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 whether it would increase the shelf life. A sample of nine jugs of chlorine had these shelf lives (in hours)
2,159
2,170
2,180
2,179
2,160
2,167
2,171
2,181
2,185
a. At the .025 level, has Holdlonger increased the shelf life of the chlorine?
b. Estimate the p-value.
What I have so far is:
H0: μ ≤ 2160
H1: μ > 2160
This is a one-tailed test because we want to see if there has been an increase in the shelf life.
The level of significance given for this problem is 0.025.
This is a t-distribution since there are less than 30 observations. Since we do not know the standard deviation of the underlying population, we use the sample standard deviation, and the value of the test statistic is:
t = (Xbar - μ)/(s/sqrt(n)) = (2172.444 - 2160)/(9.382/sqrt(9)) = 3.979
We also have that there are 9 – 1 = 8 degrees of freedom. Plugging this into the Excel formula =TDIST(3.979,8,1), we get a result of 0.002. Since this is less than 0.025, we can conclude that Holdlonger has increased the shelf life of the chlorine.
The p-value in this case is 0.998 or 99.8%.
Also note that the t-value that is associated with this example is 2.306. Since our value of 3.979 is greater than 2.306, this also confirms the rejection of H0.
The probability I calculated of 99.8% seems a little bit to certain for a homework problem asking for a 97.5% certainty. Am I on the right track here, or am I completely off base and I need to start the problem all over again?
2,159
2,170
2,180
2,179
2,160
2,167
2,171
2,181
2,185
a. At the .025 level, has Holdlonger increased the shelf life of the chlorine?
b. Estimate the p-value.
What I have so far is:
H0: μ ≤ 2160
H1: μ > 2160
This is a one-tailed test because we want to see if there has been an increase in the shelf life.
The level of significance given for this problem is 0.025.
This is a t-distribution since there are less than 30 observations. Since we do not know the standard deviation of the underlying population, we use the sample standard deviation, and the value of the test statistic is:
t = (Xbar - μ)/(s/sqrt(n)) = (2172.444 - 2160)/(9.382/sqrt(9)) = 3.979
We also have that there are 9 – 1 = 8 degrees of freedom. Plugging this into the Excel formula =TDIST(3.979,8,1), we get a result of 0.002. Since this is less than 0.025, we can conclude that Holdlonger has increased the shelf life of the chlorine.
The p-value in this case is 0.998 or 99.8%.
Also note that the t-value that is associated with this example is 2.306. Since our value of 3.979 is greater than 2.306, this also confirms the rejection of H0.
The probability I calculated of 99.8% seems a little bit to certain for a homework problem asking for a 97.5% certainty. Am I on the right track here, or am I completely off base and I need to start the problem all over again?