At issue is the proportion of people in a particular county who do not have heal

Question

At issue is the proportion of people in a particular county who do not have health insurance coverage. A simple random sample of 250 people was asked if they have insurance coverage, and 55 replied that they didn’t have coverage. a. Construct a 99% confidence interval of the population proportion of the individuals who do not have health insurance coverage. b. Based on the sample data, can we conclude that the proportion of individuals in that particular county who do not have health insurance is different from 20%? Use a level of significance of 0.01. Make sure to state the null and the alternative hypotheses and the decision rule. c. Are the results from a and b consistent? Should they be? Justify your answer.

Explanation / Answer

a) SEp = sqrt[ p * ( 1 - p ) / n ] = 0.0261992

degree of freedom, df = 250-1 = 249

tcrit = +/- 2.5957

99% CI = p +/- tcrit*SE = 0.22 +/- 2.5957*0.0261992 = (0.152, 0.288)

b) Null hypothesis: P = 0.20
Alternative hypothesis: P 0.20
= sqrt[ P * ( 1 - P ) / n ], P is the hypothesized value of population proportion in the null hypothesis,
= sqrt[0.2*0.8/250] = 0.025298

z = (p - P) / = (0.22-0.2)/0.025298 = 0.79058
p = 0.4292 > 0.01, Fail to reject null hypothesis, there is no sufficient evidence to suggest the proportion of individuals in that particular county who do not have health insurance is different from 20%

c) yes they are consistent as the confidence interval contains the 0.2 and also the hypothesis test resulted in failing to reject the null hypothesis

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