Respuesta :
Answer:
1. Point estimate Md = 2
2. The 90% confidence interval for the difference between means is (1.01, 2.99).
3. The 95% confidence interval for the difference between means is (0.82, 3.18).
Step-by-step explanation:
a) The point estimate of the difference between the two population means is the difference between sample means:
[tex]M_d=M_1-M_2=13.6-11.6=2[/tex]
2. We have to calculate a 90% confidence interval for the difference between means.
The sample 1, of size n1=50 has a mean of 13.6 and a standard deviation of 2.2.
The sample 2, of size n2=35 has a mean of 11.6 and a standard deviation of 3.
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{2.2^2}{50}+\dfrac{3^2}{35}}\\\\\\s_{M_d}=\sqrt{0.097+0.257}=\sqrt{0.354}=0.5949[/tex]
The degrees of freedom for this confidence interval are:
[tex]df=n_1+n_2-2=50+35-2=83[/tex]
The critical t-value for a 90% confidence interval is t=1.663.
The margin of error (MOE) can be calculated as:
[tex]MOE=t\cdot s_{M_d}=1.663 \cdot 0.5949=0.99[/tex]
Then, the lower and upper bounds of the confidence interval are:
[tex]LL=M_d-t \cdot s_{M_d} = 2-0.99=1.01\\\\UL=M_d+t \cdot s_{M_d} = 2+0.99=2.99[/tex]
The 90% confidence interval for the difference between means is (1.01, 2.99).
2. We have to calculate a 95% confidence interval for the difference between means.
The critical t-value for a 95% confidence interval and 83 degrees of freedom is t=1.989.
The margin of error (MOE) can be calculated as:
[tex]MOE=t\cdot s_{M_d}=1.989 \cdot 0.5949=1.18[/tex]
Then, the lower and upper bounds of the confidence interval are:
[tex]LL=M_d-t \cdot s_{M_d} = 2-1.18=0.82\\\\UL=M_d+t \cdot s_{M_d} = 2+1.18=3.18[/tex]
The 95% confidence interval for the difference between means is (0.82, 3.18).
