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In an experiment on chlorophyll inheritance in corn, for 1,103 seedlings of self-fertilized heterozygous green plants, 854 seedlings were green and 249 were yellow. Theory predicts that \(75 \%\) of the seedlings would be green. a. Specify a null hypothesis for testing the theory. b. Find the value of the chi-squared goodness-of-fit statistic and report its \(d f\). c. Report the P-value and interpret.

Short Answer

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a. Null hypothesis: 75% of seedlings are green. b. Chi-squared statistic is 13.474 with 1 degree of freedom. c. P-value < 0.001, suggesting significant deviation from the expected distribution.

Step by step solution

01

Formulate the Null Hypothesis

The null hypothesis (3H_{0}) is that the observed frequencies of green and yellow seedlings are consistent with the expected frequencies according to the theoretical prediction of 75%. This implies 3H_{0}: p = 0.75 for green seedlings, meaning that 3p, the probability for a seedling being green, is equal to 75%.
02

Calculate Expected Frequencies

To find the expected number of each type of seedling, calculate based on the total number of seedlings (1103), where 854 are expected to be green and 249 are expected to be yellow. The expected number of green seedlings is given by E( ext{Green}) = 0.75 imes 1103 = 827.25, and the expected number of yellow seedlings is E( ext{Yellow}) = 0.25 imes 1103 = 275.75.
03

Compute the Chi-Squared Statistic

The chi-squared statistic is calculated using the formula: X^2 = rac{(O_{ ext{Green}} - E_{ ext{Green}})^2}{E_{ ext{Green}}} + rac{(O_{ ext{Yellow}} - E_{ ext{Yellow}})^2}{E_{ ext{Yellow}}} Where O represents observed frequencies, and E represents expected frequencies. Substitute the numbers: O_{ ext{Green}} = 854, E_{ ext{Green}} = 827.25, O_{ ext{Yellow}} = 249, E_{ ext{Yellow}} = 275.75. Calculating gives: X^2 = rac{(854 - 827.25)^2}{827.25} + rac{(249 - 275.75)^2}{275.75} = 10.941 + 2.533 = 13.474.
04

Determine Degrees of Freedom

The degrees of freedom (df) is calculated as (df = k - 1), where k is the number of categories. Here, k = 2 (green and yellow), so df = 2 - 1 = 1.
05

Find the P-value and Interpret

Using the chi-squared statistic 13.474 and df = 1, determine the P-value from a chi-squared distribution table or calculator. The P-value is very small, less than 0.001. This means the observed deviation from the expected distribution is statistically significant at common significance levels (e.g., 0.05), suggesting that H_{0} is not supported.

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Key Concepts

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

Null Hypothesis
In any statistical test, the null hypothesis is a crucial starting point. It is a statement that what we expect to observe is actually true. In the context of our chi-squared test involving corn seedlings, the null hypothesis, denoted as \(H_{0}\), states that the observed distribution of green and yellow seedlings is consistent with theoretical expectations.
Specifically, the null hypothesis suggests that 75% of the seedlings are expected to be green. This implies \(H_{0}: p = 0.75\) for green seedlings. The purpose of setting up this hypothesis is to test whether the observed data (854 green seedlings) align with these expectations. Rejecting \(H_{0}\) would mean there is evidence that the observed and expected frequencies differ significantly.
Thus, we inspect the alignment between what we see and what we expect as per our model. If they align well, our null hypothesis stands. If not, it may be time to rethink our underlying assumptions or model.
Expected Frequencies
Expected frequencies are the predicted counts of each category based on a predefined model, such as a theoretical probability distribution. In our experiment with corn seedlings, expected frequencies are computed by multiplying the total number of seedlings by the expected proportion of each category.
For green seedlings, the expected frequency \(E_{\text{Green}}\) is found by multiplying 75% by the total 1,103 seedlings, which equals 827.25. Meanwhile, for yellow seedlings, the percentage is 25%, giving an expected count of 275.75. These calculations allow us to compare these theoretical values to the actual observed frequencies of 854 and 249, respectively.
The primary purpose of calculating expected frequencies is to have a baseline to analyze discrepancies between observed and expected data. If discrepancies are significant, it suggests that the assumption made by the theoretical model (like the null hypothesis) may not hold.
Degrees of Freedom
Degrees of freedom ( extit{df}) is a concept used to determine the number of values in a statistical calculation that are free to vary. It is an integral part of chi-squared tests and impacts the critical value needed to reject the null hypothesis.
In a chi-squared test, the formula for degrees of freedom is \(df = k - 1\), where \(k\) is the total number of categories. For our corn seedling example, we have two categories: green and yellow. Thus, \(df = 2 - 1 = 1\). Understanding degrees of freedom helps us assess how much information is available to estimate another parameter or fit a model.
The degrees of freedom calculated plays a role in determining the P-value. Different degrees of freedom correspond to different probability distributions, underscoring why this concept is fundamental in making statistical inferences.
P-value
The P-value is the probability of observing a test statistic as extreme as, or more extreme than, the statistic obtained, given that the null hypothesis is true. It helps in deciding whether to reject the null hypothesis.
For our corn seedling data, we calculated a chi-squared statistic of 13.474 with one degree of freedom. From chi-squared distribution tables or computational tools, we obtain a P-value of less than 0.001. This very low P-value suggests that the observed data differs significantly from what the null hypothesis predicts.
When the P-value is less than a chosen significance level, often 0.05, we reject the null hypothesis. In this situation, the P-value tells us that the likelihood of observing such a data set if our theoretical model is true is very small. Hence, we conclude there is strong evidence that our observed distribution of seedling colors does not fit the expected model.

