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Test the given claim. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, or critical value(s), then state the conclusion about the null hypothesis, as well as the final conclusion that addresses the original claim. Use the P-value method unless your instructor specifies otherwise. Use the normal distribution as an approximation to the binomial distribution, as described in Part 1 of this section. 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 do not reject the null hypothesis. Mendel's claim that 25% of offspring peas will be yellow is not disproven.

Step by step solution

01

- State the Hypotheses

First, identify and state the null hypothesis (H_0) and the alternative hypothesis (H_a). \(H_0\): The proportion of yellow peas is 25%, or \(p = 0.25\). \(H_a\): The proportion of yellow peas is not 25%, or \(p eq 0.25\).
02

- Determine the Test Statistic

Use the formula for the test statistic for a proportion: \[ z = \frac{\hat{p} - p}{\sqrt{\frac{p(1-p)}{n}}} \] Where: \( \hat{p} \) is the sample proportion, \( p \) is the hypothesized population proportion, and \( n \) is the sample size.Calculate the sample proportion: \[ \hat{p} = \frac{152}{428 + 152} = \frac{152}{580} \approx 0.2621 \] Then, substitute the values into the test statistic formula: \[ z = \frac{0.2621 - 0.25}{\sqrt{\frac{0.25 \times 0.75}{580}}} \approx \frac{0.0121}{0.0180} \approx 0.6722 \]
03

- Find the P-value

Use the standard normal distribution to find the P-value. The P-value is the probability that the test statistic is at least as extreme as the observed value in either direction (this is a two-tailed test). Using a standard normal distribution table (or a calculator), find the P-value for \( z = 0.6722 \):\[ P(Z > 0.6722) = 1 - 0.7499 = 0.2501 \].Since this is a two-tailed test: \[ P = 2 \times 0.2501 = 0.5002 \]
04

- Compare the P-value to the Significance Level

Compare the P-value to the significance level \( \alpha = 0.01 \). Since \( P = 0.5002 \) is greater than \( \alpha = 0.01 \), we fail to reject the null hypothesis.
05

- State the Conclusion

Since we fail to reject the null hypothesis, we do not have sufficient evidence to conclude that the proportion of yellow peas is different from 25%. Therefore, we do not reject Mendel's claim that 25% of offspring peas will be yellow.

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Key Concepts

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

Null Hypothesis
The null hypothesis, denoted as \(H_0\), is a fundamental concept in hypothesis testing. It represents a statement of no effect or no difference, and it is what we assume to be true unless we have evidence to the contrary.
If we look at the exercise, the null hypothesis is: \[ H_0: p = 0.25 \] This means that according to Mendel's original claim, 25% of the offspring peas are yellow.

In simpler terms, the null hypothesis is our starting point. We assume it is true and then use the data to determine if there is enough evidence to reject it. If we fail to reject the null hypothesis, it means we do not have enough proof to say there is a significant difference from what we expected.
Alternative Hypothesis
The alternative hypothesis, denoted as \(H_a\), is what you want to prove. It is a statement that indicates there is an effect, a difference, or some relationship present.
In the exercise example, the alternative hypothesis is: \[ H_a: p eq 0.25 \] This means that the proportion of yellow peas is not 25%.

The alternative hypothesis provides the contrast to the null hypothesis. While the null hypothesis reflects no change or no effect, the alternative hypothesis signifies that something is different from our assumption. When we conduct the test, we collect evidence to see if we can support the alternative hypothesis.
It's important to note that the alternative hypothesis can be one-tailed (indicating a specific direction, like greater than or less than) or two-tailed (indicating a difference in either direction). In this case, it is a two-tailed test because we are looking for any difference, regardless of direction.
P-value
The P-value allows us to measure the strength of the evidence against the null hypothesis. It is the probability of obtaining a test statistic at least as extreme as the one observed, assuming that the null hypothesis is true.
In the exercise, we calculate the P-value using the test statistic (z-value): \[ z = 0.6722 \]. We find that \[P(Z > 0.6722) = 1 - 0.7499 = 0.2501 \]. Since this is a two-tailed test, we double this probability: \[ P = 2 \times 0.2501 = 0.5002 \]

