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Assuming that s = 18,000 is a reasonable estimate for what sample size would be needed to ensure that we could estimate the true mean salary of all production managers with more than 15 years experience within $4200 if we wish to be 95% confident?

User Museful
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Answer:

The minimum sample size required is 71.

Explanation:

The (1 - α)% confidence interval for population mean is:


CI=\bar x\pm z_(\alpha/2)* (\sigma)/(√(n))

The margin of error for this interval is:


MOE=z_(\alpha/2)* (\sigma)/(√(n))

It is provided that the population standard deviation σ can be estimated by the sample standard deviation s.

The given information is:

s = 18000,

Margin of error = $4200.

Confidence level = 95%

The z-value for 95% confidence level is,


z_(\alpha/2)=z_(0.05/2)=z_(0.025)=1.96

*Use a z-table for the critical value.

Compute the sample size required as follows:


MOE=z_(\alpha/2)* (\sigma)/(√(n))


n=[(z_(\alpha/2)* \sigma )/(MOE)]^(2)


=[(18000* 1.96)/(4200)]^(2)


=70.56\\\approx71

Thus, the minimum sample size required is 71.

User Nucandrei
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