Question

Although the first quarter of 2002 was quite dismal on Wall Street, mutual funds investing in gold companies rose an average of 35.2%. Assume that the distribution of returns in the first quarter of 2002 for mutual funds specializing in gold companies is fairly symmetrical with a mean of 35.2% and a standard deviation of 20%. If random samples of 16 gold stock funds were selected:

Answer 1

Complete Question

Although the first quarter of 2002 was quite dismal on Wall Street, mutual funds investing in gold companies rose an average of 35.2%. Assume that the distribution of returns in the first quarter of 2002 for mutual funds specializing in gold companies is fairly symmetrical with a mean of 35.2% and a standard deviation of 20%. If random samples of 16 gold stock funds were selected:

90% of the sample mean returns are between what two values symmetrically distributed around the mean?

Answer:

The two values symmetrically distributed around the mean are

21.87\% and 43.17\%

Explanation:

From the question we are told that

The sample size is
`n = 16`

The sample mean is
`\= x = 0.352`

The standard deviation is
`\sigma = 0.20`

Generally the degree of freedom is mathematically represented as

`df = n -1`

=>
`df = 16 -1`

=>
`df = 15`

From the question we are told the confidence level is 90% , hence the level of significance is

`\alpha = (100 - 95 ) \%`

=>
`\alpha = 0.10`

Generally from the t distribution table the critical value of
`(\alpha )/(2)` at a degree of freedom of
`df = 15` is

`t_{(\alpha )/(2) , 15 } =  2.13`

Generally the margin of error is mathematically represented as

`E = Z_{(\alpha )/(2) } *  (\sigma )/(√(n) )`

=>
`E = 2.13  *  ( 0.20  )/(√(16) )`

=>
`E = 0.1065`

Generally 90% confidence interval is mathematically represented as

`0.3252  -0.1065 <  \mu < 0.3252  + 0.1065`

=>
`0.2187  <  \mu < 0.4317`

converting to percentage

`0.2187 * 100   <  \mu < 0.4317 * 100`

=>
`21.87\%   <  \mu <  43.17\%`

Hence the 90% of the sample mean returns are between 21.87\% and 43.17\%