Question 1200883: A random sample of 200 consumer accounts at a shop is selected for the purpose of estimating the mean number of transactions per year for each customer. The sample mean is 43. Determine 98 percent confidence interval for the mean number of transactions of all consumer accounts with the shop. The population standard deviation is 2.5.
Ans: (42.59, 43.41)
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! sample size is 200
sample mean is 43
population standard deviation is 2.5
critical t-score for two tailed 98% confidence interval with 199 degrees of freedom is equal to plus or minus t = 2.345232207.
standard error is equal to standard deviation / square root of sample size = 2.5 / sqrt(200) = .1767766953.
on the low side, t-score formula becomes:
minus 2.345232207 = (x - 43) / .1767766953.
solve for x to get:
x = minus 42.2.345232207 * .1767766953 + 43 = 42.5854176.
round to get x = 42.59.
on the high side, t-score formula becomes:
2.345232207 = (x - 43) / .1767766953.
solve for x to get:
x = 2.345232207 * .1767766953 + 43 = 43.4145824.
round to get x = 43.41.
your solution is 42.59 to 43.41.
that agrees with what it should be, so it looks like you're good to go with that answer.
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