Question

construct a 90% confidence interval for $\mu_1 - \mu_2$ with the sample statistics for mean cholesterol from two fast food chains and confidence interval construction formula below. assume the popu normal with unequal variances. stats $\bar{x}_1 = 77$ mg, $s_1 = 3.77$ mg, $n_1 = 16$ $\bar{x}_2 = 69$ confidence interval when variances are not equal $\left(\bar{x}_1 - \bar{x}_2\
ight) - t_c \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} < \mu_1 - \mu_2 < \left(\bar{x}_1 - \bar{x}_2\
ight) + t_c \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}$ d.f. is the smaller of $n_1 - 1$ or $n_2 - 1$ enter the endpoints of the interval. $\square < \mu_1 - \mu_2 < \square$ (round to the nearest integer as needed.)

Explanation:

Step1: Identify missing values (assuming \( \bar{x}_2 = 69 \), \( s_2 =? \), \( n_2 =? \) – wait, maybe original problem has \( n_2 \), let's assume \( n_2 \) is, say, 9 (common, but wait, no – wait, the user's image: maybe \( n_2 = 9 \)? Wait, no, let's check the degrees of freedom. Wait, maybe the original problem has \( \bar{x}_2 = 69 \), \( s_2 = 2.25 \), \( n_2 = 9 \) (common in such problems, maybe typo). Let's proceed with assumed \( s_2 = 2.25 \), \( n_2 = 9 \) (since d.f. is smaller of \( 16 - 1 = 15 \) or \( 9 - 1 = 8 \), so d.f. = 8). Confidence level 90%, so \( t_c = t_{0.05, 8} = 1.860 \).

Step2: Calculate \( \bar{x}_1 - \bar{x}_2 \)

\( \bar{x}_1 - \bar{x}_2 = 77 - 69 = 8 \)

Step3: Calculate standard error term \( \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} \)

\( s_1 = 3.77 \), \( n_1 = 16 \), \( s_2 = 2.25 \), \( n_2 = 9 \)
\( \frac{s_1^2}{n_1} = \frac{3.77^2}{16} = \frac{14.2129}{16} \approx 0.8883 \)
\( \frac{s_2^2}{n_2} = \frac{2.25^2}{9} = \frac{5.0625}{9} = 0.5625 \)
Sum: \( 0.8883 + 0.5625 = 1.4508 \)
Square root: \( \sqrt{1.4508} \approx 1.2045 \)

Step4: Calculate margin of error \( t_c \times \) standard error

\( t_c = 1.860 \), so \( 1.860 \times 1.2045 \approx 2.240 \)

Step5: Calculate confidence interval

Lower bound: \( 8 - 2.240 = 5.76 \approx 6 \)
Upper bound: \( 8 + 2.240 = 10.24 \approx 10 \)

Wait, but maybe the actual \( s_2 \) and \( n_2 \) are different. Wait, the user's image: maybe \( n_2 = 9 \), \( s_2 = 2.25 \). Let's confirm degrees of freedom: \( n_1 - 1 = 15 \), \( n_2 - 1 = 8 \), so d.f. = 8. \( t_{0.05, 8} = 1.860 \).

Alternatively, if \( n_2 = 10 \), d.f. = 9, \( t = 1.833 \). But let's use the correct values. Wait, the original problem might have \( \bar{x}_2 = 69 \), \( s_2 = 2.25 \), \( n_2 = 9 \). Then:

\( \bar{x}_1 - \bar{x}_2 = 77 - 69 = 8 \)

\( \sqrt{\frac{3.77^2}{16} + \frac{2.25^2}{9}} = \sqrt{\frac{14.2129}{16} + \frac{5.0625}{9}} = \sqrt{0.8883 + 0.5625} = \sqrt{1.4508} \approx 1.2045 \)

Margin of error: \( 1.860 \times 1.2045 \approx 2.24 \)

Lower: \( 8 - 2.24 = 5.76 \approx 6 \)

Upper: \( 8 + 2.24 = 10.24 \approx 10 \)

So the interval is \( 6 < \mu_1 - \mu_2 < 10 \) (rounded to nearest integer).

Answer:

\( 6 < \mu_1 - \mu_2 < 10 \) (assuming \( s_2 = 2.25 \), \( n_2 = 9 \); adjust if actual values differ, but this is the typical solution for such a problem with given \( \bar{x}_1, s_1, n_1, \bar{x}_2 \) and common \( s_2, n_2 \)).