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Chapter 10: Problem 31

Refer to Exercise 10.7 , in which we measured the dissolved oxygen content in river water to determine whether a stream had sufficient oxygen to support aquatic life. A pollution control inspector suspected that a river community was releasing amounts of semitreated sewage into a river. To check his theory, he drew five randomly selected specimens of river water at a location above the town, and another five below. The dissolved oxygen readings (in parts per million) are as follows: $$ \begin{array}{l|lllll} \text { Above Town } & 4.8 & 5.2 & 5.0 & 4.9 & 5.1 \\ \hline \text { Below Town } & 5.0 & 4.7 & 4.9 & 4.8 & 4.9 \end{array} $$ a. Do the data provide sufficient evidence to indicate that the mean oxygen content below the town is less than the mean oxygen content above? Test using \(\alpha=.05\). b. Suppose you prefer estimation as a method of inference. Estimate the difference in the mean dissolved oxygen contents for locations above and below the town. Use a \(95 \%\) confidence interval.

Short Answer

Expert verified
Answer: No, there is not enough evidence at the 0.05 significance level to conclude that the mean dissolved oxygen content below the town is less than that above the town. The test statistic is less than the critical value, causing us to fail to reject the null hypothesis. Additionally, the 95% confidence interval for the difference in mean dissolved oxygen content includes zero, indicating that we cannot say with 95% confidence that there is a significant difference between the dissolved oxygen contents above and below the town.

Step by step solution

01

State the hypotheses

We want to test if the mean oxygen content below the town is less than the mean oxygen content above. The null and alternative hypotheses are $$ H_0 : \mu_{Above} - \mu_{Below} = 0 \\ H_a : \mu_{Above} - \mu_{Below} > 0 $$
02

Calculate mean and variance for above and below town

We first calculate the means and variances for the above and below town samples: Mean above town: $$ \bar{x}_{Above}=\frac{4.8+5.2+5.0+4.9+5.1}{5}=5.0 $$ Mean below town: $$ \bar{x}_{Below}=\frac{5.0+4.7+4.9+4.8+4.9}{5}=4.86 $$ Variance above town: $$ s_{Above}^2 = \frac{(4.8-5.0)^2+(5.2-5.0)^2+\cdots+(5.1-5.0)^2}{5-1}=\frac{0.3}{4} $$ Variance below town: $$ s_{Below}^2 = \frac{(5.0-4.86)^2+(4.7-4.86)^2+\cdots+(4.9-4.86)^2}{5-1}=\frac{0.26}{4} $$
03

Calculate test statistic and critical value

We will calculate the t-statistic and critical value using the formulas $$ t = \frac{(\bar{x}_{Above} - \bar{x}_{Below} - D_0)}{\sqrt{\frac{s_{Above}^2}{n_{Above}}+\frac{s_{Below}^2}{ n_{Below}}}} \\ t_{critical} = t_{\alpha, df} $$ where \(D_0=0\), \(n_{Above}=n_{Below}=5\), and df is the degrees of freedom. The degrees of freedom can be approximated using the formula $$ df \approx \frac{(\frac{s_{Above}^2}{n_{Above}}+\frac{s_{Below}^2}{n_{Below}})^2}{\frac{(\frac{s_{Above}^2}{n_{Above}})^2}{n_{Above}-1}+\frac{(\frac{s_{Below}^2}{n_{Below}})^2}{n_{Below}-1}} $$ Plugging in the values, we get \(t = 1.38\), df \(\approx 7.6\), and \(t_{critical}=1.895\).
04

Compare test statistic to critical value and draw conclusion

Compare the calculated t-statistic to the critical value: $$ t = 1.38 < t_{critical} = 1.895 $$ Since the test statistic is less than the critical value, we fail to reject the null hypothesis. Thus, there is not enough evidence at the 0.05 significance level to conclude that the mean oxygen content below the town is less than the mean oxygen content above.
05

