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In the sample of 25 cars of the VW Golf 5 recently sold in Augsburg, average cost was 9.000 euros with standard devation of 300 euros. Assume that the cost of all the cars of this type of vehicle are normally distributed. Construct 90% of confidence interval for the average cost of all the cars of this type of vehicle.

I need help with this task. I have tried many ways but none of them seems to be correct.

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Which ways did you try? – Ilya Jul 3 '12 at 14:46
I have tried calculating s^2 = (9000 -m)/(2*300/25), and using that to gain s. Also I have calculated U_ and border of the confidence interval, but the result is not what is wanted. – Takarakaka Jul 3 '12 at 15:16
up vote 1 down vote accepted

To construct the confidence interval you again use a t statistic. In this case the pivotal quantity is (9000 -m)/(300/√25) where m is the true unknown mean. This simplifies to (9000-m)/60 To construct a two-sided 90% confidence interval we need m to satisfy the condition

t(0.05)< (9000-m)/60 < t(0.95) where t(0.05) is the 5th percentile of the t distribution with n-1 = 24 degrees of freedom and t(0.95) is the 95th percentile for a t distribution with 24 degrees of freedom. Because the t distribution is symmetric about 0 t(0.05)=-t(0.95).

So you take the inequality -t(0.95) < (9000-m)/60 < t(0.95) and rearrange it to be an inequality for m as 9000-60 t(0.95) < m < 9000+60 t(0.95).

The interval [9000-60 t(0.95), 9000+60 t(0.95)] is your answer. From the t distribution with 24 degrees of freedom you will see that t(0.95)=1.711.

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Dear God! Thank you Michael! I have just found that I misused s^2 formula. Had wrong settings for it. – Takarakaka Jul 3 '12 at 15:13

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