![]() H1: At least one month has a different frequency of homicides than the others. The results (do not appear) to support the expectation that the frequency for the first category is disproportionately high.ĭetermine the null (H0) and alternative (H1) hypotheses H0: Homicides occur with equal frequency in the different months. There (is not) sufficient evidence to warrant rejection of the claim that the four categories are equally likely. if you don't have StatCrunch, use and input The Test Statistic and the Degrees of freedom to find. Critical value is χ²= 6.251 The P-value is. "k" is the number of columns k-1= degrees of freedom = 3 Use the chi-square table to find that α = 0.10 and degrees of freedom = 3 gives. *O= Observed, *E= Expected Total (100)÷(number of columns) = Expected = 25 (make this chart ↓) _ Cents. The test is to determine whether the observed frequency counts agree with the claimed uniform distribution so that the frequencies for only two days are equally likely. The test is to determine whether the observed frequency counts agree with the claimed uniform distribution so that the frequencies for at least two days are equally likely. The test is to determine whether the observed frequency counts agree with the claimed uniform distribution so that the frequencies for the different days are equally likely. The test is to determine whether the observed frequency counts agree with the claimed chi-square distribution so that the frequencies for at most three days are equally likely. ![]() If we test the claim using the goodness-of-fit test, what is actually tested? Sun 529 Mon 19 Tues 9 Wed 21 Thurs 12 Fri 44 Sat 373 Choose the correct answer below. Consider the claim that the days of the week are selected with a uniform distribution so that all days have the same chance of being selected. The table below lists days of the week selected by a random sample of 1007 subjects who were asked to identify the day of the week that is best for quality family time.
0 Comments
Leave a Reply. |