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Most popular questions from this chapter

Gender gap in employment? In a town, 7600 out of 8500 graduates are males. Out of 1750 graduate employees, 1620 are males. a. For these data, \(X^{2}=23.24 .\) What is its \(d f\) value, and what is its approximate sampling distribution, if \(\mathrm{H}_{0}\) is true? b. For this test, the P-value \(<0.001\). Interpret in the context of these variables. c. What decision would you make with a 0.05 significance level? Can you accept \(\mathrm{H}_{0}\) and conclude that employment is independent of gender?

Herbs and the common cold A recent randomized experiment of a multiherbal formula (Immumax) containing echinacea, garlic, ginseng, zinc, and vitamin C was found to improve cold symptoms in adults over a placebo group. "At the end of the study, eight ( \(39 \%\) ) of the placebo recipients and \(18(60 \%)\) of the Immumax recipients reported that the study medication had helped improve their cold symptoms (chi-squared P-value \(=0.01\) )." (M. Yakoot et al., International Joumal of General Medicine, vol. 4,2011 , pp. \(45-51\) ). a. Identify the response variable and the explanatory variable and their categories for the \(2 \times 2\) contingency table that provided this particular analysis.

Degrees of freedom explained For testing independence in a contingency table of size \(r \times c,\) the degrees of freedom (df) for the chi-squared distribution equal \(d f=(r-1) \times(c-1) .\) They have the following interpretation: Given the row and column marginal totals in an \(r \times\) contingency table, the cell counts in a rectangular block of size \((r-1) \times(c-1)\) determine all the other cell counts. Consider the following table, which cross-classifies political views by whether the subject would ever vote for a female president, based on the 2010 GSS. For this \(3 \times 2\) table, suppose we know the counts in the upper left-hand \((3-1) \times(2-1)=2 \times 1\) block of the table, as shown. \begin{tabular}{lccc} \hline & \multicolumn{2}{c} { Vote for Female } & \\ & \multicolumn{2}{c} { President } & \\ \cline { 2 - 3 } Political Views & Yes & No & Total \\ \hline Extremely Liberal & 56 & & 58 \\ Moderate & 490 & & 509 \\ Extremely Conservative & & & 61 \\ \hline Total & 604 & 24 & 628 \\ \hline \end{tabular} a. Given the cell counts and the row and column totak, fill in the counts that must appear in the blank cells. b. Now, suppose instead of the preceding table, you are shown the following table, this time only revealing a \(2 \times 1\) block in the lower-right part. Find the counts in the remaining cells. \begin{tabular}{lccc} \hline & \multicolumn{2}{c} { Vote for Female } & \\ & \multicolumn{2}{c} { President } & \\ \cline { 2 - 3 } Political Views & Yes & No & Total \\ \hline Extremely Liberal & & & \\ Moderate & & 58 \\ Extremely Conservative & & 3 & 61 \\ \hline Total & & 19 & 509 \\ \hline \end{tabular} This example serves to show that once the marginal totals are fixed in a contingency table, a block of only cell counts is free to vary. Once \((r-1) \times(c-1)\) these are given (as in part a or \(\mathrm{b}\) ), the remaining cell counts follow automatically. The value for the degrees of freedom is exactly the number of cells in this block, \(d f=(r-1) \times(c-1)\) or

Study hours and grades The following table shows data on study hours per week and the effect on grades, with expected cell counts given underneath the observed counts for 200 college students in a study conducted by Washington's Public Interest Research Group (PIRG). \begin{tabular}{lcllc} \hline & \multicolumn{3}{c} { Effect on grades } & \\ \cline { 2 - 4 } Study hours per week & Positive & None & Negative & Total \\ \hline \(1-15\) & 26 & 50 & 14 & 90 \\ & 23.9 & 43.2 & 23.0 & \\ \(16-24\) & 16 & 27 & 17 & 60 \\ & 15.9 & 28.8 & 15.3 & \\ \(25-34\) & 11 & 19 & 20 & 50 \\ & 13.3 & 24.0 & 12.8 & \\ Total & 53 & 96 & 51 & 200 \\ \hline \end{tabular} 2002 ) (Source: USA Today, April 17 , a. Suppose the variables were independent. Explain what this means in this context. b. Explain what is meant by an expected cell count. Show how to get the expected cell count for the first cell, for which the observed count is 26 . c. Compare the expected cell frequencies to the observed counts. Based on this, what is the profile of subjects who tend to have (i) positive effect on grades than independence predicts and (ii) negative effect on grades than independence predicts.

Female participation in defense services? When people participating in recent surveys were asked if women should actively participate in defense services, about \(91 \%\) of females and \(91 \%\) of males answered yes and the rest answered no. a. For males and for females, report the conditional distributions on this response variable in a \(2 \times 2\) table, using outcome categories (yes, no). b. If results for the entire population are similar to these, does it seem possible that gender and opinion about having active participation of women in defense services are independent? Explain.

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