5. the chi-square test for goodness of fit - no difference from a known population suppose you are reading a study conducted in the year 2000 about welfare recipients in the united states. the authors report the following frequency data on the household size of the 1,844 welfare recipients in their random sample: observed frequencies household size 5-or-more-person 4-person 3-person 2-person 1-person 203 277 301 550 513 you wonder if welfare recipients tend to live in different-sized households than the us population at large. you obtain the following data from the 2000 census: percent distribution of us households by size household size 5-or-more-person 4-person 3-person 2-person 1-person 10.83% 14.20% 16.53% 32.63% 25.81% [source: hobbs f., & Stoops, N. (2002). Census 2000 special reports: Demographic trends in [Source: the 20th century. U.S. Census Bureau.] You use a chi-square test for goodness of fit to see how well the sample of welfare recipients fits the census data. What is the most appropriate null hypothesis? O The distribution of household sizes among welfare recipients is the same as that provided by the census data. The distribution of household sizes among welfare recipients is equal across the five household-size categories. O The distribution of household sizes among welfare recipients is different from that provided by the census data. O The distribution of household sizes among welfare recipients is not equal across the five household-size categories. Fill in the missing values in the following table indicating the expected frequencies in each category for a sample size of 1,844, if the null hypothesis is true: Expected Frequencies Household Size 5-or-more-person 4-person 3-person 2-person 1-person 475.94 304.81 601.70 The chi-square statistic has been calculated for you: x2 = 8.31. The distribution of the chi-square statistic has degrees of freedom. Use the following Distributions tool to find the critical value. Chi-Square Distribution Degrees of Freedom = 4 4 10 11 12 13 14 With the level of significance a = .01, the critical value is Because the chi-square statistic x2 is the critical value, you the null hypothesis. Therefore, you conclude that the proportions of welfare recipients in the five household-size categories differ from the corresponding proportions for all U.S. households.

The confidence interval for the given ANOVA test to calculate the average typing speed be, (144.83 , 155.17) and the margin of error of the poll be, ±5.17

On Subtracting 1 from your sample size, we get

150 – 1 = 149.

On Subtracting confidence level from 1, and then divide by two.

(1 – .95) / 2 = 0.025

Now, df = 149 and α = 0.025

from the table at df = 149 we got 2.262.

Divide your sample standard deviation by the square root of your sample size.

28 / √(150) = 2.28

Now, 2.262 × 2.28 = 5.17

So, the confidence interval be,

(150 - 5.17 , 150 + 5.17) = (144.83 , 155.17)

Hence, the confidence interval for the given ANOVA test to calculate the average typing speed be, (144.83 , 155.17) and the margin of error of the poll be, ±5.17

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