Chapter 8.3, Problem 30E

A sample of n sludge specimens is selected and the pH of each one is determined. The one-sample t test will then be used to see if there is compelling evidence for concluding that true average pH is less than 7.0. What conclusion is appropriate in each of the following situations?

1. a. n = 6, t = −2.3, α = .05
2. b. n = 15, t = −3.1, α = .01
3. c. n = 12, t = −1.3, α = .05
4. d. n = 6, t = .7, α = .05
5. e. n = 6, $\bar{x}=6.68,s/\sqrt{n}=.0820$

To determine

Find the appropriate conclusion for the given situation.

Answer to Problem 30E

There is sufficient evidence to support that true average pH is less than 7 for this situation.

Explanation of Solution

Given info:

A sample of size n sludge specimens was collected for checking pH. The t test was used for concluding that true average pH is less than 7. $n=6,t=-2.3,\alpha =0.05$

Calculation:

Hypothesis:

Null hypothesis:

$H_0:\mu =7$

Alternative hypothesis:

$H_a:\mu <7$

P-value:

Software procedure:

Step by step procedure to obtain the P-value using the MINITAB software is given below:

• Choose Graph > Probability Distribution Plot choose View Probability > OK.
• From Distribution, choose ‘t’ distribution, degrees of freedom as 5.
• Click the Shaded Area tab.
• Choose X Value and left for the region of the curve to shade.
• Enter the data value as –2.3.
• Click OK.

Output using the MINITAB software is given below:

From the MINITAB output, the P-value is 0.0348.

Decision:

If $P\text{-value}\le \alpha ,$, reject the null hypothesis $H_0$

If $P\text{-value}>\alpha ,$ fail to reject the null hypothesis $H_0$

Conclusion:

Here, the P-value is less than the level of significance.

That is, $P\text{-value}\left(=0.0348\right)<\alpha \left(0.05\right)$.

By rejection rule, reject null hypothesis.

Hence, there is sufficient evidence to support that true average pH is less than 7 for this situation.

To determine

Find the appropriate conclusion for the given situation.

Answer to Problem 30E

There is sufficient evidence to support that true average pH is less than 7 for this situation.

Explanation of Solution

Given info:

$n=15,t=-3.1,\alpha =0.01$

Calculation:

P-value:

Software procedure:

Step by step procedure to obtain the P-value using the MINITAB software is given below:

• Choose Graph > Probability Distribution Plot choose View Probability > OK.
• From Distribution, choose ‘t’ distribution, degrees of freedom as 14.
• Click the Shaded Area tab.
• Choose X Value and left for the region of the curve to shade.
• Enter the data value as –3.1.
• Click OK.

Output using the MINITAB software is given below:

From the MINITAB output, the P-value is 0.0039.

Conclusion:

Here, the P-value is less than the level of significance.

That is, $P\text{-value}\left(=0.0039\right)<\alpha \left(0.01\right)$.

By rejection rule, reject null hypothesis.

Hence, there is sufficient evidence to support that true average pH is less than 7 for this situation.

To determine

Find the appropriate conclusion for the given situation.

Answer to Problem 30E

There is no sufficient evidence to support that true average pH is less than 7 for this situation.

Explanation of Solution

Given info:

$n=12,t=-1.3,\alpha =0.05$

Calculation:

P-value:

Software procedure:

Step by step procedure to obtain the P-value using the MINITAB software is given below:

• Choose Graph > Probability Distribution Plot choose View Probability > OK.
• From Distribution, choose ‘t’ distribution, degrees of freedom as 11.
• Click the Shaded Area tab.
• Choose X Value and left for the region of the curve to shade.
• Enter the data value as –1.3.
• Click OK.

Output using the MINITAB software is given below:

From the MINITAB output, the P-value is 0.1101

Conclusion:

Here, the P-value is greater than the level of significance.

That is, $P\text{-value}\left(=0.1101\right)>\alpha \left(0.05\right)$.

By rejection rule, fail to reject null hypothesis.

Hence, there is no sufficient evidence to support that true average pH is less than 7 for this situation.

To determine

Find the appropriate conclusion for the given situation.

Answer to Problem 30E

There is no sufficient evidence to support that true average pH is less than 7 for this situation.

Explanation of Solution

Given info:

$n=6,t=0.7,\alpha =0.05$

Calculation:

P-value:

Software procedure:

Step by step procedure to obtain the P-value using the MINITAB software is given below:

• Choose Graph > Probability Distribution Plot choose View Probability > OK.
• From Distribution, choose ‘t’ distribution, degrees of freedom as 5.
• Click the Shaded Area tab.
• Choose X Value and left for the region of the curve to shade.
• Enter the data value as 0.7.
• Click OK.

Output using the MINITAB software is given below:

From the MINITAB output, the P-value is 0.7424.

Conclusion:

Here, the P-value is greater than the level of significance.

That is, $P\text{-value}\left(=0.7434\right)>\alpha \left(0.05\right)$.

By rejection rule, fail to reject null hypothesis.

Hence, there is no sufficient evidence to support that true average pH is less than 7 for this situation.

To determine

Find the appropriate conclusion for the given situation.

Answer to Problem 30E

There is sufficient evidence to support that true average pH is less than 7 at 5% level of significance.

Explanation of Solution

Given info:

$n=6,\overline{x}=6.68,\frac{s}{\sqrt{n}}=0.820$

Calculation:

Step 1: Parameter of interest:

Let $\mu$ be the true average pH.

Step 2: Null hypothesis:

$H_0:\mu =7$

Step 3: Alternative hypothesis:

$H_a:\mu <7$.

Step 4: Test statistic:

Software procedure:

Step by step procedure to obtain the test-statistic value using the MINITAB software:

• Choose Stat > Basic Statistics > 1-Sample Z.
• In Summarized data, enter the sample size as 6 and mean as 6.68.
• In Standard deviation, enter 0.2.
• In Perform hypothesis test, enter the test mean as 7
• Check Options; enter Confidence level as 95%.
• Choose less than in alternative.
• Click OK in all dialogue boxes.

Output using the MINITAB software is given below:

From the MINITAB output, the test statistic value is –3.92

Step 5: P-value:

From the output of step 4, the P-value is given by 0.006.

Step 5: Decision:

If $P\text{-value}\le \alpha ,$, reject the null hypothesis $H_0$

If,$P\text{-value}>\alpha ,$ fail to reject the null hypothesis $H_0$

Conclusion:

Here, the P-value is less than the level of significance.

That is, $P\text{-value}\left(=0.006\right)<\alpha \left(0.05\right)$.

By rejection rule, reject null hypothesis.

Step 6: Interpretation:

Thus, it can be concluded that there is sufficient evidence to support that true average pH is less than 7 at 5% level of significance.