Question

When a scientist conducted a genetics experiments with​ peas, one sample of offspring consisted of 927 ​peas, with 722 of them having red flowers. If we​ assume, as the scientist​ did, that under these​ circumstances, there is a 3/4 probability that a pea will have a red​ flower, we would expect that 695.25 ​(or about 695​) of the peas would have red​ flowers, so the result of 722 peas with red flowers is more than expected.

a. If the​ scientist's assumed probability is​ correct, find the probability of getting 722 or more peas with red flowers.
b. Is 722 peas with red flowers significantly​ high?
c. What do these results suggest about the​ scientist's assumption that 3/4 of peas will have red​ flowers?

Answer 1

A.

P(x ≥ 722) = 1 - P(x < 722)

where P is the probability of a pea having a red flower, x is the number of peas with red flowers, and 1 - P(x < 722) represents the probability of getting 722 or more peas with red flowers.

Using the binomial probability formula, we get:

P(x ≥ 722) = 1 - P(x < 722)

= 1 - Σ (nCx)(P^x)(1-P)^(n-x), where n = 927, P = 3/4, and x ranges from 0 to 721.

P(x < 722) = 0.00019032

P(x ≥ 722) = 1 - P(x < 722) = 1 - 0.00019032 = 0.99980968 (rounded to 5 decimal places).

So the probability of getting 722 or more peas with red flowers is approximately 0.9998 or 99.98%.

B.

To determine if 722 peas with red flowers is significantly high, we need to perform a hypothesis test. The null hypothesis (H0) is that the true probability of a pea having a red flower is 3/4, while the alternative hypothesis (Ha) is that the true probability is greater than 3/4.

We can use a one-tailed z-test to test this hypothesis, where the test statistic is given by:

z = (x - μ) / (σ / sqrt(n)), where x = 722, μ = np = 927 * 3/4 = 695.25, σ = sqrt(np(1-p)) = sqrt(927 * 3/4 * 1/4) = 11.15 (rounded to 2 decimal places), and n = 927.

Substituting the values, we get:

z = (722 - 695.25) / (11.15 / sqrt(927))

z = 6.43 (rounded to 2 decimal places)

we find that the p-value (the probability of getting a test statistic as extreme or more extreme than 6.43) is very small, approximately 0. Therefore, we can reject the null hypothesis at the 5% level of significance and conclude that the true probability of a pea having a red flower is significantly higher than 3/4.

C.

Given that the observed proportion of peas with red flowers (722/927 = 0.779) is much greater than the predicted proportion (3/4 = 0.75), it appears that the scientist's assumption that 3/4 of peas will have red flowers is incorrect. The peas' blossom color may be influenced by various elements, including genetic variation and environmental influences.