The average level of dissolved oxygen in a stream or river is an important indicator of the water's ability to support aquatic life. A researcher measures the DO level at 43 randomly chosen locations along a stream. Below are the results in mg/l:

3.844.084.245.394.04
5.75.064.942.216.03
5.834.34.063.234.94
4.62.675.126.753.08
2.984.894.684.525.9
2.54.125.894.12.98
5.33.644.434.853.48
3.063.593.514.894.15
4.344.221.77
An average level of dissolved oxygen less than 4.8 mg/l puts aquatic life at risk. At 5% significance level can we conclude that an average level of dissolved oxygen different from 4.8 mg/l? (Note: The average and the standard deviation of the data are respectively 4.28 mg/l and 1.1 mg/l.)

Procedure:

Assumptions: (select everything that applies)

Step 1. Hypotheses Set-Up:

 H0:\displaystyle {H}_{{0}}:  = , where is the and the units are
 Ha:\displaystyle {H}_{{a}}:  , and the test is

Step 2. The significance level α=\displaystyle \alpha= %

Step 3. Compute the value of the test statistic: = (Round the answer to 3 decimal places)

Step 4. Testing Procedure: (Round the answers to 3 decimal places)

CVA PVA
Provide the critical value(s) for the Rejection Region: Compute the P-value of the test statistic:
left CV is and right CV is P-value is

Step 5. Decision:

CVA PVA
Is the test statistic in the rejection region? Is the P-value less than the significance level?

Conclusion:

Step 6. Interpretation:

At 5% significance level we have sufficient evidence to reject the null hypothesis in favor of the alternative hypothesis.