/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 18 Mendelian Genetics When Mendel c... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Mendelian Genetics When Mendel conducted his famous genetics experiments with peas, one sample of offspring consisted of 428 green peas and 152 yellow peas. Use a \(0.01\) significance level to test Mendel's claim that under the same circumstances, \(25 \%\) of offspring peas will be yellow. What can we conclude about Mendel's claim?

Short Answer

Expert verified
We fail to reject Mendel's claim; there is not enough evidence to say the proportion of yellow peas is not 25%.

Step by step solution

01

State the Hypotheses

Define the null and alternative hypotheses. The null hypothesis (ull) is that the proportion of yellow peas is 0.25. The alternative hypothesis (ull) is that the proportion of yellow peas is not 0.25. ull: \( p = 0.25 \) ull: \( p eq 0.25 \)
02

Calculate the Test Statistic

Use the sample data to calculate the test statistic. The formula for the test statistic in hypothesis testing for a proportion is: \[ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0 (1 - p_0)}{n}}} \] where \( \hat{p} \) is the sample proportion, \( p_0 \) is the hypothesized proportion, and \( n \) is the sample size. First, calculate \( \hat{p} \) as the sample proportion of yellow peas: \( \hat{p} = \frac{152}{428 + 152} = \frac{152}{580} = 0.262 \). Then use \( p_0 = 0.25 \) and \( n = 580 \) to calculate the test statistic: \[ z = \frac{0.262 - 0.25}{\sqrt{\frac{0.25 (1 - 0.25)}{580}}} \] \[ z \approx 0.60 \]
03

Determine the Critical Value

Determine the critical value for a two-tailed test at the \(0.01\) significance level. Using a Z-table, the critical values for \(0.01\) significance level (two-tailed) are \( \pm 2.576 \).
04

Compare the Test Statistic to the Critical Value

Compare the calculated test statistic to the critical value to decide whether to reject the null hypothesis. Since \( |0.60| < 2.576 \), the test statistic does not fall in the critical region. Therefore, we do not reject the null hypothesis.
05

Draw a Conclusion

Based on the comparison in Step 4, we conclude that there is not enough evidence at the \(0.01\) significance level to reject Mendel's claim. Thus, we fail to reject the null hypothesis that \(25\%\) of the offspring peas will be yellow.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Mendelian Genetics
Mendelian genetics form the foundation of classical genetics. They are principles derived from the experiments of Gregor Mendel, who studied the inheritance of traits in pea plants. Mendel's work established the concepts of dominant and recessive alleles, segregation, and independent assortment.
For example, in the context of pea plants, green and yellow colors are determined by different alleles. Green is the dominant trait, while yellow is recessive.
Understanding Mendelian genetics is crucial for interpreting the results of genetic crosses. It helps predict the proportions of offspring with specific traits, like in the provided exercise.
Significance Level
The significance level, denoted by alpha (α), is a threshold used in hypothesis testing to determine whether to reject the null hypothesis. It represents the probability of making a Type I error, which occurs when the null hypothesis is wrongly rejected.
In the given problem, the significance level is set to 0.01, or 1%. This means that there is a 1% risk of concluding that the proportion of yellow peas is different from 0.25 when it is not.
  • This threshold helps control the likelihood of incorrect conclusions.
  • A lower significance level, like 0.01, leads to stricter criteria for rejecting the null hypothesis.
This ensures that the conclusions drawn are more robust and reliable.
Null Hypothesis
The null hypothesis (Hâ‚€) is a statement that there is no effect or no difference, and it serves as the starting point for hypothesis testing.
In our exercise, the null hypothesis is that the proportion of yellow peas (p) is 0.25: \( H_0: p = 0.25\).
By setting up the null hypothesis, we are essentially assuming that Mendel's claim is true unless the evidence strongly suggests otherwise.
  • This hypothesis is tested using a statistical method (like the z-test).
  • If the data significantly deviate from what is expected under the null hypothesis, we may reject it.
However, if the data support the null hypothesis, we do not reject it, implying there is insufficient evidence against Mendel's original claim.
Test Statistic
The test statistic is a value computed from sample data during a hypothesis test. It helps determine whether to reject the null hypothesis.
In the scenario given, the test statistic is calculated using the formula for a proportion z-test:
\[ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0 (1 - p_0)}{n}}} \]
where \( \hat{p} = \frac{152}{580} = 0.262 \) is the sample proportion of yellow peas, \( p_0 = 0.25 \) is the hypothesized proportion, and \( n = 580 \) is the sample size.
This calculation yielded \( z ≈ 0.60 \).
Comparing this to the critical value of ±2.576 for a 0.01 significance level, we see that 0.60 does not fall in the critical region. Thus, we do not reject the null hypothesis as the test statistic indicates that the deviation is not significant.
Understanding these core concepts ensures a solid grasp of hypothesis testing in genetics, making it easier to draw accurate conclusions from data.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Bank Lines The Jefferson Valley Bank once had a separate customer waiting line at each teller window, but it now has a single waiting line that feeds the teller windows as vacancies occur. The standard deviation of customer waiting times with the old multiple-line configuration was \(1.8\) min. Listed below is a simple random sample of waiting times (minutes) with the single waiting line. Use a \(0.05\) significance level to test the claim that with a single waiting line, the waiting times have a standard deviation less than \(1.8 \mathrm{~min}\). What improvement occurred when banks changed from multiple waiting lines to a single waiting line? $$ \begin{array}{llllllllll} 6.5 & 6.6 & 6.7 & 6.8 & 7.1 & 7.3 & 7.4 & 7.7 & 7.7 & 7.7 \end{array} $$

Bias in Jury Selection In the case of Casteneda v. Partida, it was found that during a period of 11 years in Hidalgo County, Texas, 870 people were selected for grand jury duty and \(39 \%\) of them were Americans of Mexican ancestry. Among the people eligible for grand jury duty, \(79.1 \%\) were Americans of Mexican ancestry. Use a \(0.01\) significance level to test the claim that the selection process is biased against Americans of Mexican ancestry. Does the jury selection system appear to be biased?

Online Data Claim: Most adults would erase all of their personal information online if they could. A GFI Software survey of 565 randomly selected adults showed that \(59 \%\) of them would erase all of their personal information online if they could.

Mickey D's In a study of the accuracy of fast food drive-through orders, McDonald's had 33 orders that were not accurate among 362 orders observed (based on data from \(Q S R\) magazine). Use a \(0.05\) significance level to test the claim that the rate of inaccurate orders is equal to \(10 \%\). Does the accuracy rate appear to be acceptable?

Medication Usage In a survey of 3005 adults aged 57 through 85 years, it was found that \(81.7 \%\) of them used at least one prescription medication (based on data from "Use of Prescription and Over-the-Counter Medications and Dietary Supplements Among Older Adults in the United States," by Qato et al., Journal of the American Medical Association, Vol. 300 , No. 24). Use a \(0.01\) significance level to test the claim that more than \(3 / 4\) of adults use at least one prescription medication. Does the rate of prescription use among adults appear to be high?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.