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The article "Linkage Studies of the Tomato" (Transactions of the Royal Canadian Institute [1931]:1-19) reported the accompanying data on phenotypes resulting from crossing tall cut-leaf tomatoes with dwarf potato-leaf tomatoes. There are four possible phenotypes: (1) tall cut-leaf, (2) tall potato-leaf, (3) dwarf cutleaf, and (4) dwarf potato-leaf. $$ \begin{array}{c|cccc} & {\text { Phenotype }} \\ & 1 & 2 & 3 & 4 \\ \hline \text { Frequency } & 926 & 288 & 293 & 104 \\ \hline \end{array} $$ Mendel's laws of inheritance imply that \(p_{1}=9 / 16, p_{2}=\) \(3 / 16, p_{3}=3 / 16\), and \(p_{4}=1 / 16\). Are the data from this experiment consistent with Mendel's laws? Use a .01 significance level.

Short Answer

Expert verified
To determine whether the data is consistent with Mendel's laws, a chi-square goodness-of-fit test needs to be conducted and the calculated chi-square test statistic should be compared with the critical value from the chi-square distribution table.

Step by step solution

01

Define Null and Alternative Hypotheses

The null hypothesis (H0) states that the data is consistent with Mendel's laws, i.e. the observed proportions match the expected proportions. The alternative hypothesis (H1) states that the data is not consistent with Mendel's laws, i.e. the observed proportions differ from the expected proportions.
02

Calculate Expected Frequencies

To calculate the expected frequencies, multiply the total number of observations by the expected proportions, \(p_1\), \(p_2\), \(p_3\) and \(p_4\). The total number of observations is the sum of the observed frequencies (926+288+293+104 = 1611). Therefore, \(E_1 = 1611 * (9/16)\), \(E_2 = 1611 * (3/16)\), \(E_3 = 1611 * (3/16)\), \(E_4 = 1611 * (1/16)\).
03

Calculate Test Statistic

The chi-square test statistic is calculated using the formula \(\chi^2 = \sum( (O-E)^2 / E )\), where O are the observed frequencies and E are the expected frequencies. Plug in the above values to get: \(\chi^2 = ((926-E_1)^2 / E_1) + ((288-E_2)^2 / E_2) + ((293-E_3)^2 / E_3) + ((104-E_4)^2 / E_4)\)
04

Determine Significance

Refer to a chi-square distribution table with 3 degrees of freedom (since there are 4 categories, the degrees of freedom is 4-1=3). The critical value for a 0.01 significance level is approximately 11.34. If the calculated chi-square statistic is greater than 11.34, reject the null hypothesis.

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Key Concepts

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

Mendel's Laws of Inheritance
Mendel's Laws form the foundation of genetic inheritance studies. Gregor Mendel, through his experiments with pea plants, discovered that traits are passed from parents to offspring in predictable ways. These laws are primarily:
  • **Law of Segregation**: Each organism carries two alleles for each trait, which separate during the formation of gametes, resulting in offspring inheriting one allele from each parent.
  • **Law of Independent Assortment**: Alleles of different traits are distributed to gametes independently of one another, which means the inheritance of one trait generally doesn’t affect the inheritance of another.
In the textbook problem, Mendel's laws predict specific proportions for each of the four phenotypes of tomatoes, based on the inheritance patterns of two pairs of alleles. The expected ratios provided are a direct application of Mendel's insights. For example, 9/16 for the first phenotype reflects complex interactions of dominant and recessive alleles.
Understanding the Null Hypothesis
The null hypothesis ( H_0 ) is a fundamental concept in statistics, serving as a default position that there is no effect or no difference. In the context of the chi-square test, it allows us to assess whether any observed deviations from expected frequencies are due to random chance.
In this exercise, the null hypothesis posits that the observed tomato phenotypes occur in proportions predicted by Mendel's laws, suggesting no real departure from these theoretical ratios.
The alternative hypothesis ( H_1 ) would claim the opposite, that the observed data do not fit the expected distribution, potentially indicating influences beyond simple Mendelian genetics.
Testing the null hypothesis helps determine whether any observed discrepancies are statistically significant or if they could simply be the result of random variation in the sample data.
Calculating Expected Frequencies
Expected frequencies are calculated under the assumption that the null hypothesis is true, providing a theoretical distribution to compare against observed data. For this problem, you calculate expected frequencies using the total number of observations and Mendel's expected proportions:
  • For phenotype 1, expect: \( E_1 = 1611 \times \frac{9}{16} \)
  • For phenotype 2, expect: \( E_2 = 1611 \times \frac{3}{16} \)
  • For phenotype 3, expect: \( E_3 = 1611 \times \frac{3}{16} \)
  • For phenotype 4, expect: \( E_4 = 1611 \times \frac{1}{16} \)
These computations give you the expected frequencies or counts for each phenotype if Mendel's law applies perfectly to the sample. Comparing these expected frequencies with observed ones helps determine if deviations are significant.
Significance Level in Hypothesis Testing
The significance level, often denoted as \( \alpha \), is a threshold set by the researcher to decide if an observed effect is statistically significant. A common significance level is 0.01, as used in this problem. It defines the probability of rejecting the null hypothesis when it is actually true.
Here's a simple breakdown:
  • A significance level of 0.01 means there's a 1% risk of concluding that Mendel's laws do not apply if they actually do.
  • The critical value, for example, 11.34 in this case, is derived from the chi-square distribution with appropriate degrees of freedom. It marks the boundary beyond which the null hypothesis would be rejected.
By comparing the calculated chi-square statistic to this critical value, the researcher determines whether the deviation from expected frequencies is significant enough to reject H_0. Understanding and correctly applying this concept is crucial for valid statistical inference.

