Question

A mutual fund manager is trying to estimate the expected fund flows for the next quarter. To make the estimate, the manager calls 15 clients and asks each of them about their planned deposits/withdrawals from the fund. The asset-weighted average of the sample is a 4.2 % change in assets under management, with a standard deviation of 10 %. What is the width of the 95% confidence interval for next quarter's fund flows? Enter answer in percents, accurate to two decimal places. The answer is 11.8 please help me get to that number

Answer 1

Here sample size (n) = 15 clients=15

mean of the sample (x) = 4.2% change in assets under management = 4.2%

Standard deviation of sample (stdev) = 10%

From statistics, we can say that there is a 95% probability that the next quarter's fund flows will be between

[ x - t (2.5%, 14)* stdev / sqrt (n) , x +t (2.5%, 14)* stdev / sqrt (n) ]

[ 4.2%- t(2.5%, 14) * 10%/ sqrt(15) , 4.2%+ t(2.5%, 14) * 10%/ sqrt(15) ]

The t value from the t distribution table for two sided 95% confidence (5% significance) and n-1 i.e. 14 degrees of freedom is equal to 2.145

So, the interval is

(4.2% - 2.145*10% / 3.873 , 4.2% + 2.145*10%/ 3.873)

=(4.2% - 5.5384% , 4.2% + 5.5384%)

= (-1.34% , 9.74%)

The width of this interval = 9.74 % -- 1.34% = 11.08%

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