The level of dissolved oxygen (DO) 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 15 randomly chosen locations along a stream. Here are the results in milligrams per liter (mg/l): 4.53, 5.04, 3.29,5.23,4.13,5.50, 4.83, 4.40,5.42,6.38,4.01,4.66, 2.87,5.73,5.55 A dissolved oxygen level below 5 mg/1 puts aquatic life at risk.

(a) A frequency histogram of the data is given below. Explain why it is most appropriate to apply t procedures with this sample

(b) Do we have convincing evidence at the 0.05 significance level that aquatic life in this stream is at risk?

c) Given your conclusions of part (b), which kind of error - Type 1 vs Type II - could you have made? Explain what this mistake would mean in context?

a) It is most appropiate to apply t procedures with this sample because the population standard deviation is unknown. We have to estimate it from the sample standard deviation.

b) There is no enough evidence to claim that disolved oxygen level is below 5 mg/l.

c) As the null hypothesis was not rejected, the only error we could have made is a Type II error.

This Type II error is made when fail to reject the null hypotesis even when the alternative hypothesis is true.

In this context, it would mean that the aquatic life is really at risk (dissolved oxygen levels below 5 mg/l), but because the sample results combined with its size were not giving enough evidence to prove the claim.

Step-by-step explanation:

b) We have to perform a hypothesis test on the mean, at a significance level of 0.05.

The null and alternative hypothesis are:

The sample mean is 4.77 and the sample standard deviation is 0.94.

The sample size is 15, so the degrees of freedom are:

The t-statistic is:

For a t=0.95 and df=14, the P-value is P=0.18.

This P-value is bigger than the significance level, so the null hypothesis is failed to be rejected.

There is no enough evidence to claim that disolved oxygen level is below 5 mg/l.

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