Starting salaries of 64 college graduates who have taken a statistics course have a mean of $44,500 with a standard deviation of $6,800. Find a 99.7% confidence interval for μ. (NOTE: Do not use commas or dollar signs in your answers. Round each bound to three decimal places.)
Lower bound: Upper bound:
Solution :
Given that,
Point estimate = sample mean =
= $44500
Population standard deviation =
= $6800
Sample size = n =64
At 99.7% confidence level the z is ,
= 1 - 99.7% = 1 - 0.997 = 0.003
/ 2 = 0.003 / 2 = 0.0015
Z_{/2} = Z_{0.0015} = 2.97
Margin of error = E = Z/2
* (
/n)
= 2.97 * ( 6800 / 64
)
= 2524.5
At 99.7% confidence interval estimate of the population mean
is,
- E <
<
+ E
44500 - 2524.5 <
< 44500 + 2524.5
41975.500 <
< 47024.500
( 41975.500 ,47024.500)
Starting salaries of 64 college graduates who have taken a statistics course have a mean of...
(10 points) Starting salaries of 64 college graduates who have taken a statistics course have a mean of $44,500 with a standard deviation of $6,800. Find a 99.7% confidence interval for u. (NOTE: Do not use commas or dollar signs in your answers. Round each bound to three decimal places.) Lower-bound: Upper-bound:
(10 points) Starting salaries of 64 college graduates who have taken a statistics course have a mean of $44,500 with a standard deviation of $6,800. Find an 90% confidence interval for u. (NOTE: Do not use commas or dollar signs in your answers. Round each bound to three decimal places.) Lower-bound: Upper-bound:
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