SOLUTION: The average hourly wage(in dollars) for a sample 36 employees at a factory id $10.67 and the standard deviation of the wages is $2.97. Estimate the average hourly wage for all wor
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Question 1141171: The average hourly wage(in dollars) for a sample 36 employees at a factory id $10.67 and the standard deviation of the wages is $2.97. Estimate the average hourly wage for all workers at this factory with 90% confidence.
Type of Interval
Estimate
z or t value
margin of error
interval
summary
Answer by rothauserc(4718) (Show Source): You can put this solution on YOUR website!
Since the sample size is > 30, we can use a normal distribution(z value)
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Use a 90% confidence interval
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the average hourly wage($10.67) is our sample statistic
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The standard error(SE) = 2.97/square root(36) = 0.495
:
alpha(a) = 1 - (90/100) = 0.10
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critical probability(p*) = 1 - (0.10/2) = 0.95
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The critical value(CV) is the z-score associated with p* = 0.95. The z-score for 0.95 is 1.64
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Margin of error(ME) = CV * SE = 1.64 * 0.495 = 0.8118 is approximately 0.81
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The range of the confidence interval is defined by the sample statistic + or minus the ME
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The 90% confidence interval is 10.67 + or - 0.81 which is ($9.86, $11.48)
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The 90% confidence interval says that the average hourly wage is greater than $9.86 and less than $11.48.
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A 90% confidence level means that we would expect 90% of the confidence interval estimates to include the average hourly wage as long as we use the same sampling method for each confidence interval estimate
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