## Statistics

A sample of 81 observations is taken from a normal population with a standard deviation of .5. The sample mean is 40. Determine the 95% confidence interval for the population mean.

• The formula for a confidence interval:

xbar ± z * (s / sqrt(n))
Where:
xbar = sample mean
s = sample standard deviation
n = number of observations
z = 1.96 (from table lookup)

Since n >30, we assume a normal distribution and the sample standard deviation can be substituted for the population standard deviation.

The confidence interval is :
= xbar ± z * (s / sqrt(n))
= 40 ± 1.96 * (5 / sqrt(81))
= 40 ± 1.96 * (5 / 9)
= 40 ± 1.089
= 38.911 to 41.089

~ Mitch ~

• Z 95% = 1.96

So confidence interval = (40 - 1.96 * 0.5 / sqrt(81),40 + 1.96 * 0.5 / sqrt(81)) = (39.891,40.109)

confidence interval: xbar ± z * (s / sqrt(n))
Where:
xbar = sample mean
s = sample standard deviation
n = number of observations
z = 1.96 (from table lookup)

Since n >30, we assume a normal distribution and the sample standard deviation can be substituted for the population standard deviation.

The confidence interval is :
= xbar ± z * (s / sqrt(n))
= 40 ± 1.96 * (5 / sqrt(81))
= 40 ± 1.96 * (5 / 9)
= 40 ± 1.089
= 38.911 to 41.089

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