Question

In: Statistics and Probability

. As health insurance costs and the number of disabled employees increase, more companies are firing...

. As health insurance costs and the number of disabled employees increase, more companies are firing these employees. A survey of 723 employers found that 195 dismiss employees as soon as they go on long-term disability.

  1. Construct a 95% confidence interval for the proportion of employers who dismiss employees as soon as they go on long-term disability.
  2. Construct a 99% confidence interval for the proportion of employers who dismiss employees as soon as they go on long-term disability.
  3. Which interval is wider? Explain why this is true.

Solutions

Expert Solution

a) = 195/723 = 0.2697

At 95% confidence interval the critical value is z* = 1.96

The 95% confidence interval for population proportion is

+/- z* * sqrt((1 - )/n)

= 0.2697 +/- 1.96 * sqrt(0.2697 * (1 - 0.2697)/723)

= 0.2697 +/- 0.0324

= 0.2373, 0.3021

b) At 99% confidence interval the critical value is z* = 2.58

The 95% confidence interval for population proportion is

+/- z* * sqrt((1 - )/n)

= 0.2697 +/- 2.58 * sqrt(0.2697 * (1 - 0.2697)/723)

= 0.2697 +/- 0.0426

= 0.2271, 0.3123

c) The 99% confidence interval is wider. Because if the confidence level increases, the margin of error also increases, so that the confidence interval becomes wider.


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