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The coloration of male guppies may affect the mating preference of the female guppy. To test this hypothesis, scientists first identified two types of guppies, Yarra and Paria, that display different colorations ("Evolutionary Mismatch of Mating Preferences and Male Colour Patterns in Guppies," Animal Behaviour \([1997]: 343-51\) ). The relative area of orange was calculated for fish of each type. A random sample of 30 Yarra guppies resulted in a mean relative area of \(.106\) and a standard deviation of .055. A random sample of 30 Paria guppies resulted in a mean relative area of \(.178\) and a standard deviation \(.058\). Is there evidence of a difference in coloration? Test the relevant hypotheses to determine whether the mean area of orange is different for the two types of guppies.

Short Answer

Expert verified
The solution depends on the value of the calculated test statistic. If it is larger than the critical value or smaller than the negative of the critical value, there is evidence of a significant difference in coloration between the two types of guppies, and thus we would reject the null hypothesis. Conversely, if the test statistic is within the critical values, we would fail to reject the null hypothesis, indicating no significant difference in coloration.

Step by step solution

01

Set up the hypotheses

The null hypothesis, denoted \( H_0 \), states that the means are equal. Therefore, \( H_0: \mu_1 = \mu_2 \).\nThe alternative hypothesis, denoted \( H_1 \), states that the means are not equal. Therefore, \( H_1: \mu_1 \neq \mu_2 \).
02

Calculate the test statistic

The formula for the test statistic in a two-sample t-test is \[ t = \frac{\bar{x_1} - \bar{x_2}}{s_p \sqrt{1/n_1 + 1/n_2}} \] where \( \bar{x} \) is the sample mean, \( s_p \) is the pooled standard deviation, \( n \) is the sample size and subscripts 1 and 2 distinguish the two samples.\nGiven data:\nSample 1 (Yarra guppies): \( \bar{x_1} = 0.106 \), \( s_1 = 0.055 \), and \( n_1 = 30 \)\nSample 2 (Paria guppies): \( \bar{x_2} = 0.178 \), \( s_2 = 0.058 \), and \( n_2 = 30 \)\nThe pooled standard deviation, \( s_p \), is calculated as \[ s_p = \sqrt{\frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1+n_2-2}} \].\nSubstitute these values into the formula to get the test statistic.
03

Identify the critical value

The critical value for this test can be found from the t distribution table. As it's a two-tailed test and we have 30+30-2=58 degrees of freedom and assuming a level of significance of 0.05, the critical value is approximately 2.00.
04

Compare test statistic with critical value

If the test statistic is larger than the critical value or smaller than the negative of the critical value, we reject the null hypothesis.
05

Conclusion

Based on the comparison between the test statistic and the critical value, we determine whether or not to reject the null hypothesis. If we reject the null hypothesis, there is significant evidence to conclude that there is a difference in the mean color area between the two types of guppies.

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Key Concepts

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

Two-sample t-test
A two-sample t-test is a statistical technique used to determine if there is a significant difference between the means of two independent groups. It helps in understanding whether the observed differences in sample means are enough to reflect a true difference in the population.
This test is particularly useful when comparing two distinct sets of data, like the relative area of orange color in two types of guppies.
  • First, ensure you have two independent groups; in this case, Yarra and Paria guppies.
  • These groups should be sampled randomly, and the samples should have approximately equal size.
  • The t-test checks whether the means of these groups are statistically different, using the difference between sample means, pooled standard deviation, and sample sizes.
Two-sample t-tests are widely used in research to check hypotheses where distinctions in characteristics between two groups are under investigation.
Null and Alternative Hypotheses
In hypothesis testing, setting up null and alternative hypotheses is a critical step. They form the foundation of a statistical test.
The null hypothesis (\( H_0 \)) usually posits that there is no effect or difference. For our guppy example: \( H_0: \mu_1 = \mu_2 \). This means the means of the orange areas for Yarra and Paria guppies are equal.
  • The null hypothesis serves as the default or starting assumption.
  • It's what we attempt to find evidence against.
Conversely, the alternative hypothesis (\( H_1 \)) is what you might conclude if there is enough evidence to reject the null hypothesis. In our example: \( H_1: \mu_1 eq \mu_2 \).
  • This indicates that there is a significant difference in the mean orange area between the two guppy types.
  • The alternative hypothesis represents the effect or difference you're testing for.
Understanding these hypotheses is crucial, as they guide your entire statistical test.
Pooled Standard Deviation
The pooled standard deviation is an essential component of a two-sample t-test. It provides a weighted average of the standard deviations from two samples, helping you estimate the variability within the combined samples.
Calculating the pooled standard deviation allows you to:
  • Account for differences in sample sizes between groups.
  • Provide a more accurate measure of variability when the population variance is assumed equal for both groups.
The formula is given by: \[s_p = \sqrt{\frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1+n_2-2}}\]Here, \(n_1\) and \(n_2\) are the sample sizes, and \(s_1\) and \(s_2\) are the standard deviations of each sample.
Using the pooled standard deviation is crucial as it standardizes the test statistic calculation, allowing meaningful comparison between groups.
Critical Value
The critical value is a threshold that helps determine the significance of your test results in hypothesis testing. It acts as a benchmark against which we compare our test statistic.
To find the critical value, one must:
  • Choose a significance level (commonly 0.05).
  • Determine degrees of freedom, typically the total number of samples minus two in a two-sample t-test.
  • Use a t-distribution table to find the critical value for the selected significance level and degrees of freedom.
In our guppy test example, with 58 degrees of freedom (30 Yarra + 30 Paria - 2), and assuming a 0.05 significance level, the critical value would be approximately 2.00.
If the calculated test statistic exceeds this critical value, in either direction in a two-tailed test, the null hypothesis is rejected, suggesting a significant difference in group means.

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Most popular questions from this chapter

O British health officials have expressed concern about problems associated with vitamin D deficiency among certain immigrants. Doctors have conjectured that such a deficiency could be related to the amount of fiber in a person's diet. An experiment was designed to compare the vitamin D plasma half-life for two groups of healthy individuals. One group was placed on a normal diet, whereas the second group was placed on a high-fiber diet. The accompanying table gives the resulting data (from "Reduced Plasma Half-Lives of Radio- Labeled \(25(\mathrm{OH}) \mathrm{D} 3\) in Subjects Receiving a High-Fibre Diet," British Journal of Nutrition \([1993]: 213-216\) ). \(\begin{array}{lllllll}\text { Normal diet } & 19.1 & 24.0 & 28.6 & 29.7 & 30.0 & 34.8\end{array}\) \(\begin{array}{llllllll}\text { High-fiber diet } & 12.0 & 13.0 & 13.6 & 20.5 & 22.7 & 23.7 & 24.8\end{array}\) Use the following MINITAB output to determine whether the data indicate that the mean half-life is higher for those on a normal diet than those on a high- fiber diet. Assume that treatments were assigned at random and the two plasma half-life distributions are normal. Test the appropriate hypotheses using \(\alpha=.01\). Two-sample T for normal ws high \(\begin{array}{lllrr} & \text { N } & \text { Mean } & \text { StDev } & \text { SE Mean } \\ \text { Normal } & 6 & 27.70 & 5.44 & 2.2 \\ \text { High } & 7 & 18.61 & 5.55 & 2.1 \\ 95 \% & \text { C.I. for mu normal }-m u \text { high: }(2.3,15.9) & \end{array}\) T-Test mu normal = mu high(vs >): \(\mathrm{T}=2.97 \mathrm{P}=0.0070 \mathrm{DF}=10\)

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