Question 1205508: If you solve the question below, I'll be appreciated.
A statistics practitioner took a random sample of 48 observations from a population whose standard deviation is 24 and computed the sample mean to be 98.
Note: For each confidence interval, enter your answer in the form (LCL, UCL). You must include the parentheses and the comma between the confidence limits.
A. Estimate the population mean with 95% confidence.
Confidence Interval = ?
B. Estimate the population mean with 95% confidence, changing the population standard deviation to 49;
Confidence Interval = ?
C. Estimate the population mean with 95% confidence, changing the population standard deviation to 6;
Confidence Interval = ?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! A statistics practitioner took a random sample of 48 observations from a population whose standard deviation is 24 and computed the sample mean to be 98.
Note: For each confidence interval, enter your answer in the form (LCL, UCL). You must include the parentheses and the comma between the confidence limits.
since the standard deviation is taken from the population, the z-score is used.
A. Estimate the population mean with 95% confidence.
critical z-score for two tailed 95% confidence interval is z = plus or minus 1.96.
z-score formula is z = (x-m)/s
z is the critical z-score.
x is the critical raw score
m is the population mean
s is the standard error when you're looking for the mean of a sample.
sample size is 48.
standard error = standard deviation / sqrt(sample size) = 24 / sqrt(48) = 3.4641 rounded to 4 decimal places.
on the high side of the confidence interval, z = (x-m)/s becomes 1.96 = (x-98)/3.4641.
solve for x to get x = 1.96 * 3.4641 + 98 = 104.79 rounded to 2 decimal places.
on the low side of the confidence interval, z = (x-m)/s becomes -1.96 = (x-98)/3.4641.
solve for x to get x = -1.96 * 3.4641 + 98 = 91.21 rounded to 2 decimal places.
your 95% confidence limit is equal to (91.21,104.79).
B. Estimate the population mean with 95% confidence, changing the population standard deviation to 49;
population standard deviation = 49
standard error = standard deviation / sqrt(sample size) = 49 / sqrt(48) = 7.0725 rounded to 4 decimal places.
critical z-score for two tailed 95% confidence interval is z = plus or minus 1.96.
on the high side of the confidence interval, z-score formula becomes 1.96 = (x-98)/7.0725.
solve for x to get x = 1.96 * 7.0725 + 98 = 111.86 rounded to 2 decimal places.
on the low side of the confidence interval, z-score formula becomes -1.96 = (x-98)/7.0725.
solve for x to get x = -1.96 * 7.0725 + 98 = 84.14 rounded to 2 decimal places.
your 95% confidence limit is equal to (84.14,111.86).
C. Estimate the population mean with 95% confidence, changing the population standard deviation to 6;
critical z-score for two tailed 95% confidence interval is z = plus or minus 1.96.
standard error = standard deviation / sqrt(ample size) = 6 / sqrt(48) = .8660 rounded to 4 decimal places.
on the high side of the confidence interval, z-score formula becomes 1.96 = (x-98)/.8660.
solve for x to get x = 1.96 * .8660 + 98 = 99.70 rounded to 2 decimal places.
on the low side of the confidence interval, z-score formula becomes -1.96 = (x-98)/.8660.
solve for x to get x = -1.96 * .8660 + 98 = 96.30 rounded to 2 decimal places.
your 95% confidence interval is equal to (96.30,99.70).
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