The degrees-of-freedom for a chi-square test for a 3 by 6 cross-tab is: 18 12 10
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- The degrees of freedom for a 4 x 3 cross-tabulation table for the chi-square test of independence equal 4 3 6 5if the results of a study using the chi-square tests are summarized as x2 ( 1,N=360) = 10.00,p<.01,then you would know that it was a what study?If the chi-square test is x2 (4, N=71) = 10, if the alpha is .05, then we ___.
- In the chi-square tests, the test statistics formula uses O and E notations. Explain the circumstancesunder which the chi-square test statistic values will be smaller or largerWhich of the following accurately describes the observed frequencies for a chi-square test? a. They are always whole numbers. b. They can contain fractions or decimal values. c. They can contain both positive and negative values. d. They can contain fractions and negative numbers.A veterinary study of horses looked at water sources for horses, and the investigators found that horses received water from a well, from city water, or from a stream. The investigators wanted to know if horses are equally likely to get water from each of those three sources. They collected data and observed the given values. Water Source Well City Water Stream Total Observed count 62 43 31 136 The cell that contributed least to the chi-square goodness-of-fit statistic is: - The city water cell with = 0.1201 - The well water cell with = 6.13 - The stream water cell with = 0.1201 - The city water cell with = 1/3
- You obtain a sample chi-square value of x2= -5.49. On the basis of this value, you know thatA local restaurant claims that it gets 45 percent of its customers from Monday through Thursday, 20 percent on Friday, 20 percent on Saturday, and 15 percent on Sunday. How many degrees of freedom should be used to conduct a chi-square goodness-of-fit test of the claim? A) 3B) 4C) 6D) 7E) It is not possible to determine the degrees of freedom without knowing the sample size.Fill the chi square for Cross 1 and Cross 2 from the following data in the attached image: Cross 1: Phenotypes Ratio Observed Expected (O-E)2/E Totals X2 = Cross 2: Phenotypes Ratio Observed Expected (O-E)2/E Totals X2 =
- The toll plaza on Route 1 in Dover has four tollbooths open to all vehicles. Ms. Timlin conducted a chi-square goodness of fit test to see if all of the booths are used in equally. A chi-square value of 23.4 was calculated, resulting in a p-value of 0.006. Which of the following is correct? a. There is sufficient evidence to suggest that the tollbooths are not used in equal proportions. Selected:b. There is sufficient evidence to suggest that the tollbooths are used in equal proportions.This answer is incorrect. c. The tollbooths are used in equal proportions. d. There is insufficient evidence to suggest that the tollbooths are used in equal proportions. e. There is insufficient evidence to suggest that the tollbooths are not used in equal proportions.15. For each of the following examples, state whether the chi-square goodness-of-fit or the chi-square test for independence is appropriate. 15d. A public health employee evaluates the proportion of lean, healthy, overweight, and obese students at a local college. A. goodness of fit test B. test for independenceFrom a test cross you get 152 green, hairless; 23 green, hairy; 148 white, hairy, and 27 white, hairless. Calculate the chi squared to test the independent assortment hypothesis and determine if these genes assort independently.
