Question

In one of Mendel's famous hybridization experiments, 8023 offspring peas were obtained, and 24.94% of them had green flowers. The others had white flowers. Consider a hypothesis test that uses a 0.05 significance level to test the claim that green-flowered peas occur at a rate of 25%. Round to 3 decimal places.

a) What is the test statistic?
b) What are the critical values?
c) What is the p-value?

Answer 1

To test the claim that green-flowered peas occur at a rate of 25%, we can use a hypothesis test. The test statistic is calculated using the standard error and z-score. The critical values and p-value can be found using a z-table.

Step-by-step explanation:

To test the claim that green-flowered peas occur at a rate of 25%, we can use a hypothesis test. Let's define the null hypothesis (H0) as the claim being tested, which states that green-flowered peas occur at a rate of 25%. The alternative hypothesis (Ha) would be that green-flowered peas do not occur at a rate of 25%.

a) To find the test statistic, we need to calculate the standard error (SE) and the z-score. The formula for SE is sqrt(p*(1-p)/n), where p is the proportion of green-flowered peas and n is the sample size. SE = sqrt(0.2494*(1-0.2494)/8023) = 0.0034. The z-score is calculated as (sample proportion - population proportion) / SE. z = (0.2494 - 0.25) / 0.0034 = -0.1481.

b) The critical values for a significance level of 0.05 can be found using a z-table. Since it's a two-tailed test, we divide the significance level by 2 to get 0.025 for each tail. Using the z-table, the critical value is approximately -1.96 and 1.96.

c) To find the p-value associated with the test statistic, we can use the z-table. The p-value is the probability of observing a test statistic as extreme as the one calculated if the null hypothesis is true. Since the test statistic is -0.1481, the p-value is the area to the left of -0.1481 (since it's a two-tailed test). Using the z-table, the p-value is approximately 0.4416.