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Chapter 11: Problem 11

The Fish and Game Department stocked Lake Lulu with fish in the following proportions: \(30 \%\) catfish, \(15 \%\) bass, \(40 \%\) bluegill, and \(15 \%\) pike. Five years later it sampled the lake to see if the distribution of fish had changed. It found that the 500 fish in the sample were distributed as follows. \(\begin{array}{cccc}\text { Catfish } & \text { Bass } & \text { Bluegill } & \text { Pike } \\ 120 & 85 & 220 & 75\end{array}\) In the 5 -year interval, did the distribution of fish change at the \(0.05\) level?

Short Answer

Expert verified
The fish distribution has changed significantly in five years.

Step by step solution

01

Define the Null and Alternative Hypotheses

The null hypothesis ( H_0) states that the distribution of fish proportions in the lake has not changed from the initial distribution. The alternative hypothesis ( H_1) suggests that the distribution has changed after five years.
02

Organize the Observed Frequencies

The observed frequencies of the fish sample taken from the lake are: Catfish = 120, Bass = 85, Bluegill = 220, Pike = 75. These correspond to a total of 500 fish.
03

Calculate Expected Frequencies

Determine the expected frequency for each fish species if no change occurred. Expected frequencies are calculated based on the initial proportions:- Catfish: \(0.30 \times 500 = 150\)- Bass: \(0.15 \times 500 = 75\)- Bluegill: \(0.40 \times 500 = 200\)- Pike: \(0.15 \times 500 = 75\).
04

Perform Chi-Square Test

Calculate the chi-square statistic using the formula:\[\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i}\]Where \(O_i\) is the observed frequency and \(E_i\) is the expected frequency for each species:- Catfish: \(\frac{(120 - 150)^2}{150} = 6\)- Bass: \(\frac{(85 - 75)^2}{75} = 1.33\)- Bluegill: \(\frac{(220 - 200)^2}{200} = 2\)- Pike: \(\frac{(75 - 75)^2}{75} = 0\).The total \(\chi^2\) statistic is \(6 + 1.33 + 2 + 0 = 9.33\).
05

Determine the Critical Value and Compare

To determine if the distribution of fish has changed, compare the calculated \(\chi^2\) value against the critical value from the chi-square distribution table with 3 degrees of freedom (since there are 4 categories, degrees of freedom = 4 - 1) at a significance level of \(\alpha = 0.05\). The critical value is 7.815.
06

Conclusion

Since the calculated chi-square value of 9.33 exceeds the critical value of 7.815, we reject the null hypothesis. There is significant evidence to say that the distribution of fish has changed over the 5-year timeframe.

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Key Concepts

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

Null Hypothesis
In a statistical hypothesis test, the null hypothesis serves as a starting point. It is essentially a default position that suggests there is no effect or no difference. In our context, we are asserting that the fish distribution in Lake Lulu has remained unchanged after five years.
The null hypothesis is denoted by the symbol \( H_0 \). Declaring this hypothesis involves stating that the original proportions鈥�30% catfish, 15% bass, 40% bluegill, and 15% pike鈥攁re still valid.
  • You rely on the null hypothesis to establish a baseline.
  • It is crucial as it provides a basis for comparison to see if observed outcomes differ significantly.
  • A hypothesis test often revolves around determining whether there is enough evidence to reject the null hypothesis.
Alternative Hypothesis
Contrasting the null hypothesis, the alternative hypothesis posits that there is indeed an effect or a change. For the fish sample from Lake Lulu, the alternative hypothesis suggests a shift in the fish proportions from what was initially stocked five years prior.
This hypothesis is symbolized as \( H_1 \) and often provides the direction of change or difference we suspect:
  • In this scenario, \( H_1 \) claims that the fish populations are different now.
  • We typically aim to find evidence in data to support the alternative hypothesis rather than the null.
  • The hypothesis tells us what we are trying to prove: that the fish distribution has changed.
Understanding the difference between the null and the alternative hypothesis is key to correctly performing statistical tests.
Expected Frequency
Expected frequencies are central to performing a chi-square test, as they represent the counts we would expect if the null hypothesis were true. In our fish distribution exercise, these frequencies are derived from the original stocking proportions.
The basic idea is this:
  • For each category (fish type), multiply the total number of sampled fish by the relevant stocking proportion.
  • Example calculation: For catfish, the expected frequency is \(0.30 \times 500 = 150\).
  • Repeat these calculations for bass, bluegill, and pike.
Having expected frequencies on hand is crucial as it helps to determine the degree of deviation exhibited by the observed data, allowing for meaningful comparisons in a chi-square test.
Degrees of Freedom
In statistical tests, degrees of freedom (df) help determine the reliability of test results and influence the critical value from statistical tables. For the chi-square test performed here, degrees of freedom are determined by the formula: \(df = ext{number of categories} - 1\).
Applying this to our example:
  • We have 4 categories (the different types of fish).
  • Thus, the degrees of freedom are \(4 - 1 = 3\).
  • This influences which chi-square distribution reference value we use to interpret the test statistic.
Degrees of freedom are an important concept because they adjust for the number of categories involved, hence providing a standardized value for comparing against chi-square test results. A correct understanding of degrees of freedom ensures that our conclusions about the test are accurate.

