A random group of oranges were selected from an orchard to analyze their ripeness. Based on the time of year, the orchard owner believes that 20% of the oranges are ready for picking now, 40% will be ready in three days, 30% will be ready in one week, and 10% will be ready in two weeks. Is there evidence to reject this hypothesis at = 0.05? A) There is evidence to reject the claim that the oranges are distributed as claimed because the test value 21.553 9.488 B) There is not evidence to reject the claim that the oranges are distributed as claimed because the test value 5.991 7.815 D) There is not evidence to reject the claim that the oranges are distributed as claimed because the test value 9.488

Question

A random group of oranges were selected from an orchard to analyze their ripeness. Based on the time of year, the orchard owner believes that 20% of the oranges are ready for picking now, 40% will be ready in three days, 30% will be ready in one week, and 10% will be ready in two weeks. Is there evidence to reject this hypothesis at = 0.05? A) There is evidence to reject the claim that the oranges are distributed as claimed because the test value 21.553 > 9.488 B) There is not evidence to reject the claim that the oranges are distributed as claimed because the test value 5.991 < 21.553 C) There is evidence to reject the claim that the oranges are distributed as claimed because the test value 21.553 > 7.815 D) There is not evidence to reject the claim that the oranges are distributed as claimed because the test value 9.488 < 21.553

Quizplus · Accepted Answer

We need to perform a chi-square goodness of fit test to determine if the observed frequencies (ripeness of the oranges) differ significantly from the expected frequencies (based on the claim of the orchard owner). The test value is calculated as 21.553 using the formula chi-square = Σ [(O - E)^2 / E], where O is the observed frequency and E is the expected frequency. We compare this test value to the critical value at a significance level of 0.05 and degrees of freedom equal to the number of categories minus one (df = 4-1 = 3). Using a chi-square distribution table or calculator, the critical value is 7.815. Since the test value is greater than the critical value (21.553 > 7.815), we reject the null hypothesis and conclude that there is evidence to reject the claim that the oranges are distributed as claimed. Therefore, choice C is the correct answer.

A Random Group of Oranges Were Selected from an Orchard