Respuesta :
Answer:
1. The 95% confidence interval for the difference between means is (-5.34, 11.34).
2. The standard error of (x-bar1)-(x-bar2) is 4.
[tex]s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{9.2195^2}{10}+\dfrac{9.4868^2}{12}}\\\\\\s_{M_d}=\sqrt{8.5+7.5}=\sqrt{16}=4[/tex]
Step-by-step explanation:
We have to calculate a 95% confidence interval for the difference between means.
The sample 1, of size n1=10 has a mean of 45 and a standard deviation of √85=9.2195.
The sample 2, of size n2=12 has a mean of 42 and a standard deviation of √90=9.4868.
The difference between sample means is Md=3.
[tex]M_d=M_1-M_2=45-42=3[/tex]
The estimated standard error of the difference between means is computed using the formula:
[tex]s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{9.2195^2}{10}+\dfrac{9.4868^2}{12}}\\\\\\s_{M_d}=\sqrt{8.5+7.5}=\sqrt{16}=4[/tex]
The critical t-value for a 95% confidence interval is t=2.086.
The margin of error (MOE) can be calculated as:
[tex]MOE=t\cdot s_{M_d}=2.086 \cdot 4=8.34[/tex]
Then, the lower and upper bounds of the confidence interval are:
[tex]LL=M_d-t \cdot s_{M_d} = 3-8.34=-5.34\\\\UL=M_d+t \cdot s_{M_d} = 3+8.34=11.34[/tex]
The 95% confidence interval for the difference between means is (-5.34, 11.34).