Estimate the difference using a confidence interval

To estimate the difference between the means using a 95% confidence interval, we will use the formula: $$ (\bar{x}_{Above} - \bar{x}_{Below}) \pm t_{0.025, df} \cdot \sqrt{\frac{s_{Above}^2}{n_{Above}}+\frac{s_{Below}^2}{n_{Below}}} $$ Using the calculated values, we find the confidence interval to be: $$ (5.0 - 4.86) \pm 2.365 \cdot \sqrt{\frac{0.3}{5}+\frac{ 0.26}{5}} \\ (5.0 - 4.86) \pm 2.365 \cdot \sqrt{0.112} \\ 0.14 \pm 0.787 $$
06

Interpret the confidence interval

The 95% confidence interval for the difference in mean dissolved oxygen content is \((0.14-0.787, 0.14+0.787)\) or \((-0.647, 0.927)\). This interval includes zero, meaning we cannot say with 95% confidence that there is a significant difference between the dissolved oxygen contents above and below the town.

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Key Concepts

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

Confidence Interval
When we talk about a confidence interval in statistics, we're referring to a range of values that likely includes the true difference between two population means. In simpler terms, it provides an estimate of the uncertainty around the mean difference we calculated from our samples. By using a 95% confidence interval, we're saying that if we were to draw many samples and calculate the interval each time, 95% of those intervals would contain the true mean difference. This offers a solid level of probability.

In our exercise, we estimated the difference in mean dissolved oxygen content between water samples taken above and below a town. The calculated confidence interval ranged from -0.647 to 0.927. Because this interval includes zero, it suggests that we cannot confidently say the means are different鈥攖here could be no real difference at all. Confidence intervals are highly useful as they provide a sense of how precise our estimate is and allow us to visually see whether there might be an overlap between the two population means we're examining.

In hypothesis testing, if the confidence interval for the difference in means encompasses zero, we often fail to reject the null hypothesis, implying no significant difference between these means.
T-Statistic
The t-statistic is a value used in hypothesis testing to help decide whether to reject the null hypothesis. It is a way of standardizing the difference between sample means, given the variation in your data.

To calculate it, you subtract the hypothesized value of the mean difference (often zero) from the observed mean difference, and then divide by the standard error. This gives you the t-statistic. In our case, the calculated t-statistic was 1.38.

We then compare the t-statistic to a critical value that corresponds to a predetermined probability level, usually found in a t-distribution table. If the t-statistic is greater than the critical value, we have enough evidence to reject the null hypothesis. However, in our exercise example with a t-statistic of 1.38 and a critical value of 1.895, since 1.38 is less, we failed to find sufficient evidence to reject the null hypothesis.

The t-statistic is essential for determining the likelihood that observed differences occurred by chance, helping researchers make informed decisions based on statistical analyses.
Null Hypothesis
In hypothesis testing, the null hypothesis is a statement we aim to test. It assumes that there is no effect or difference and is typically denoted as \(H_0\). For our exercise, the null hypothesis stated that the mean dissolved oxygen content below the town is equal to the mean content above, \(H_0: \mu_{Above} - \mu_{Below} = 0\).

The role of the null hypothesis is to serve as a baseline or a status quo for the test. We assume it to be true until we have convincing evidence to contradict it. The goal of the test is either to reject the null hypothesis or fail to reject it based on the data.

Ultimately, failure to reject \(H_0\) in our example exercise indicated that there was not enough evidence to suggest a difference in dissolved oxygen levels. It's important to understand the null hypothesis does not prove anything is true but rather checks if there is enough evidence against it.
Mean Difference
The mean difference is exactly what it sounds like: the difference between the averages or means of two groups of data. In statistical analyses, comparing mean differences helps us investigate if there is an impact or effect between two groups.

Calculating the mean difference is straightforward. We subtract the mean of one group from the mean of the other. In our exercise, the mean above town was 5.0, and the mean below town was 4.86, giving us a mean difference of 0.14.

This metric is particularly useful in hypothesis testing, where we look at whether the observed difference in means is likely to have occurred by chance. If the mean difference is statistically significant (beyond what we would expect from random sampling variability), it suggests a true difference between the groups. In our case, however, the mean difference's confidence interval included zero, so we couldn't say with certainty that a true difference existed between the two groups.

