A genetic experiment involving peas yielded one sample of offspring consisting o

Question

A genetic experiment involving peas yielded one sample of offspring consisting of 440 green peas and 172 yellow peas. Use a 0.05 significance level to test the claim that under the same circumstances, 23% 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.

a. what is the test statistic? z=?? (round to two decimal places as needed)

b. what is the P-value?? P-value =?? (round to four decimal places as needed)

Explanation / Answer

Solution:-

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: P = 0.23

Alternative hypothesis: P 0.23

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 ]

= 0.017

a) z = (p - P) /

z = 3.00

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 -3.00 or greater than 3.00.

b) Thus, the P-value = 0.0026

Interpret results. Since the P-value (0.0026) is less than the significance level (0.05), we have to reject the null hypothesis.

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