The proportion of U.S. of citizens lacking healthcare insurance is known to be 25%. A healthcare...

The proportion of U.S. of citizens lacking healthcare insurance is known to be 25%. A healthcare economist suspects the rate of uninsured in Louisiana is higher than the national average. A survey was conducted of 200 Louisiana residents who were asked if they had healthcare insurance. Seventy-two responded that they did not have healthcare insurance. Based on the results of the survey, construct a 95% CI for p, the population proportion of Louisiana residents who do not have healthcare insurance. Based on the confidence interval results, can you conclude that the proportion of uninsured in Louisiana is different (higher) from the national proportion?

Expert Solution

Sample proportion = 72 / 200 = 0.36

95% confidence interval for p is

- Z/2 * sqrt [ (1 - ) / n ] < p < + Z/2 * sqrt [ (1 - ) / n ]

0.36 - 1.96 * sqrt [ 0.36 ( 1 - 0.36) / 200 ] < p < 0.36 + 1.96 * sqrt [ 0.36 ( 1 - 0.36) / 200 ]

0.293 < p < 0.427

95% CI is ( 0.293 , 0.427 )

Since Claimed proportion 0.25 not contained in the confidence interval, we have sufficient evidence to

support the claim that proportion is different from the national proportion.

orchestra answered 2 years ago

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The proportion of U.S. of citizens lacking healthcare insurance is known to be 25%. A healthcare...

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