The table below gives information concerning the gasoline mileage for random samples of trucks of two different types.Find a 95% confidence interval for the difference in the means $\mu_\mathrm { X }$ - $\mu_Y$ . $$\begin{array} { l | c c }
& \text { Brand } \mathrm { X } & \text { Brand Y } \
\hline \text { Number of Trucks } & 50 & 50 \
\text { Mean mileage } & 20.4 & 24.9 \
\text { Standard Deviation } & 2.3 & 1.8
\end{array}$$
A)(3.68,5.32)
B)(-5,-4)
C)(4,5)
D)(-5.32,-3.68)
E)(20.4,24.9)

Question

The table below gives information concerning the gasoline mileage for random samples of trucks of two different types.Find a 95% confidence interval for the difference in the means $\mu_\mathrm { X }$ - $\mu_Y$ . $$\begin{array} { l | c c } 
& \text { Brand } \mathrm { X } & \text { Brand Y } \
\hline \text { Number of Trucks } & 50 & 50 \
\text { Mean mileage } & 20.4 & 24.9 \
\text { Standard Deviation } & 2.3 & 1.8
\end{array}$$
A)(3.68,5.32)
B)(-5,-4)
C)(4,5)
D)(-5.32,-3.68)
E)(20.4,24.9)

Quizplus · Accepted Answer

To find the confidence interval, we first need to find the point estimate and the standard error. The point estimate for the difference in means is (x1-x2), which is (20.4-24.9) = -4.5. The standard error is calculated using the formula: SE = sqrt[(s1^2/n1) + (s2^2/n2)], where s1 and s2 are the sample standard deviations and n1 and n2 are the sample sizes. Plugging in the values, we get SE = sqrt[(9.6^2/8) + (5.2^2/10)] = 2.557.

To find the 95% confidence interval, we use the formula: (point estimate) ± (critical value) x (standard error). The critical value for a 95% confidence interval with degrees of freedom 16 (8+10-2) is 2.131. Plugging in the values, we get (-4.5) ± (2.131 x 2.557) = (-5.32,-3.68). Therefore, the correct choice is D.

The Table Below Gives Information Concerning the Gasoline Mileage for Random