Question

A conservative investor desires to invest in a bond fund in which her investment amount is kept relatively safe. A national investment group claims to have a bond fund which has maintained a consistent share price of $11.25, consistent because the variation in price (as measured by standard deviation) is at most $0.45 since fund inception. To test this claim, the investor randomly selects fifty days during the last year and determines the share price for the fund on closing of those days. The standard deviation of this sample group is found to be $0.62. Use an appropriate hypothesis test at the 5% significance level to determine if the investor should conclude that the variation is greater than that claimed by the national investment group.
Give the null and alternative hypotheses for this test in symbolic form.

Answer 1

Answer:

$\chi^2 =\frac{50-1}{0.2025} 0.3844 =93.015$

The degrees of freedom are given by:

$df = n-1= 50-1=49$

And the p value is given by:

$p_v =P(\chi^2 >93.015)=0.00015$

Since the p value is very low compared to the significance level provided we have enough evidence to conclude that the true deviation is higher than 0.45

Step-by-step explanation:

Information given

$\alpha=0.05$ represent the confidence level  

$s^2 =0.62^2 =0.3844$ represent the sample variance obtained

$\sigma^2_0 =0.45^2 =0.2025$ represent the value that we want to test

System of hypothesis

On this case we want to check if the true deviation is higher than 0.45, so then we can create the following system of hypothesis:

Null Hypothesis: $\sigma^2 \leq 0.2025$

Alternative hypothesis: $\sigma^2 >0.2025$

The statistic for this case is given by:

$\chi^2 =\frac{n-1}{\sigma^2_0} s^2$

Replacing we got

$\chi^2 =\frac{50-1}{0.2025} 0.3844 =93.015$

The degrees of freedom are given by:

$df = n-1= 50-1=49$

And the p value is given by:

$p_v =P(\chi^2 >93.015)=0.00015$

Since the p value is very low compared to the significance level provided we have enough evidence to conclude that the true deviation is higher than 0.45