The P-value is then compared to the significance level \(\alpha\), which is often set at 0.05 or 0.01. In our example, \(\alpha = 0.01\). Since the P-value \[ P = 0.5002 \] is greater than \(\alpha \), we fail to reject the null hypothesis.
  • A P-value less than \(\alpha\) indicates that the observed data is unlikely under the null hypothesis, hence we reject the null hypothesis.
  • A P-value greater than or equal to \(\alpha\) suggests that the data is consistent with the null hypothesis, so we fail to reject it.
In summary, a P-value quantifies the evidence against the null hypothesis. If it is very low, we have strong evidence to reject the null hypothesis in favor of the alternative hypothesis.

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Most popular questions from this chapter

Identify the indicated values or interpret the given display. Use the normal distribution as an approximation to the binomial distribution, as described in Part I of this section. Use a 0.05 significance level and answer the following: a. Is the test two-tailed, left-tailed, or right-tailed? b. What is the test statistic? c. What is the \(P\) -value? d. What is the null hypothesis, and what do you conclude about it? e. What is the final conclusion? The drug Lipitor (atorvastatin) is used to treat high cholesterol. In a clinical trial of Lipitor, 47 of 863 treated subjects experienced headaches (based on data from Pfizer). The accompanying TI-83/84 Plus calculator display shows results from a test of the claim that fewer than \(10 \%\) of treated subjects experience headaches.

Test the given claim. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, or critical value(s), then state the conclusion about the null hypothesis, as well as the final conclusion that addresses the original claim. Use the P-value method unless your instructor specifies otherwise. Use the normal distribution as an approximation to the binomial distribution, as described in Part 1 of this section. In a study of store checkout-scanners, 1234 items were checked for pricing accuracy; 20 checked items were found to be overcharges, and 1214 checked items were not overcharges (based on data from "UPC Scanner Pricing Systems: Are They Accurate?" by Goodstein, Journal of Marketing, Vol. 58). Use a 0.05 significance level to test the claim that with scanners, \(1 \%\) of sales are overcharges. (Before scanners were used, the overcharge rate was estimated to be about \(1 \% .\) Based on these results, do scanners appear to help consumers avoid overcharges?

Test the given claim. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, or critical value(s), then state the conclusion about the null hypothesis, as well as the final conclusion that addresses the original claim. Assume that a simple random sample is selected from a normally distributed population. Coffee Vending Machines The Brazil vending machine dispenses coffee, and a random sample of 27 filled cups have contents with a mean of 7.14 oz and a standard deviation of 0.17 oz. Use a 0.05 significance level to test the claim that the machine dispenses amounts with a standard deviation greater than the standard deviation of 0.15 oz specified in the machine design.

Assume that a simple random sample has been selected and test the given claim. Unless specified by your instructor, use either the P-value method or the critical value method for testing hypotheses. Identify the null and alternative hypotheses, test statistic, P-value (or range of P-values), or critical value(s), and state the final conclusion that addresses the original claim. Data Set 26 "Cola Weights and Volumes" in Appendix B includes volumes (ounces) of a sample of cans of regular Coke. The summary statistics are \(n=36, \bar{x}=12.19\) oz, \(s=0.11\) oz. Use a 0.05 significance level to test the claim that cans of Coke have a mean volume of 12.00 ounces. Does it appear that consumers are being cheated?

Use a significance level of \(\alpha=0.05\) and use the given information for the following: a. State a conclusion about the null hypothesis. (Reject \(H_{0}\) or fail to reject \(H_{0 .}\) ) b. Without using technical terms or symbols, state a final conclusion that addresses the original claim. Original claim: More than \(58 \%\) of adults would erase all of their personal information online if they could. The hypothesis test results in a \(P\) -value of 0.3257.

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