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A simple random sample of size n equals 400 individuals who are currently employed is asked if they work at home at least once per week. Of the 400 employed individuals? surveyed, 38 responded that they did work at home at least once per week. Construct a? 99% confidence interval for the population proportion of employed individuals who work at home at least once per week. The lower bound is? ?(Round to three decimal places as? needed.) The upper bound is ? ?(Round to three decimal places as? needed.)

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Answer:

Lower bound: 0.057 < P

Upper bound: P < 0.133

0.057 < P < 0.133

Step-by-step explanation:

The sample proportion of individuals who work at home at least once per week (s) is:

[tex]p= \frac{38}{400} \\p=0.095[/tex]

For a 99% confidence interval, the margin of error is determined as:

[tex]ME = 2.575\sqrt{\frac{p(1-p)}{n}}[/tex]

[tex]ME = 2.575\sqrt{\frac{0.095(1-0.095)}{400}}\\ME = 0.038[/tex]

The lower bound for 99% confidence interval is

[tex]0.095-0.038 < P\\0.057 < P[/tex]

The upper bound for 99% confidence interval is

[tex]P < 0.095+0.038\\P < 0.133[/tex]

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