Answer to Question #127217 in Statistics and Probability for mimi

Question

Year Return
1985 -5.03
1986 49.88
1987 38.73
1988 -14.19
1989 30.4
1990 -19.86
1991 4.37
1992 -19.41
1993 93.36
1994 -5.42
1995 -4.48
1996 -0.75
1997 -38.92
1998 -3.91
1999 28.82
2000 -7.34
2001 18.33
2002 33.35
2003 59.45
2004 8.09
2005 43.79
2006 34.3
2007 36.13
2008 -56.02
2009 76.46
Using data from the past 25 years, an investor wants to test whether the average return of Vanguard’s Precious Metals and Mining Fund is greater than 12%. Assume returns are normally distributed with a population standard deviation of 30%.
a. State the null and the alternative hypotheses for the test.
b. Calculate the value of the test statistic and the p-value.
c. At

Answer 1

The provided sample mean is $\bar X = 29.2316$ using excel with command =average(data)  and the known population standard deviation is $\sigma = 30$

and the sample size is n = 25

The following null and alternative hypotheses need to be tested:

Ho:μ=12

Ha: μ>12

This corresponds to a right-tailed test, for which a z-test for one mean, with known population standard deviation will be used.

The z-statistic is computed as follows:

$z= \frac{\overline{X}-\mu }{\frac{\sigma}{\sqrt{n}}}$

$z= \frac{29.2316-12 }{\frac{30}{\sqrt{25}}}$ = 2.872

Using the P-value approach: Using z table

p value = P(Z$\geq$2.872) = 1- 0.998= 0.002

The p-value is p = 0.002, and since p=0.002<0.05, it is concluded that the null hypothesis is rejected.