Question 1083604
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Answer: <font color=red> Choice B) 273.18 &#8804; &#956; &#8804; 296.82 </font>
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Explanation:


At 95% confidence, the critical z value is z = 1.960
See <a href="http://www.sjsu.edu/faculty/gerstman/StatPrimer/t-table.pdf">this table</a>. 
Look at the bottom where it shows "95%" in the confidence level section. Then look at the value just above the "95%" and you'll see 1.960 


So in short, z = 1.960


The sample mean is xbar = 285 which is given


The sample standard deviation is s = 60. 


The sample size is given to be n = 99.


We don't know the population standard deviation sigma, but since n > 30 we can use the standard normal Z distribution.
The sample standard deviation s approximates the population standard deviation.


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Let's use the four values mentioned above (z = 1.960, xbar = 285, s = 60, n = 99) to compute the upper and lower endpoints of the confidence interval


L = lower endpoint of confidence interval
L = xbar - z*s/sqrt(n)
L = 285 - 1.96*60/sqrt(99)
L = 285 - 117.6/sqrt(99)
L = 285 - 117.6/9.95
L = 285 - 11.819
L = 273.181
which rounds to 273.18


U = upper endpoint of confidence interval
U = xbar + z*s/sqrt(n)
U = 285 + 1.96*60/sqrt(99)
U = 285 + 117.6/sqrt(99)
U = 285 + 117.6/9.95
U = 285 + 11.819
U = 296.819
which rounds to 296.82


The confidence interval is therefore (L, U) = (273.18, 296.82)


Put another way, we are 95% confident that the population mean mu is between 273.18 and 296.82. We can write it like so


<font color=red> 273.18 &#8804; &#956; &#8804; 296.82 </font>


which is the final answer.
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