The liquid chlorine added to swimming pools to combat algae has a relatively short shelf life before it loses its effectiveness. Records indicate that the mean shelf life of a 5-gallon jug of chlorine is 2,160 hours (90 days). As an experiment, Holdlonger was added to the chlorine to find whetehr it would increase the shelf life. A sample of nine jugs of chlorine had hese shelf lives (in hours):

2159, 2170, 2180, 2179, 2160, 2167, 2171, 2181, 2185

At the .025 level, has Holdlonger increased the shelf life of the chlorine? (estimate the p-value). Show your detailed work.

If we compare the p value and the significance level given [tex]\alpha=0.025[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is higher than 2160 hours at 2.5% of signficance.

Step-by-step explanation:

Data given and notation

2159, 2170, 2180, 2179, 2160, 2167, 2171, 2181, 2185

We can calculate the sample mean and deviation with these formulas:

[tex]\bar X = \sum_{i=1}^n \frac{X_i}{n}[/tex]

[tex]s =\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}[/tex]

[tex]\bar X=2172.44[/tex] represent the sample mean

[tex]s=9.382[/tex] represent the sample standard deviation

[tex]n=9[/tex] sample size

[tex]\mu_o =2160[/tex] represent the value that we want to test

[tex]\alpha=0.025[/tex] represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

[tex]p_v[/tex] 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 meanis higher than 2160, the system of hypothesis would be:

Null hypothesis:[tex]\mu \leq 2160[/tex]

Alternative hypothesis:[tex]\mu > 2160[/tex]

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (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:

[tex]t=\frac{2172.44-2160}{\frac{9.382}{\sqrt{9}}}=3.978[/tex]

P-value

The first step is calculate the degrees of freedom, on this case:

[tex]df=n-1=9-1=8[/tex]

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

[tex]p_v =P(t_{(8)}>3.978)=0.00204[/tex]

Conclusion

If we compare the p value and the significance level given [tex]\alpha=0.025[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is higher than 2160 hours at 2.5% of signficance.