Question

One water quality standard for water that is discharged into a particular type of stream or pond is that the average daily water temperature be at most 18°C. Six samples taken throughout the day gave the data: 16.8 21.5 19.1 12.8 18.0 20.7 The sample mean x - = 18.15 exceeds 18, but perhaps this is only sampling error. Determine whether the data provide sufficient evidence, at the 10% level of significance, to conclude that the mean temperature for the entire day exceeds 18°C.

Answer 1

`t=(18.15-18)/((3.134)/(√(6)))=0.117`

`p_v =P(t_(5)>0.117)=0.456`

If we compare the p value and the significance level given
`\alpha=0.1` we see that
`p_v>\alpha` so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can conclude that the mean is not significantly higher than 18 at 10% of signficance.

Explanation:

Data given and notation

16.8 21.5 19.1 12.8 18.0 20.7

We can calculate the mean and deviation with the following formulas:

`\bar X = (\sum_(i=1)^n X_i)/(n)`

`s= \sqrt{(\sum_(i=1)^n (X_i -\bar X)^2)/(n-1)}`

And the results are:

`\bar X=18.15` represent the sample mean

`s=3.134` represent the sample standard deviation

`n=6` sample size

`\mu_o =18` represent the value that we want to test

`\alpha=0.1` represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)

`p_v` represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the true mean is higher than 18, the system of hypothesis would be:

Null hypothesis:
`\mu \leq 18`

Alternative hypothesis:
`\mu > 18`

Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

`t=(\bar X-\mu_o)/((s)/(√(n)))` (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

`t=(18.15-18)/((3.134)/(√(6)))=0.117`

P-value

The degrees of freedom are given by:

`df = n-1 = 6-1= 5`

Since is a right tailed test the p value would be:

`p_v =P(t_(5)>0.117)=0.456`

Conclusion

If we compare the p value and the significance level given
`\alpha=0.1` we see that
`p_v>\alpha` so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can conclude that the mean is not significantly higher than 18 at 10% of signficance.