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In: Statistics and Probability

Survey of 85 company employees shows that the mean length of the Christmas vacation was 4.5...

Survey of 85 company employees shows that the mean length of the Christmas vacation was 4.5 days, with a standard deviation of 1.2 days. Construct a 92% confidence interval for the population mean length of vacation [5 points]. Interpret the results using complete sentences. [5 points]

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Expert Solution

Solution :


Given that,

Point estimate = sample mean =     = 4.5

Population standard deviation =    = 1.2

Sample size n =85

At 92% confidence level the z is ,

Z/2 = Z0.04 = 1.75   ( Using z table )

Margin of error = E =   Z/2    * ( /n)
= 1.75 * ( 1.2 / 85)

E= 0.2278
At 92% confidence interval estimate of the population mean
is,

- E < < + E

4.5 - 0.2278 <   <4.5 + 0.2278

4.2722 <   < 4.7278
( 4.2722 , 4.7278 )

orchestra answered 2 years ago
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