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Most popular questions from this chapter

For the study regarding mean cadence (see Problem 1), two-way ANOVA was used. Recall that the two factors were walking device (none, standard walker, rolling walker) and dual task (being required to respond vocally to a signal or no dual task required). Results of two-way ANOVA showed that there was no evidence of interaction between the factors. However, according to the article, "the ANOVA conducted on the cadence data reyealed a main effect of walking device." When the hypothesis regarding no difference in mean cadence according to which, if any, walking device was used, the sample \(F\) was \(30.94\), with d.f. \(\mathrm{N}=2\) and \(d . f \cdot \mathrm{D}=18\). Further, the \(P\) -value for the result was reported to be less than \(0.01\). From this information, what is the conclusion regarding any difference in mean cadence according to the factor "walking device used"?

The following problem is based on information taken from Academe, Bulletin of the American Association of University Professors. Let \(x\) represent the average annual salary of college and university professors (in thousands of dollars) in the United States. For all colleges and universities in the United States, the population variance of \(x\) is approximately \(\sigma^{2}=47.1 .\) However, a random sample of 15 colleges and universities in Kansas showed that \(x\) has a sample variance \(s^{2}=83.2 .\) Use a \(5 \%\) level of significance to test the claim that the variance for colleges and universities in Kansas is greater than 47.1. Find a \(95 \%\) confidence interval for the population variance.

For chi-square distributions, as the number of degrees of freedom increases, does any skewness increase or decrease? Do chi-square distributions become more symmetric (and normal) as the number of degrees of freedom becomes larger and larger?

A set of solar batteries is used in a research satellite. The satellite can run on only one battery, but it runs best if more than one battery is used. The variance \(\sigma^{2}\) of lifetimes of these batteries affects the useful lifetime of the satellite before it goes dead. If the variance is too small, all the batteries will tend to die at once. Why? If the variance is too large, the batteries are simply not dependable. Why? Engineers have determined that a variance of \(\sigma^{2}=23\) months (squared) is most desirable for these batteries. A random sample of 22 batteries gave a sample variance of \(14.3\) months (squared). (i) Using a \(0.05\) level of significance, test the claim that \(\sigma^{2}=23\) against the claim that \(\sigma^{2}\) is different from 23 . (ii) Find a \(90 \%\) confidence interval for the population variance \(\sigma^{2}\). (iii) Find a \(90 \%\) confidence interval for the population standard deviation \(\sigma .\)

A new fuel injection system has been engineered for pickup trucks. The new system and the old system both produce about the same average miles per gallon. However, engineers question which system (old or new) will give better consistency in fuel consumption (miles per gallon) under a variety of driving conditions. A random sample of 31 trucks were fitted with the new fuel injection system and driven under different conditions. For these trucks, the sample variance of gasoline consumption was 58.4. Another random sample of 25 trucks were fitted with the old fuel injection system and driven under a variety of different conditions. For these trucks, the sample variance of gasoline consumption was \(31.6 .\) Test the claim that there is a difference in population variance of gasoline consumption for the two injection systems. Use a \(5 \%\) level of significance. How could your test conclusion relate to the question regarding the consistency of fuel consumption for the two fuel injection systems?
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