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\%
Step-by-step explanation:
From the question we are told that
The sample size is [tex]n = 16[/tex]
The sample mean is [tex]\= x = 0.352[/tex]
The standard deviation is [tex]\sigma = 0.20[/tex]
Generally the degree of freedom is mathematically represented as
[tex]df = n -1[/tex]
=> [tex]df = 16 -1[/tex]
=> [tex]df = 15[/tex]
From the question we are told the confidence level is 90% , hence the level of significance is
[tex]\alpha = (100 - 95 ) \%[/tex]
=> [tex]\alpha = 0.10[/tex]
Generally from the t distribution table the critical value of [tex]\frac{\alpha }{2}[/tex] at a degree of freedom of [tex]df = 15[/tex] is
[tex]t_{\frac{\alpha }{2} , 15 } = 2.13 [/tex]
Generally the margin of error is mathematically represented as
[tex]E = Z_{\frac{\alpha }{2} } * \frac{\sigma }{\sqrt{n} }[/tex]
=> [tex]E = 2.13 * \frac{ 0.20 }{\sqrt{16} }[/tex]
=> [tex]E = 0.1065 [/tex]
Generally 90% confidence interval is mathematically represented as
[tex]0.3252 -0.1065 < \mu < 0.3252 + 0.1065 [/tex]
=> [tex]0.2187 < \mu < 0.4317 [/tex]
converting to percentage
[tex]0.2187 * 100 < \mu < 0.4317 * 100 [/tex]
=> [tex] 21.87\% < \mu < 43.17\% [/tex]
Hence the 90% of the sample mean returns are between 21.87\% and 43.17\%
