Question

A bond analyst is analyzing the interest rates for equivalent municipal bonds issued by Georgia and Ohio. A sample of 60 bonds issued by Georgia had a mean interest rate of 3.6% with a standard deviation of 0.03%. A sample of 80 bonds issued by Ohio had a mean interest rate of 4.15% with a standard deviation of 0.01%. At = .05, is there a difference in the interest rates paid by the two states? Group of answer choices Yes, because the test value –22.00 is outside the interval (-1.96, 1.96) Yes, because the test value –8.80 is outside the interval (-1.96, 1.96) Yes, because the test value 484.00 is outside the interval (-1.96, 1.96) No, because the test value 0.00 is inside the interval (-1.96, 1.96)

Answer 1

Answer:

Since our calculated value is outside than our critical value,$z_{calc}=-136.438$, we have enough evidence to reject the null hypothesis at 5% of significance. So there is not nough evidence to conclude that mean for Ohio it's significantly different than the mean for Georgia.

Step-by-step explanation:

1) Data given and notation  

$\bar X_{1}=3.6$ represent the mean for Georgia  

$\bar X_{2}=4.15$ represent the mean for Ohio  

$s_{1}=0.03$ represent the population standard deviation for Georgia

$\sigma_{2}=0.01$ represent the population standard deviation for Ohio

$n_{1}=60$ sample size for Georgia  

$n_{2}=80$ sample size for Ohio  

z would represent the statistic (variable of interest)

$p_v$ represent the p value  

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to check if the means are different for both groups, the system of hypothesis would be:

H0:$\mu_{1} = \mu_{2}$  

H1:$\mu_{1} \neq \mu_{2}$  

If we analyze the size for the samples both are higher than 30, we can assume that the normal distribution is the correct disirbution for the statistic, so for this case is better apply a z test to compare means, and the statistic is given by:  

$z=\frac{\bar X_{1}-\bar X_{2}}{\sqrt{\frac{\sigma^2_{1}}{n_{1}}+\frac{\sigma^2_{2}}{n_{2}}}}$ (1)  

z-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.  

3) Calculate the statistic  

We have all in order to replace in formula (1) like this:  

$z=\frac{3.6-4.15}{\sqrt{\frac{0.03^2}{60}+\frac{0.01^2}{80}}}=-136.438$  

4) Find the critical value

In order to find the critical value we need to take in count that we are conducting a two tailed test, so we are looking on the normal standard distribution a value that accumulates 0.025 of the area on the right and 0.975 of the area on the left. We can us excel or a table to find it, for example the code in Excel is:

"=NORM.INV(1-0.025,0,1)", and we got $z_{critical}=\pm 1.96$

5) Statistical decision

Since our calculated value is outside than our critical value,$z_{calc}=-136.438$, and outside of (-1.96,1.96), we have enough evidence to reject the null hypothesis at 5% of significance. So there is not nough evidence to conclude that mean for Ohio it's significantly different than the mean for Georgia.