Afinancial analyst wanted to estimate the mean annual return on mutual funds. a random sample of funds' returns shows an average rate of 12%. if the population standard deviation is assumed to be 4%, the 95% confidence interval estimate for the annual return on all mutual funds is

a. 0.037773 to 0.202227
b. 3.7773% to 20.2227%
c. 59.98786% to 61.01214%
d. 51.7773% to 68.2227%
e. 10.988% to 13.012%

A financial analyst wanted to estimate the mean annual return on mutual funds. A random sample of 60 funds' returns shows an average rate of 12%. If the population standard deviation is assumed to be 4%, the 95% confidence interval estimate for the annual return on all mutual funds is

A. 0.037773 to 0.202227

B. 3.7773% to 20.2227%

C. 59.98786% to 61.01214%

D. 51.7773% to 68.2227%

E. 10.988% to 13.012%

E. 10.988% to 13.012%

Step-by-step explanation:

Given;

Mean x= 12%

Standard deviation r = 4%

Number of samples tested n = 60

Confidence interval is 95%

Z' = t(0.025)= 1.96

Confidence interval = x +/- Z'(r/√n)

= 12% +/- 1.96(4%/√60)

= 12% +/- 0.01214%

Confidence interval= (10.988% to 13.012%)

the mistake in the pattern is that :

b. the decimal point was moved one too many place values each time

the correct pattern could be made by :

b. moving the decimal point to the right the same numbers of place values as the exponents

55.44 inches

step-by-step explanation:

part a. assuming a perfectly linear relationship, we can find the slope from the first two data points.

slope = m = (change in rainfall)/(change in years)

= (61.80 -62.33)/(2005 -2004) = -0.53/1 = -0.53

then the point-slope form of the equation of the line can be written as

y = m(x -h) +k . . . . . m = -0.53, (h, k) = (0, 62.33)

y = -0.53x +62.33 . . . x = years after 2004

part b. in 2017, x = 2017 -2004 = 13. then the predicted rainfall is

y = -0.53·13 +62.33 = 55.44 . . inches

the predicted rainfall in 2017 is 55.44 in.