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Most popular questions from this chapter

The paper referenced in the previous exercise also gave the accompanying data on the age at which smoking started for a sample of 1031 men who smoked low- tar cigarettes. $$ \begin{array}{cc} \text { Age } & \text { Frequency } \\ \hline<16 & 237 \\ 16-17 & 258 \\ 18-20 & 320 \\ \geq 21 & 216 \\ \hline \end{array} $$ a. Use a chi-square goodness-of-fit test to test the null hypothesis \(H_{0}: p_{1}=.25, p_{2}=.2, p_{3}=.3, p_{4}=.25\) where \(p_{1}=\) proportion of male low-tar cigarette smokers who started smoking before age 16, and \(p_{2}\), \(p_{3}\), and \(p_{4}\) are defined in a similar way for the other three age groups. b. The null hypothesis from Part (a) specifies that half of male smokers of low-tar cigarettes began smoking between the ages of 16 and \(20 .\) Explain why \(p_{2}=.2\) and \(p_{3}=.3\) is consistent with the ages between 16 and 20 being equally likely to be when smoking started.

A survey was conducted in the San Francisco Bay area in which each participating individual was classified according to the type of vehicle used most often and city of residence. A subset of the resulting data are given in the accompanying table (The Relationship of Vehicle Type Choice to Personality. Lifestyle, Attitudinal and Demographic Variables, Technical Report UCD-ITS-RRO2-06, DaimlerCrysler Corp., 2002). $$ \begin{array}{l|crc} & & \text { City } \\ \hline \begin{array}{l} \text { Vehide } \\ \text { Type } \end{array} & \text { Concord } & \begin{array}{c} \text { Pleasant } \\ \text { Hills } \end{array} & \begin{array}{c} \text { North San } \\ \text { Francisco } \end{array} \\ \hline \text { Small } & 68 & 83 & 221 \\ \text { Compact } & 63 & 68 & 106 \\ \text { Midsize } & 88 & 123 & 142 \\ \text { Large } & 24 & 18 & 11 \\ \hline \end{array} $$ Do the data provide convincing evidence of an association between city of residence and vehicle type? Use a significance level of \(.05 .\) You may assume that it is reasonable to regard the sample as a random sample of Bay area residents.

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Each boy in a sample of Mexican American males, age 10 to 18 , was classified according to smoking status and response to a question asking whether he likes to do risky things. The following table is based on data given in the article "The Association Between Smoking and Unhealthy Behaviors Among a National Sample of Mexican-American Adolescents" (Journal of School Health [1998]: 376-379): $$ \begin{array}{l|cr} & \text { Smoking Status } \\ \hline & \text { Smoker } & \text { Nonsmoker } \\ \hline \text { Likes Risky Things } & 45 & 46 \\ \text { Doesn't Like Risky Things } & 36 & 153 \\ \hline \end{array} $$ Assume that it is reasonable to regard the sample as a random sample of Mexican-American male adolescents. a. Is there sufficient evidence to conclude that there is an association between smoking status and desire to do risky things? Test the relevant hypotheses using \(\alpha=.05 .\) b. Based on your conclusion in Part (a), is it reasonable to conclude that smoking causes an increase in the desire to do risky things? Explain.

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