Question 1199107: We sampled 40 NY teachers for their annual income. In our sample,
the average annual income is $60,000 and the standard deviation of annual
incomes is $7,500. Based on that sample, we want to predict the mean annual
income μ of all the NY teachers with the confidence level of 0.99 (99 percent).
Part a) What is the error bound/margin E in our prediction about μ?
Part b) What prediction do we make about μ?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! the sample size is 40
the sample mean is 60000.
the sample standard deviation is 7500.
you would use the t-score to help solve this.
the formula is t = (x-m)/s
t is the t-score
x is the sample mean
m is the population mean
s is the standard error.
the standard error is equal to the standard deviation / square root of sample size = 7500 / sqrt(40) = 1185.854123.
at 99% confidence interval, you would calculate the two-tailed critical t-score with 39 degrees of freedom (sample size minus 1) to be plus or minus 2.707913179.
on the high side of the confidence interval, your t-score formula becomes:
2.707913179 = (60000 - m) / 1185.854123.
your margin of error on the high side is equal to (60000 - m) which is equal to 2.707913179 * 1185.854123 = 3211.190008.
this indicates that your mean is 60000 - 3211.190008 = 56788.80999.
your solution should be:
Part a) What is the error bound/margin E in our prediction about μ?
margin of error = 3211.19 rounded to 2 decimal places.
Part b) What prediction do we make about μ?
population mean is assumed to be 56788.81 rounded to 2 decimal places.
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