Question

On Jul 17 2018, a UVAToday article headline stated that “Only Half of Americans Believe Elections are Fair and Open”. The following statement can be found in the article: “In the Ipsos poll, which surveyed more than 1,000 American adults on July 5 and 6, 51 percent of the respondents agreed with the statement that American elections are fair and open.” The poll was based on a sample of 1,006 adults. For this question, refer only to the information provided in the preceding paragraph. (a) What is the variable used in this inference? Is it categorical or quantitative? What inferential procedure should be used for this inference? 1 (b) We want to investigate if more than half of Americans agree that American elections are fair and open. Let p denote the population proportion of Americans that agree. Carry out the appropriate hypothesis test. Be sure to write the null and alternative hypotheses, calculate the test statistic, obtain the p-value and critical value, and write a conclusion in context. Use a significance level of 0.1. (c) Verify that the conditions to carry out the hypothesis test in 2b are met. (d) Construct a 90% confidence interval for the proportion of Americans who agree that American elections are fair and open. What is the margin of error? Does this interval support the claim that a majority of Americans agree with the statement? (e) Verify that the conditions to construct the confidence interval in 2d are met.

Answer 1

(a) The variable is the proportion of all Americans who think the election is fair and open. It is a quantitative variable. We should use a test for one proportion (upper-tailed test)

(b)

Data:    

n = 1006   

p = 0.5   

p' = 0.51   

Hypotheses:   

Ho: p ≤ 0.5   

Ha: p > 0.5   

Decision Rule:   

α = 0.1   

Critical z- score = 1.281551566

Reject Ho if z > 1.281551566

Test Statistic:   

SE = √{(p (1 - p)/n} = √(0.5 * (1 - 0.5)/√1006) = 0.015764167

z = (p' - p)/SE = (0.51 - 0.5)/0.0157641665279858 = 0.634350061

p- value = 0.2629262   

Decision (in terms of the hypotheses):

Since 0.6343501 < 1.281551566 we fail to reject Ho

Conclusion (in terms of the problem):

There is no sufficient evidence that more than half the Americans think the elections are fair and open

(c) The sample is a random sample, the sample size is > 30, and the sampling units are independent. So, all the assumptions are met.

(d)

n = 1006    

p = 0.51    

% = 90    

Standard Error, SE = √{p(1 - p)/n} =    √(0.51(1 - 0.51))/1006 = 0.015761013

z- score = 1.644853627    

Margin of error = z * SE =     1.64485362695147 * 0.0157610133793338 = 0.02592456

Lower Limit of the confidence interval = P - width =     0.51 - 0.0259245600214279 = 0.48407544

Upper Limit of the confidence interval = P + width =     0.51 + 0.0259245600214279 = 0.53592456

The confidence interval is [0.484, 0.536]

The above interval contains values which are less than 0.5. So it does not support the claim that more than half the Americans think the elections are fair and open.

(e) The sample is a random sample, the sample size is > 30, and the sampling units are independent. Also,
np = np(1 - p) ≥ 10. So, all the assumptions are met.

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