Exam 11: Other Chi-Square Tests

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Two allergists recorded the main area of allergy for new patients during a month. At α=0.05\alpha = 0.05 , test the claim that the allergy diagnosis and the doctor that treated the patients at independent. Allergies Pollen Food Mold Pets Doctor 1 7 7 8 8 Doctor 2 26 7 10 24

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A random group of students was selected from a large student conference to analyze their class in school. Is there evidence to reject the hypothesis that the number of students is equally distributed between the four classes, at α=0.05?\alpha = 0.05 ? Year in school Freshman Sophomore Junior Senior Number of students 12 12 13 23

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The computational procedure for the homogeneity of proportions differs from that of the test of independence in that it takes into account the standard deviation.

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The formula for finding the chi-square test value is .

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The following table presents the numbers of cucumber seeds, by brand, that germi failed to germinate 14 days after planting. Brand A Brand B Brand C Brand D Germinated 40 37 41 42 Failed to Germinate 17 14 15 7 Can you conclude that the the germination rate is related to the brand of seed? Usi α=0.01\alpha = 0.01 level of significance.

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A biologist had mice and rats run through a maze and recorded the number that finished the maze successfully and the number that did not. The table lists the results of the study. At α=0.05\alpha = 0.05 , test the claim that the rodent type and the success of finishing the maze are nc related. Mice Rats Finished 22 28 Did not Finish 30 11

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The method that uses the formula χ2=∑(∣O−E∣−0.5)2E\chi ^ { 2 } = \sum \frac { ( | O - E | - 0.5 ) ^ { 2 } } { E } is called the________ correction for continuity.

(Short Answer)
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A contingency table has 5 rows and 4 columns. How many degrees of freedom are there?

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A study was performed at a university to analyze whether the preference for hamburgers or fried chicken is related to the gender of the student. This table lists the results of the study. At α=0.05\alpha = 0.05 , test the claim that the meal preference and the gender of the student a related. Hamburgers Fried Chicken Female 11 26 Male 15 20

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If the null hypothesis is not rejected, it can be assumed the proportions are 67) and the differences in them are due to chance.

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An assumption for the goodness-of-fit test is that the frequency for each category must be five or more.

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A biology professor claims that, on the average, 15%15 \% of her students get a grade of AA , 25%25 \% get a B, 40%40 \% get a C , 10%10 \% get a D, and 10%10 \% get an F\mathrm { F } . The grades of a random sample of 131 students were recorded. The following table presents the results. Grade 1 2 3 4 5 Observed 13 21 73 11 13 What is the value of χ2\chi ^ { 2 } ?

(Multiple Choice)
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Following are observed frequencies. The null hypothesis is H0:p1=0.35,p2=0.2H _ { 0 } : p _ { 1 } = 0.35 , p _ { 2 } = 0.2 , p3=0.15,p4=0.15,p5=0.15p _ { 3 } = 0.15 , p 4 = 0.15 , p _ { 5 } = 0.15 . Category 1 2 3 4 5 Observed 38 26 22 4 16 Compute the value of χ2\chi ^ { 2 } .

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A product testing laboratory selected a sample of 20 printers from 2 different manufacturers to determine if they could print more than 100,000 pages without breaking down. At α=0.05\alpha = 0.05 , test the claim that the proportions lasting more than 100,000 pages are equal. Manufacturer 1 Manufacturer 2 Lasted past 100,000 14 7 Did not last to 100,000 6 13 Total 20 20

(Multiple Choice)
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In a chi-square goodness-of-fit test when there is close agreement between the observed frequency and the expected frequency, the chi-square test value will be small.

(True/False)
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A quality control supervisor selected a sample of 100 electronic components from each of three different production chains to determine if the production chains had the same rates of meeting their quality standards. At α=0.05\alpha = 0.05 , test the claim that the proportions ar equal. Components Chain 1 Chain 2 Chain 3 Meets the standard 80 87 74 Does not meet the standard 20 13 26 Total 100 100 100

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The chi-square distribution is continuous, whereas the goodness-of-fit test is .

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A research firm reported that 15% of those surveyed described their health as poor, 26% as good, 40% as very good, and 19% as excellent. A health professional in Chicago wanted to determine if people in Chicago had similar feelings toward their health. In a sample of 610 people in Chicago, 70 described their health as poor, 180 as good, 220 as very good, and 140 as excellent. Compute the test value.

(Short Answer)
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 The null hypothesis should not be rejected for the data given in the figure below. \text { The null hypothesis should not be rejected for the data given in the figure below. } \text { The null hypothesis should not be rejected for the data given in the figure below. }

(True/False)
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Following are observed frequencies. The null hypothesis is H0:p1=0.35,p2=0.25H _ { 0 } : p _ { 1 } = 0.35 , p _ { 2 } = 0.25 , p3=0.1,p4=0.1,p5=0.2p _ { 3 } = 0.1 , p _ { 4 } = 0.1 , p _ { 5 } = 0.2 . Category 1 2 3 4 5 Observed 43 20 10 4 31 Compute the expected frequencies.

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