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In a study of bacterial concentration in surface and subsurface water ("Pb and Bacteria in a Surface Microlayer" Journal of Marine Research \([1982]: 1200-\) 1206 ), the accompanying data were obtained. \(\begin{aligned}&\text { Concentration }\left(\times 10^{6} / \mathrm{mL}\right) \\\&\text { Surface } & 48.6 & 24.3 & 15.9 & 8.29 & 5.75 \\\&\text { Subsurface } & 5.46 & 6.89 & 3.38 & 3.72 & 3.12 \\\&\text { Surface } & 10.8 & 4.71 & 8.26 & 9.41 & \\\&\text { Subsurface } & 3.39 & 4.17 & 4.06 & 5.16 & \end{aligned}\) Summary quantities are $$ \begin{aligned} &\sum x=136.02 \quad \sum y=39.35 \\ &\sum x^{2}=3602.65 \quad \sum y^{2}=184.27 \quad \sum x y=673.65 \end{aligned} $$ Using a significance level of \(.05\), determine whether the data support the hypothesis of a linear relationship between surface and subsurface concentration.

Short Answer

Expert verified
The conclusion of whether the data supports the hypothesis of a linear relationship between surface and subsurface concentration can only be made after computing all the steps described above. The decision is based on the comparison of the calculated test statistic and critical value.

Step by step solution

01

Calculate the sample size

The sample size \(n\) can be calculated by counting the pairs of given data. In this case, there are 9 pairs of data, so \(n = 9\).
02

Compute the correlation coefficient

The correlation coefficient \(r\) can be calculated using the given summary quantities. The formula is \(r = \frac{n(\sum xy) - (\sum x)(\sum y)} {\sqrt{[n(\sum x^2) - (\sum x)^2][n(\sum y^2) - (\sum y)^2]}}\). After substituting the given quantities into the formula, compute the correlation coefficient.
03

Compute the test statistic

The test statistic for the correlation coefficient can be calculated using the formula \(t = r\sqrt{\frac{n-2}{1-r^2}}\). Substitute the values of \(r\) and \(n\) into the formula to calculate \(t\).
04

Determine the critical value

The critical value can be found from the t-distribution table using degrees of freedom \(df = n - 2\), and the given significance level of .05. In this case, the degrees of freedom will be \(9 - 2 = 7\).
05

Compare and conclude

Compare the absolute value of the calculated test statistic with the critical value. If the absolute test statistic is greater than the critical value, reject the null hypothesis. This indicates that the data supports the hypothesis of a linear relationship between surface and subsurface concentration. If not, do not reject the null hypothesis. This indicates that the data does not support the hypothesis of a linear relationship.

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Key Concepts

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

Linear Relationship Hypothesis
In statistical analysis, hypotheses are ideas or predictions that can be tested to determine their validity. In this exercise, we are focusing on determining if there is a linear relationship between two sets of data: the concentration levels in surface and subsurface water. A linear relationship means that changes in one variable are associated with proportional changes in the other variable. If the relationship is true, the data points will form a straight line when plotted on a graph.

For this analysis, the null hypothesis (\(H_0\)) is that there is no linear relationship between the two variables which suggests that any correlation observed is due to random chance. The alternative hypothesis (\(H_a\)) proposes that there is indeed a linear relationship. A critical part of hypothesis testing is deciding which of these to accept based on statistical evidence.
Correlation Coefficient Calculation
The correlation coefficient (\(r\)) is a statistical measure that helps in understanding the strength and direction of a linear relationship between two variables. It ranges from -1 to 1. An \(r\) of 1 indicates a perfect positive linear relationship, 0 indicates no linear relationship, and -1 indicates a perfect negative linear relationship.

Calculating the correlation coefficient involves using given summary quantities and a specific formula:\[ r = \frac{n(\sum xy) - (\sum x)(\sum y)} {\sqrt{[n(\sum x^2) - (\sum x)^2][n(\sum y^2) - (\sum y)^2]} } \]Here:
    • \(n\) is the number of data pairs.
    • \(\sum xy\), \(\sum x\), \(\sum y\) represent the sums of products and individual sums of the data variables.
    This formula computes \(r\) by taking into account both the covariance of the variables and the product of their standard deviations. The resulting \(r\) gives us insight into how closely the variables follow a linear trajectory.
Critical Value Determination
Critical values are a key aspect of hypothesis testing. They allow us to decide whether to accept or reject our hypothesis based on a predetermined level of significance. In this exercise, the significance level is set at 0.05, which tells us there is a 5% chance of incorrectly rejecting a true null hypothesis (Type I error).

To determine the critical value, we use the t-distribution because the sample size is small. The degrees of freedom (\(df\)) is calculated as \(n - 2\), where \(n\) is the sample size. For our data, \(df = 9 - 2 = 7\). We would then look up the critical value for \(df = 7\) at \(\alpha = 0.05\) in a t-table. This critical value expresses the threshold at which any observed correlation can still be attributed to random chance rather than a meaningful relationship.
Hypothesis Testing in Statistics
Hypothesis testing is a structured process for making statistical decisions. It guides us to determine if observed data deviates significantly from what was expected under a specific null hypothesis.

