =20-1\\ =19\end{array}$

The P -value using Ti-83 calculator can be computed as:

  

The above value is for one tailed.

The P -value for two tailed can be calculated as:

  $\begin{array}{c}P-value=2\times 0.0431\\ =0.086\end{array}$

Here, also the P -value is 0.086.

Now, P - value using table can be calculated as:

  $\begin{array}{c}P-\text{value}=2\times P\left(t>\left|t\right|\right)\\ =2\times P\left(t>\left|1.81\right|\right)\\ =2\times 0.0431\\ =0.086\end{array}$

Thus, the P -value is 0.086.

Here, P -value (0.086) >$\alpha \left(0.01\right)$ and $\alpha \left(0.05\right)$ . Thus, the null hypothesis is not rejected at $\alpha =\left(0.01\right)\text{ and }\alpha =\left(0.05\right)$ .

Conclusion:

Thus, at 1% and 5% significance levels there are sufficient evidence to conclude that $\mu =5$ .