As part of an annual review of its accounts, a discount brokerage selects a rand
ID: 3021339 • Letter: A
Question
As part of an annual review of its accounts, a discount brokerage selects a random sample of 27 customers. Their accounts are reviewed for total account valuation, which showed a mean of $34,500, with a sample standard deviation of $8,800. (Use t Distribution Table.)
What is a 98% confidence interval for the mean account valuation of the population of customers? (Round your answers to the nearest dollar amount.)
As part of an annual review of its accounts, a discount brokerage selects a random sample of 27 customers. Their accounts are reviewed for total account valuation, which showed a mean of $34,500, with a sample standard deviation of $8,800. (Use t Distribution Table.)
Explanation / Answer
Given mean =34500, SD =8,800 and n =27
The two sided t-value at 98% confidence is at 27-1 degrees of freedom is t*= 2.4786
SE= SD/Sqrt(n) = 1693.561
98% confidence interval is [Sample mean- t* SE, Sample mean- t* SE]=(30302.29,38697.71)
round to the nearest dollar 98% confidence interval is ( $ 30302, $38698)
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