In 2006, 9.8 % of all U.S. families had incomes below the poverty level, accordi
ID: 3226412 • Letter: I
Question
In 2006, 9.8 % of all U.S. families had incomes below the poverty level, according to the U.S. Census Bureau. During that same year, out or400 randomly selected Wyoming families, 25 had incomes below the poverty level. If we set alpha = 0 01 (a 1% significance level), conduct a hypothesis nest to determine if the data provide sufficient evidence to conclude that, in 2006, the percentage of families with incomes below the poverty was lower among those living in Wyoming than among all U.S. families? (You do not need to check conditions for problem.) Null hypothesis: Alternative hypothesis: P-value: Conclusion: The USDA limit for salmonella contamination for chicken is 20%. A meat inspector reports that the chicken produced by a company exceeds the USDA limit. You perform a hypothesis test to determine whether he mean inspector's claim is true. a. What are the null and alternative hypotheses in this situation? b. What would a Type l error be? c. What would a Type Il error be? d. Which error would be worse for the chicken producer? Give a reason why. e. Which error would be worse for the consumers? Give a reason why. Which error do you think is more serious? Why?Explanation / Answer
Solution:-
P = 0.098
x = 25, n = 400
p = 25/400
p = 0.0625
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: P > 0.098
Alternative hypothesis: P < 0.098
Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected only if the sample proportion is too small.
Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method, shown in the next section, is a one-sample z-test.
Analyze sample data. Using sample data, we calculate the standard deviation () and compute the z-score test statistic (z).
= sqrt[ P * ( 1 - P ) / n ]
= 0.01487
z = (p - P) /
z = - 2.39
where P is the hypothesized value of population proportion in the null hypothesis, p is the sample proportion, and n is the sample size.
Since we have a one-tailed test, the P-value is the probability that the z-score is less than - 2.39. We use the Normal Distribution Calculator to find P(z < - 2.39) = 0.0084
Interpret results. Since the P-value (0.0084) is less than the significance level (0.05), we cannot accept the null hypothesis.
From the above test we have sufficient evidence in the favor of the claim that proportion of family below poverty line is less than the all U.S families.
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