Question

A financial analyst would like to evaluate the outcome of passive investing, which is a “buy-and-hold” portfolio strategy for long-term investment horizons with minimal trading in the market. The analyst collected the annual rate of return in 2019 from 25 clients who had taken such an investment strategy. The sample mean rate of return is 3% and the sample standard deviation is 0.5%.

The financial analyst does not know the population standard deviation of the annual rate of return for passive investing in 2019. To construct a 90% confidence interval for the mean annual rate of return, the critical value needed here is ____________ .

The small sample size made the financial analyst concerned. Fortunately, the normal probability plot has justified that the sample is approximately normal. Hence, a 90% confidence interval for the mean annual rate of return can be constructed as______________% (lower limit) and____________% (upper limit).

In scenario analysis, the probability of extreme events are worth analyzing. The sample mean of annual rate of return based on the 25 sample values will be larger than____________% with a probability of 2%, and will be smaller than ____________% with a probability of 1%.

Using this sample as a pilot sample, the financial analyst may find out that a sample size ____________will give a sampling error of 0.1% for the 90% confidence interval he constructed for the mean annual rate of return.

Answer 1

x̅ = 3, s = 0.5, n = 25

At α = 0.1 and df = n-1 = 24, two tailed critical value, t-crit = T.INV.2T(0.1, 24) = 1.711

90% Confidence interval :

Lower Bound = x̅ - t-crit*s/√n = 3 - 1.711 * 0.5/√25 = 2.8289

Upper Bound = x̅ + t-crit*s/√n = 3 + 1.711 * 0.5/√25 = 3.1711

2.8289 < µ < 3.1711

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Z score at p = 0.02 using excel = NORM.S.INV(0.02) = -2.0537

Value of X = µ + z*σ = 3 + (-2.0537)*0.5 = 1.97

Z score at p = 1-0.01 = 0.99 using excel = NORM.S.INV(0.99) = 2.3263

Value of X = µ + z*σ = 3 + (2.3263)*0.5 = 4.16

The probability of extreme events are worth analyzing. The sample mean of annual rate of return based on the 25 sample values will be larger than 1.97% with a probability of 2%, and will be smaller than 4.16% with a probability of 1%.

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Population standard deviation, σ = 0.5

Margin of error, E = 0.1

Confidence interval, CL = 0.9

Significance level, α = 1-CL = 0.1

Critical value, z = NORM.S.INV(0.1/2) = 1.6449

Sample size, n = (z * σ / E)² = (1.6449 * 0.5 / 0.1)² = 67.64 = 68