question 12 (1 point) saved an investor believes that investing in domestic and international stocks will…

question 12 (1 point) saved 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 4 decimal places. average group 1 = 2.1234, sd group 1 = 4.8765, n1 = 15 average group 2 = 3.0945, sd group 2 = 5.1115, n2 = 15 _<μ1 - μ2<_

question 12 (1 point) saved 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 4 decimal places. average group 1 = 2.1234, sd group 1 = 4.8765, n1 = 15 average group 2 = 3.0945, sd group 2 = 5.1115, n2 = 15 _<μ1 - μ2<_

Answer

Explanation:

Step1: Determine degrees of freedom

We use the formula for degrees of freedom for two - sample t - test with unequal variances: [df=\frac{\left(\frac{s_1^{2}}{n_1}+\frac{s_2^{2}}{n_2}\right)^{2}}{\frac{(s_1^{2}/n_1)^{2}}{n_1 - 1}+\frac{(s_2^{2}/n_2)^{2}}{n_2 - 1}}] where (s_1 = 4.8765), (n_1=15), (s_2 = 5.1115), (n_2 = 15). [ \frac{s_1^{2}}{n_1}=\frac{4.8765^{2}}{15}=\frac{23.7803}{15}\approx1.5854] [ \frac{s_2^{2}}{n_2}=\frac{5.1115^{2}}{15}=\frac{26.1274}{15}\approx1.7418] [ \left(\frac{s_1^{2}}{n_1}+\frac{s_2^{2}}{n_2}\right)^{2}=(1.5854 + 1.7418)^{2}=(3.3272)^{2}=11.0703] [ \frac{(s_1^{2}/n_1)^{2}}{n_1 - 1}=\frac{1.5854^{2}}{14}=\frac{2.5135}{14}\approx0.1795] [ \frac{(s_2^{2}/n_2)^{2}}{n_2 - 1}=\frac{1.7418^{2}}{14}=\frac{3.0348}{14}\approx0.2168] [df=\frac{11.0703}{0.1795+0.2168}=\frac{11.0703}{0.3963}\approx27.9342\approx28]

Step2: Find the critical value

For a two - tailed test with (\alpha=0.10), the critical value (t_{\alpha/2}) with (df = 28) is (t_{0.05,28}=1.7011)

Step3: Calculate the confidence interval

The formula for the confidence interval for the difference in means (\mu_1-\mu_2) is ((\bar{x}_1-\bar{x}2)\pm t{\alpha/2}\sqrt{\frac{s_1^{2}}{n_1}+\frac{s_2^{2}}{n_2}}) where (\bar{x}_1 = 2.1234), (\bar{x}_2=3.0945), (s_1 = 4.8765), (n_1 = 15), (s_2 = 5.1115), (n_2 = 15) (\bar{x}_1-\bar{x}_2=2.1234 - 3.0945=- 0.9711) (\sqrt{\frac{s_1^{2}}{n_1}+\frac{s_2^{2}}{n_2}}=\sqrt{1.5854 + 1.7418}=\sqrt{3.3272}\approx1.823) The lower limit is (-0.9711-1.7011\times1.823=-0.9711 - 3.0911=-4.0622) The upper limit is (-0.9711 + 1.7011\times1.823=-0.9711+3.0911 = 2.12)

Answer:

(-4.0622<\mu_1-\mu_2<2.1200)