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Most popular questions from this chapter

To properly treat patients, drugs prescribed by physicians must not only have a mean potency value as specified on the drug's container, but also the variation in potency values must be small. Otherwise, pharmacists would be distributing drug prescriptions that could be harmfully potent or have a low potency and be ineffective. A drug manufacturer claims that his drug has a potency of \(5 \pm .1\) milligram per cubic centimeter \((\mathrm{mg} / \mathrm{cc})\). A random sample of four containers gave potency readings equal to 4.94,5.09 , 5.03, and \(4.90 \mathrm{mg} / \mathrm{cc} .\) a. Do the data present sufficient evidence to indicate that the mean potency differs from \(5 \mathrm{mg} / \mathrm{cc} ?\) b. Do the data present sufficient evidence to indicate that the variation in potency differs from the error limits specified by the manufacturer? (HINT: It is sometimes difficult to determine exactly what is meant by limits on potency as specified by a manufacturer. Since he implies that the potency values will fall into the interval \(5 \pm .1 \mathrm{mg} / \mathrm{cc}\) with very high probability - the implication is almost always-let us assume that the range \(.2 ;\) or 4.9 to \(5.1,\) represents \(6 \sigma\) as suggested by the Empirical Rule).

Refer to Exercise 10.94. Suppose that the word-association experiment is conducted using eight people as blocks and making a comparison of reaction times within each person; that is, each person is subjected to both stimuli in a random order. The reaction times (in seconds) for the experiment are as follows: $$ \begin{array}{ccc} \text { Person } & \text { Stimulus } 1 & \text { Stimulus 2 } \\ \hline 1 & 3 & 4 \\ 2 & 1 & 2 \\ 3 & 1 & 3 \\ 4 & 2 & 1 \\ 5 & 1 & 2 \\ 6 & 2 & 3 \\ 7 & 3 & 3 \\ 8 & 2 & 3 \end{array} $$ Do the data present sufficient evidence to indicate a difference in mean reaction times for the two stimuli? Test using \(\alpha=.05 .\)

A random sample of size \(n=7\) from a normal population produced these measurements: \(1.4,3.6,1.7,\) 2.0,3.3,2.8,2.9 a. Calculate the sample variance, \(s^{2}\). b. Construct a \(95 \%\) confidence interval for the population variance, \(\sigma^{2}\). c. Test \(H_{0}: \sigma^{2}=.8\) versus \(H_{\mathrm{a}}: \sigma^{2} \neq .8\) using \(\alpha=.05 .\) State your conclusions. d. What is the approximate \(p\) -value for the test in part c?

As part of a larger pilot study, students at a Riverside, California middle school, will compare the learning of algebra by students using iPads versus students using the traditional algebra textbook with the same author and publisher. \(^{20}\) To remove teacher-to-teacher variation, the same teacher will teach both classes, and the iPad and textbook material are both provided by the same author and publisher. Suppose that after 1 month, 10 students were selected from each class and their scores on an algebra advancement test recorded. The summarized data follow. $$ \begin{array}{lcc} & \text { iPad } & \text { Textbook } \\ \hline \text { Mean } & 86.4 & 79.7 \\ \text { Standard Deviation } & 8.95 & 10.7 \\ \text { Sample Size } & 10 & 10 \end{array} $$ a. Use the summary data to test for a significant difference in advancement scores for the two groups using \(\alpha=.05 .\) b. Find a \(95 \%\) confidence interval for the difference in mean scores for the two groups. c. In light of parts a and b, what can we say about the efficacy of using an iPad versus a traditional textbook in learning algebra at the middle school level?

To test the comparative brightness of two red dyes, nine samples of cloth were taken from a production line and each sample was divided into two pieces. One of the two pieces in each sample was randomly chosen and red dye 1 applied; red dye 2 was applied to the remaining piece. The following data represent a "brightness score" for each piece. Is there sufficient evidence to indicate a difference in mean brightness scores for the two dyes? Use \(\alpha=.05 .\) $$ \begin{array}{lrlrrrrrrr} \text { Sample } & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ \hline \text { Dye 1 } & 10 & 12 & 9 & 8 & 15 & 12 & 9 & 10 & 15 \\ \text { Dye 2 } & 8 & 11 & 10 & 6 & 12 & 13 & 9 & 8 & 13 \end{array} $$
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