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To ensure adequate customer service, a bus company wants to estimate the average number of passengers on a bus during morning rush hours. A number of bus passengers is a random variable with standard deviation of $9.2$. According to records, the average number of bus passengers at $36$ randomly selected times is $40.5$.

(a) Construct a $90%$ confidence interval for the expectation of the number of bus passengers. (b) Suppose that the true population mean number of bus passengers is $41$. If you are to collect another sample of passenger counts at $36$ randomly selected times, what is the probability that a sample average will be within the interval computed in (a)?

Now for (a) I found that the interval is $37,9778$, $43,0223$.

However I am not sure how to go about (b). I tried to calculate the interval using the mean of $41$ $(38,4778, 43,5223)$ and then calculate how much these intervals overlap, which is about $90%$, but I'm am hesitant that this is the right answer.

tien lee
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Onno Weerts
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    For the second part of the question, I suggest that you re-read the definition of confidence interval in your text book. – Brad S. May 26 '18 at 02:05
  • "A range of values constructed from sample data so that the population parameter is likely to occur within that range at a specified probability" However I'm not looking for the population parameter here, but a random sample average, does it still work the same? Is it just 90% because I constructed a 90% confidence interval? – Onno Weerts May 26 '18 at 02:30
  • What is the percentage overlap a percentage of? Also, what about the 5% tails, are any of them part of the overlap? – Phil H May 26 '18 at 02:42
  • The percentage overlap is the part that both intervals share, without including any tails, so the interval of the sample mean is 37,9778, 43,0223 and the interval of the actual population mean is 38,4778, 43,5223. The total shared amount between the intervals is 4,5445 and one total interval is 5,0445, so the shared amount is 4,5445/5,0445 which is roughly 90% – Onno Weerts May 26 '18 at 03:06
  • @OnnoWeerts The second part says "suppose the true population mean"....how is that not the population parameter? The second part is precisely a direct application of the definition of confidence interval. – Brad S. May 26 '18 at 03:52
  • @BradS. You have me confused here, according to the definition of confidence interval, in repeated samples of the same sample size, the true population mean should fall into 90% of all the possible intervals. However I have to find the probability that the sample mean of the next sample falls into the confidence interval of the first sample. Now excuse if I don't quite understand it, but shouldn't that be different? – Onno Weerts May 26 '18 at 04:51
  • Calculate the Z scores of 39.9778 and 43.0223 for a mean of 41 and determine the area which lies between them. Hence the probability of being in that confidence interval. – Phil H May 26 '18 at 05:03
  • Make that 37.9778 – Phil H May 26 '18 at 05:10

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Calculate the Z scores of $37.9778$ and $43.0223$ for a mean of $41$ and determine the area which lies between them. Hence the probability of being in that confidence interval. You can look up the Z scores on a Z table to get the p values.

Phil H
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