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Chapter 10: Problem 24

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

Short Answer

Expert verified
Holdlonger did not significantly increase the shelf life (p > 0.025).

Step by step solution

01

State the Hypotheses

The null hypothesis (\(H_0\)) is that the mean shelf life with Holdlonger is equal to 2,160 hours. The alternative hypothesis (\(H_1\)) is that the mean shelf life with Holdlonger is greater than 2,160 hours. So:\[ H_0: \mu = 2160 \]\[ H_1: \mu > 2160 \]
02

Calculate the Sample Mean

First, sum up all the sample values:\[2159 + 2170 + 2180 + 2179 + 2160 + 2167 + 2171 + 2181 + 2185 = 19452\]Then, find the sample mean:\[\bar{x} = \frac{19452}{9} = 2161.333\]
03

Calculate the Sample Standard Deviation

First, calculate the deviations of each sample value from the mean:\[(2159 - 2161.333)^2, (2170 - 2161.333)^2, \, \ldots, \, (2185 - 2161.333)^2\]Then, calculate the sample standard deviation using:\[ s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n - 1}} \]Calculating these steps gives us:\[ s = 9.138 \]
04

Calculate the Test Statistic

The test statistic (\(t\)) is calculated using:\[ t = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}} \]Where:- \(\bar{x} = 2161.333\) (sample mean)- \(\mu = 2160\) (hypothesized mean)- \(s = 9.138\) (sample standard deviation)- \(n = 9\) (sample size)Plugging in these values gives:\[ t = \frac{2161.333 - 2160}{\frac{9.138}{\sqrt{9}}} = 0.438 \]
05

Determine the Critical Value

Using a t-distribution table with \(n - 1 = 8\) degrees of freedom and \(\alpha = 0.025\) for a one-tailed test, the critical value \(t_{critical}\) is approximately 2.306.
06

Compare Test Statistic and Critical Value

The calculated test statistic \(t = 0.438\) is less than the critical value \(t_{critical} = 2.306\). Therefore, we do not reject the null hypothesis.
07

Estimate the p-value

Since the calculated \(t\)-value is lower than the critical value, the \(p\)-value is greater than 0.025 (significance level). Using a t-distribution table, the \(p\)-value is around 0.34.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Sample Mean
The sample mean is a useful measure in statistical analysis as it summarizes a set of data points by indicating the central tendency of those points. To find it, we first add together all individual values from our sample, which consists of the shelf lives of chlorinated water jugs in this case. Given values such as 2159, 2170, and so forth, the sum of these observations is 19452. Then, the sum is divided by the total number of observations in the sample, yielding a result of 2161.333. This value represents the average shelf life in our sample and is crucial in testing hypotheses. Use the sample mean when you have data that you want to describe with a single center point.
  • Add all sample observations together.
  • Divide the total by the number of observations.
  • The result is the sample mean.
Understanding how to compute the sample mean helps make informed decisions about statistical data, as it leads the way in comparing sample data against predetermined hypotheses.
Sample Standard Deviation
The sample standard deviation calculates the amount of variation or dispersion within a set of data points. It is particularly relevant when determining how spread out the values in a sample are around the mean. It starts by finding each value's deviation from the mean. For example, the deviation for the first value would be: \( (2159 - 2161.333)^2 \).
Keep repeating this and summing all squared differences. Once summed, these are divided by one less than the number of observations (\(n-1\)), followed by the square root of this whole expression, which gives us the sample standard deviation, here calculated as approximately 9.138.
  • Calculate each deviation by \( (x_i - \bar{x})^2 \).
  • Sum all squared deviations.
  • Divide the sum by \( (n-1) \), followed by taking its square root.
This tells us that most chlorine jugs' shelf lives are around approximately 9.138 hours away from the mean.
t-test
A t-test is a statistical method to determine if a sample mean significantly differs from a known value or another sample mean. It is immensely useful in doing hypothesis tests, for example, observing whether Holdlonger has truly impacted chlorine's longevity. To perform a t-test, you first need the sample mean, the standard deviation, and the number of observations.
The test statistic formula is: \( t = \frac{\bar{x} - \mu}{s/\sqrt{n}} \). Here, \( \bar{x} \) is the sample mean, \( \mu \) is the population mean, \( s \) is the sample standard deviation, and \( n \) is the sample size. In this instance, the calculated t-value was 0.438. This value helps decide whether to reject the null hypothesis by comparing it with the critical value from the t-distribution table. If \( t \) is less than \( t_{critical} \), we do not reject \( H_0 \).
  • Calculate \( t \) using the known statistics.
  • Compare \( t \) with the critical value.
  • Decide on the null hypothesis.
The t-test guides decisions about your hypothesis with empirical backing.

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Most popular questions from this chapter

A statewide real estate sales agency, Farm Associates, specializes in selling farm property in the state of Nebraska. Their records indicate that the mean selling time of farm property is 90 days. Because of recent drought conditions, they believe that the mean selling time is now greater than 90 days. A statewide survey of 100 farms sold recently revealed that the mean selling time was 94 days, with a standard deviation of 22 days. At the .10 significance level, has there been an increase in selling time?

An urban planner claims that, nationally, 20 percent of all families renting condominiums move during a given year. A random sample of 200 families renting condominiums in Dallas Metroplex revealed that 56 had moved during the past year. At the .01 significance level, does this evidence suggest that a larger proportion of condominium owners moved in the Dallas area? Determine the \(p\) -value.

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Listed below is the rate of return for one year (reported in percent) for a sample of 12 mutual funds that are classified as taxable money market funds. $$\begin{array}{lllllllllll}4.63 & 4.15 & 4.76 & 4.70 & 4.65 & 4.52 & 4.70 & 5.06 & 4.42 & 4.51 & 4.24 & 4.52\end{array}$$ Using the .05 significance level is it reasonable to conclude that the mean rate of return is more than 4.50 percent?

Many grocery stores and large retailers such as Wal-Mart and K-Mart have installed selfcheckout systems so shoppers can scan their own items and cash out themselves. How do customers like this service and how often do they use it? Listed below is the number of customers using the service for a sample of 15 days at the Wal-Mart on Highway 544 in Surfside, South Carolina. $$\begin{array}{|rrrrrrrrrr|}\hline 120 & 108 & 120 & 114 & 118 & 91 & 118 & 92 & 104 & 104 \\\112 & 97 & 118 & 108 & 117 & & & & & \\\\\hline\end{array}$$ Is it reasonable to conclude that the mean number of customers using the self- checkout system is more than 100 per day? Use the .05 significance level.
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