Question

An investor believes that investing in domestic and international stocks will give a difference in the mean rate of return. They take two random samples of 15 months over the past 30 years and find the following rates of return from a selection of domestic (Group 1) and international (Group 2) investments. Can they conclude that there is a difference at the 0.10 level of significance? Assume the data is normally distributed with unequal variances. Use a confidence interval method. Round to 3 decimal places. Average Group 1 = 2.0233, SD Group 1 = 4.893387, n1 = 15 Average Group 2 = 3.048, SD Group 2 = 5.12399, n2 = 15

Answer 1

Answer:

So on this case the 90% confidence interval would be given by $-4.137 \leq \mu_1 -\mu_2 \leq 2.087$  

For this case since the confidence interval for the difference of means contains the 0 we can conclude that we don't have significant differences at 10% of significance between the two means analyzed.

Step-by-step explanation:

Previous concepts  

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

$\bar X_1 =2.0233$ represent the sample mean 1  

$\bar X_2 =3.048$ represent the sample mean 2  

n1=15 represent the sample 1 size  

n2=15 represent the sample 2 size  

$s_1 =4.893387$ population sample deviation for sample 1  

$s_2 =5.12399$ population sample deviation for sample 2  

$\mu_1 -\mu_2$ parameter of interest at 0.1 of significance so the confidence would be 0.9 or 90%

We want to test:

H0: $\mu_1 = \mu_2$

H1: $\mu_1 \neq \mu_2$

And we can do this using the confidence interval for the difference of means.

Solution to the problem  

The confidence interval for the difference of means is given by the following formula:  

$(\bar X_1 -\bar X_2) \pm t_{\alpha/2}\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}$ (1)  

The point of estimate for $\mu_1 -\mu_2$ is just given by:  

$\bar X_1 -\bar X_2 =2.0233-3.048=-1.0247$

The degrees of freedom are given by:

$df = n_1 +n_2 -2 = 15+15-2=28$  

Since the Confidence is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,28)".And we see that $t_{\alpha/2}=\pm 1.701$  

Now we have everything in order to replace into formula (1):  

$-1.0247-1.701\sqrt{\frac{4.893387^2}{15}+\frac{5.12399^2}{15}}=-4.137$  

$-1.0247+1.701\sqrt{\frac{4.893387^2}{15}+\frac{5.12399^2}{15}}=2.087$  

So on this case the 90% confidence interval would be given by $-4.137 \leq \mu_1 -\mu_2 \leq 2.087$  

For this case since the confidence interval for the difference of means contains the 0 we can conclude that we don't have significant differences at 10% of significance between the two means analyzed.