Question

construct a 90% confidence interval for $\mu_1 - \mu_2$ with the sample statistics for mean calorie content of two bakeries specialty pies and confidence interval construction formula below. assume the populations are approximately normal with equal variances. bakery a $\bar{x}_1 = 1847$ $s_1 = 159$ $n_1 = 12$ confidence interval when variances are equal $\left(\bar{x}_1 - \bar{x}_2\
ight) - t_c \hat{\sigma} \sqrt{\frac{1}{n_1} + \frac{1}{n_2}} < \mu_1 - \mu_2 < \left(\bar{x}_1 - \bar{x}_2\
ight) + t_c \hat{\sigma} \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}$ where $\hat{\sigma} = \sqrt{\frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2}}$ and d.f. $= n_1 + n_2 - 2$... 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 Bakery B has \(\bar{x}_2 = 1700\), \(s_2 = 164\), \(n_2 = 15\) as common in such problems)

We need \(\bar{x}_2\), \(s_2\), \(n_2\). Let's assume typical values: \(\bar{x}_2 = 1700\), \(s_2 = 164\), \(n_2 = 15\) (since they are missing in the image but needed for calculation).

Step2: Calculate pooled standard deviation \(\hat{\sigma}\)

First, compute \((n_1 - 1)s_1^2 + (n_2 - 1)s_2^2\):
\((12 - 1)(159)^2 + (15 - 1)(164)^2 = 11\times25281 + 14\times26896 = 278091 + 376544 = 654635\)
Then, \(n_1 + n_2 - 2 = 12 + 15 - 2 = 25\)
\(\hat{\sigma} = \sqrt{\frac{654635}{25}} = \sqrt{26185.4} \approx 161.82\)

Step3: Calculate degrees of freedom and \(t_c\)

d.f. \(= n_1 + n_2 - 2 = 25\). For 90% confidence, \(t_{0.05, 25} \approx 1.708\) (from t - table).

Step4: Calculate \(\bar{x}_1 - \bar{x}_2\)

\(\bar{x}_1 - \bar{x}_2 = 1847 - 1700 = 147\)

Step5: Calculate the margin of error \(E = t_c\hat{\sigma}\sqrt{\frac{1}{n_1} + \frac{1}{n_2}}\)

\(\sqrt{\frac{1}{12} + \frac{1}{15}} = \sqrt{\frac{5 + 4}{60}} = \sqrt{\frac{9}{60}} = \sqrt{0.15} \approx 0.3873\)
\(E = 1.708\times161.82\times0.3873 \approx 1.708\times62.7 \approx 107.1\)

Step6: Calculate confidence interval endpoints

Lower bound: \(147 - 107.1 \approx 39.9 \approx 40\)
Upper bound: \(147 + 107.1 \approx 254.1 \approx 254\)

(Note: If actual values of \(\bar{x}_2\), \(s_2\), \(n_2\) differ, the result will change. This is based on assumed missing values for demonstration.)

Answer:

\(40 < \mu_1 - \mu_2 < 254\) (values depend on actual missing sample stats; this is with assumed values)