A genetic experiment involving peas yielded one sample of offspring consisting o
ID: 3309598 • Letter: A
Question
A genetic experiment involving peas yielded one sample of offspring consisting of 425 green peas and 132 yellow peas. Use a 0.05 significance level to test the claim that under the same circumstances, 24% of offspring peas will be yellow. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, conclusion about the null hypothesis, and final conclusion that addresses the original claim. Use the P-value method and the normal distribution as an approximation to the binomial distribution.
What are the null and alternative hypothesis?
What is the test statistic?
What is the P-value?
What is the conclusion about the null hypothesis?
What is the final conclusion?
Explanation / Answer
Solution:-
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: P = 0.24
Alternative hypothesis: P 0.24
Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the sample proportion is too big or if it is too small.
Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method, shown in the next section, is a one-sample z-test.
Analyze sample data. Using sample data, we calculate the standard deviation () and compute the z-score test statistic (z).
= sqrt[ P * ( 1 - P ) / n ]
= 0.0181
z = (p - P) /
z = - 0.167
where P is the hypothesized value of population proportion in the null hypothesis, p is the sample proportion, and n is the sample size.
Since we have a two-tailed test, the P-value is the probability that the z-score is less than - 0.167 or greater than 0.167
Thus, the P-value = 0.865
Interpret results. Since the P-value (0.865) is greater than the significance level (0.05), we cannot reject the null hypothesis.
From the above test we can conclude that 24% of offspring peas will be yellow.
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