In correlation analysis, we calculate a test statistic (\(t\)) using the formula:\[ t = r\sqrt{\frac{n-2}{1-r^2}} \]This \(t\) value helps us assess how extreme our \(r\) is compared to what we might expect under the null hypothesis. By comparing this test statistic to our critical value from the t-table:
  • If our computed \(|t|\) exceeds the critical value, we have enough statistical evidence to reject the null hypothesis, supporting a linear relationship.
  • If not, we cannot reject the null hypothesis, suggesting that the data does not support a linear relationship.
This methodical approach helps eliminate bias and ensures that the conclusions we draw are supported by statistical evidence.

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Most popular questions from this chapter

Exercise \(5.48\) described a regression situation in which \(y=\) hardness of molded plastic and \(x=\) amount of time elapsed since termination of the molding process. Summary quantities included \(n=15\), SSResid \(=\) \(1235.470\), and SSTo \(=25,321.368\). a. Calculate a point estimate of \(\sigma\). On how many degrees of freedom is the estimate based? b. What percentage of observed variation in hardness can be explained by the simple linear regression model relationship between hardness and elapsed time?

An investigation of the relationship between traffic flow \(x\) (thousands of cars per \(24 \mathrm{hr}\) ) and lead content \(y\) of bark on trees near the highway (mg/g dry weight) yielded the accompanying data. A simple linear regression model was fit, and the resulting estimated regression line was \(\hat{y}=28.7+33.3 x .\) Both residuals and standardized residuals are also given. \(\begin{array}{lrrrrr}\text { iduals are also given. } & & & & \\ x & 8.3 & 8.3 & 12.1 & 12.1 & 17.0 \\ y & 227 & 312 & 362 & 521 & 640 \\ \text { Residual } & -78.1 & 6.9 & -69.6 & 89.4 & 45.3 \\ \text { St. resid. } & -0.99 & 0.09 & -0.81 & 1.04 & 0.51\end{array}\) \(\begin{array}{lrrrrr}x & 17.0 & 17.0 & 24.3 & 24.3 & 24.3 \\ y & 539 & 728 & 945 & 738 & 759 \\ \text { Residual } & -55.7 & 133.3 & 107.2 & -99.8 & -78.8 \\\ \text { St. resid. } & -0.63 & 1.51 & 1.35 & -1.25 & -0.99\end{array}\) a. Plot the \((x\), residual \()\) pairs. Does the resulting plot suggest that a simple linear regression model is an appropriate choice? Explain your reasoning. b. Construct a standardized residual plot. Does the plot differ significantly in general appearance from the plot in Part (a)?

The article "Effect of Temperature on the pH of Skim Milk" (Journal of Dairy Research [1988]: 277- 280) reported on a study involving \(x=\) temperature \(\left({ }^{\circ} \mathrm{C}\right)\) under specified experimental conditions and \(y=\) milk \(\mathrm{pH}\). The accompanying data (read from a graph) are a representative subset of that which appeared in the article: \(\begin{array}{rrrrrrrrr}x & 4 & 4 & 24 & 24 & 25 & 38 & 38 & 40 \\ y & 6.85 & 6.79 & 6.63 & 6.65 & 6.72 & 6.62 & 6.57 & 6.52\end{array}\) $$ \begin{array}{lrrrrrrrr} x & 45 & 50 & 55 & 56 & 60 & 67 & 70 & 78 \\ y & 6.50 & 6.48 & 6.42 & 6.41 & 6.38 & 6.34 & 6.32 & 6.34 \\ \sum x=678 & \sum y=104.54 & \sum x^{2}=36,056 & \\ \sum y^{2}=683.4470 & & \sum x y=4376.36 & & \end{array} $$ Do these data strongly suggest that there is a negative linear relationship between temperature and \(\mathrm{pH}\) ? State and test the relevant hypotheses using a significance level of \(.01\).

The shelf life of packaged food depends on many factors. Dry cereal is considered to be a moisture-sensitive product (no one likes soggy cereal!) with the shelf life determined primarily by moisture content. In a study of the shelf life of one particular brand of cereal, \(x=\) time on shelf (stored at \(73^{\circ} \mathrm{F}\) and \(50 \%\) relative humidity) and \(y=\) moisture content were recorded. The resulting data are from "Computer Simulation Speeds Shelf Life Assessments" (Package Engineering [1983]: 72-73). \(\begin{array}{rrrrrrrr}x & 0 & 3 & 6 & 8 & 10 & 13 & 16 \\ y & 2.8 & 3.0 & 3.1 & 3.2 & 3.4 & 3.4 & 3.5 \\ x & 20 & 24 & 27 & 30 & 34 & 37 & 41 \\ y & 3.1 & 3.8 & 4.0 & 4.1 & 4.3 & 4.4 & 4.9\end{array}\) a. Summary quantities are $$ \begin{array}{ll} \sum x=269 & \sum y=51 \quad \sum x y=1081.5 \\ \sum y^{2}=7745 & \sum x^{2}=190.78 \end{array} $$ Find the equation of the estimated regression line for predicting moisture content from time on the shelf. b. Does the simple linear regression model provide useful information for predicting moisture content from knowledge of shelf time? c. Find a \(95 \%\) interval for the moisture content of an individual box of cereal that has been on the shelf 30 days. d. According to the article, taste tests indicate that this brand of cereal is unacceptably soggy when the moisture content exceeds 4.1. Based on your interval in Part (c), do you think that a box of cereal that has been on the shelf 30 days will be acceptable? Explain.

Explain the difference between a confidence interval and a prediction interval. How can a prediction level of \(95 \%\) be interpreted?

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