Problem 18 - Basic Statistics for Business and Economics with Student

Question

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): $$\begin{array}{|lllllllll|}\hline 2,159 & 2,170 & 2,180 & 2,179 & 2,160 & 2,167 & 2,171 & 2,181 & 2,185 \\\\\hline\end{array}$$ At the .025 level, has Holdlonger increased the shelf life of the chlorine? Estimate the \(p\) -value.

Step-by-Step Solution

Answer

Holdlonger increased the shelf life; the p-value is below 0.025.

Step 1: State Hypotheses

To determine if Holdlonger has increased the shelf life, state the null hypothesis (\(H_0\) : \(\mu = 2160\)) and the alternative hypothesis (\(H_a\) : \(\mu > 2160\)), where \(\mu\) is the mean shelf life with Holdlonger.

Step 2: Calculate Sample Mean

Calculate the mean of the sample shelf lives: \(\bar{x} = \frac{2159 + 2170 + 2180 + 2179 + 2160 + 2167 + 2171 + 2181 + 2185}{9} = 2172.44\).

Step 3: Calculate Sample Standard Deviation

Find the sample standard deviation (s) using the formula:\[ s = \sqrt{\frac{1}{n-1}\sum_{i=1}^{n}(x_i - \bar{x})^2} \approx 9.30 \]

Step 4: Conduct t-test

Calculate the t-statistic using the formula:\[ t = \frac{\bar{x} - \mu}{s/\sqrt{n}} = \frac{2172.44 - 2160}{9.30/\sqrt{9}} \approx 4.11 \]

Step 5: Determine Critical Value and Decision

For \( n = 9 \) and \( \alpha = 0.025 \), the critical value \( t_{critical} \) for a one-tailed test with 8 degrees of freedom is 2.306. Since \( t = 4.11 \) is greater than \( t_{critical} = 2.306 \), reject \( H_0 \).

Step 6: Estimate p-value

Using a t-distribution table, locate 4.11 with 8 degrees of freedom. The \( p \)-value is below 0.005, much smaller than 0.025, confirming the rejection of \( H_0 \).

Key Concepts

T-Test

The T-Test is a statistical method used to determine if there is a significant difference between the means of two groups. In the context of our exercise, the T-Test helped us decide whether the Holdlonger additive had a significant effect on the chlorine's shelf life. The process involves setting up two hypotheses: the null hypothesis (H_0), which assumes that there is no difference in mean shelf life, and the alternative hypothesis (H_a), which expects that the mean shelf life with Holdlonger is greater.

• Null Hypothesis (H_0): \( \mu = 2160 \)
• Alternative Hypothesis (H_a): \( \mu > 2160 \)

The calculation of the T-statistic is central to the test. We compare the calculated T value to a critical value from the T-distribution table to make our decision. If the T-statistic exceeds the critical value, we reject the null hypothesis, suggesting that there is significant evidence that the mean shelf life has increased.

P-Value

The P-Value helps quantify the evidence against the null hypothesis. It's a probability measure that tells us the likelihood of observing the test results if the null hypothesis were true. In simple terms, the smaller the P-Value, the stronger the evidence against H_0.In our exercise, we found a P-Value well below 0.025, indicating a statistically significant result. This P-Value is below our level of significance (\( \alpha = 0.025 \)), leading us to reject the null hypothesis and conclude that Holdlonger likely increased the shelf life of chlorine.The P-Value is essential to making data-driven decisions. Along with the T-statistic, it offers a clear picture of what the data is suggesting in terms of hypothesis testing.

Sample Mean

The Sample Mean is an essential concept in statistics, representing the average value of a sample. It provides insight into the overall trend of a dataset and acts as a foundation for further statistical analysis. In our exercise, the sample mean is crucial for determining the effect of Holdlonger on chlorine's shelf life.We calculate the sample mean, \( \bar{x} \), by summing all sample values and dividing by the number of samples:\[ \bar{x} = \frac{2159 + 2170 + 2180 + 2179 + 2160 + 2167 + 2171 + 2181 + 2185}{9} = 2172.44 \]This value then serves as a comparison point against the hypothesized population mean to see if a significant difference exists. A higher sample mean compared to the null hypothesis suggests that the Holdlonger treatment may have extended the chlorine's shelf life. It's through this lens that statistical inferences about populations are often made.

Sample Standard Deviation

The Sample Standard Deviation is a statistic that reflects the dispersion or spread of a sample dataset. It provides insight into how much individual data points deviate from the sample mean. In hypothesis testing, understanding the data's variability is crucial as it impacts the reliability of the sample mean.To calculate the sample standard deviation (\( s \)), we use the formula:\[ s = \sqrt{\frac{1}{n-1}\sum_{i=1}^{n}(x_i - \bar{x})^2} \approx 9.30 \]This measure helps us comprehend the extent to which the shelf lives differ from the average shelf life of 2172.44 hours. A smaller standard deviation signifies that the data points are closer to the mean, indicating more consistency. The standard deviation directly influences the T-statistic, thus affecting the outcome of the hypothesis test. Understanding its role enables more accurate data analyses and result interpretations.