In constructing a 95% confidence interval estimate for the difference between the means of two normally distributed populations, where the unknown population variances are assumed not to be equal, summary statistics computed from two independent samples are: , , , , , and . The upper confidence limit is:
A) 18.78
B) 6.78
C) 5.76
D) 77.3
E) 89.3

Question

In constructing a 95% confidence interval estimate for the difference between the means of two normally distributed populations, where the unknown population variances are assumed not to be equal, summary statistics computed from two independent samples are:  ,  ,  ,  ,  , and  . The upper confidence limit is:
A) 18.78 
B) 6.78 
C) 5.76 
D) 77.3 
E) 89.3

Quizplus · Accepted Answer

Since the sample sizes are small and the population variances are assumed to be unequal, we need to use the t-distribution. 
The formula for the confidence interval is:
$$(\bar{x}_1 - \bar{x}_2) \pm t_{\alpha/2,df} \cdot \sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}$$
where $\bar{x}_1$ and $\bar{x}_2$ are the sample means, $s_1$ and $s_2$ are the sample standard deviations, $n_1$ and $n_2$ are the sample sizes, $df$ is the degrees of freedom, and $t_{\alpha/2,df}$ is the critical value from the t-distribution with $df$ degrees of freedom and a significance level of $\alpha/2$. 
Plugging in the given values, we get: 
$$\bar{x}_1 - \bar{x}_2 = 8.74$$
$$s_1 = 0.52, s_2 = 3.84$$
$$n_1 = n_2 = 5$$
To find the degrees of freedom, we use the following formula: 
$$df = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{(s_1^2/n_1)^2}{n_1-1} + \frac{(s_2^2/n_2)^2}{n_2-1}}$$
Plugging in the values, we get $df = 6.69$. 
Using a t-table or calculator, we find the critical value to be $t_{\alpha/2,df} = 2.447$. 
Plugging all the values into the formula, we get: 
$$(8.74) \pm 2.447 \cdot \sqrt{\frac{0.52^2}{5}+\frac{3.84^2}{5}}$$
Simplifying, we get: 
$$(8.74) \pm 6.78$$
Therefore, the upper confidence limit is $8.74 + 6.78 = 15.52$, which is closest to choice B) 6.78.

In Constructing a 95% Confidence Interval Estimate for the Difference