Answer :
Answer:
The 98% confidence interval for the mean is (8.84, 18.96).
Step-by-step explanation:
We have to calculate a 98% confidence interval for the mean.
The sample 1, of size n1=11 has a mean of 77.5 and a standard deviation of 4.5.
The sample 1, of size n1=9 has a mean of 63.6 and a standard deviation of 5.1.
The difference between sample means is Md=.
[tex]M_d=M_1-M_2=77.5-63.6=13.9[/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{4.5^2}{11}+\dfrac{5.1^2}{9}}\\\\\\s_{M_d}=\sqrt{1.841+2.89}=\sqrt{4.731}=2.175[/tex]
The t-value for a 98% confidence interval is t=2.326.
The margin of error (MOE) can be calculated as:
[tex]MOE=t\cdot s_M=2.326 \cdot 2.175=5.06[/tex]
Then, the lower and upper bounds of the confidence interval are:
[tex]LL=M-t \cdot s_M = 13.9-5.06=8.84\\\\UL=M+t \cdot s_M = 13.9+5.06=18.96[/tex]
The 98% confidence interval for the mean is (8.84, 18.96).