Construct a 90 % confidence interval for mu 1 minus mu 2 with the sample statistics for mean cholesterol content of a hamburger from two fast food chains and confidence interval construction formula below. Assume the populations are approximately normal with unequal variances.
Stats
x overbar 1 equals 57 mg comma s 1 equals 3.62 mg comma n 1 equals 13
x overbar 2 equals 45 mg comma s 2 equals 2.03 mg comma n 2 equals 17
Confidence interval when variances are not equal
left parenthesis x overbar 1 minus x overbar 2 right parenthesis minus t Subscript c Baseline StartRoot StartFraction s Subscript 1 Superscript 2 Over n 1 EndFraction plus StartFraction s Subscript 2 Superscript 2 Over n 2 EndFraction EndRoot less than mu 1 minus mu 2 less than left parenthesis x overbar 1 minus x overbar 2 right parenthesis plus t Subscript c Baseline StartRoot StartFraction s Subscript 1 Superscript 2 Over n 1 EndFraction plus StartFraction s Subscript 2 Superscript 2 Over n 2 EndFraction EndRoot
d.f. is the smaller of n 1minus1 or n 2minus1

To construct a 90% confidence interval for \(\mu_1 - \mu_2\) with the given sample statistics, we will follow these steps:

1. **Calculate the difference between the sample means**:
   \[
   \overline{x}_1 - \overline{x}_2 = 57 \, \text{mg} - 45 \, \text{mg} = 12 \, \text{mg}
   \]

2. **Calculate the standard error of the difference between means**:
   \[
   SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}
   \]
   Substituting the given values:
   \[
   SE = \sqrt{\frac{3.62^2}{13} + \frac{2.03^2}{17}}
   \]
   \[
   SE = \sqrt{\frac{13.1044}{13} + \frac{4.1209}{17}}
   \]
   \[
   SE = \sqrt{1.008 + 0.2424} = \sqrt{1.2504} \approx 1.1182 \, \text{mg}
   \]

3. **Determine the degrees of freedom**:
   The degrees of freedom is the smaller of \(n_1 - 1\) or \(n_2 - 1\):
   \[
   df = \min(13 - 1, 17 - 1) = \min(12, 16) = 12
   \]

4. **Find the critical value \(t_c\) for a 90% confidence interval with \(df = 12\)**:
   From the t-distribution table, for a two-tailed test with \(df = 12\) and a 90% confidence level, the critical value \(t_c\) is approximately 1.782.

5. **Construct the confidence interval**:
   \[
   (\overline{x}_1 - \overline{x}_2) \pm t_c \cdot SE
   \]
   Substituting the values:
   \[
   12 \, \text{mg} \pm 1.782 \cdot 1.1182 \, \text{mg}
   \]
   \[
   12 \, \text{mg} \pm 1.994 \, \text{mg}
   \]

   Therefore, the 90% confidence interval for \(\mu_1 - \mu_2\) is:
   \[
   (12 - 1.994, 12 + 1.994) = (10.006 \, \text{mg}, 13.994 \, \text{mg})
   \]

### Summary:
The 90% confidence interval for the difference in mean cholesterol content of a hamburger between the two fast food chains is approximately \( (10.006 \, \text{mg}, 13.994 \, \text{mg}) \).

Would you like more details or have any questions about this calculation? Here are some related questions for further exploration:

1. How do you interpret the confidence interval in the context of this problem?
2. What assumptions are necessary for the validity of this confidence interval?
3. How would the confidence interval change if the sample sizes were larger?
4. What effect would a higher or lower confidence level have on the interval width?
5. How can we check the assumption of normality in the populations?
6. What are some potential sources of error in this type of analysis?
7. How would the confidence interval differ if the population variances were assumed equal?
8. Can you calculate the confidence interval using a different method, such as bootstrapping?

**Tip:** When interpreting confidence intervals, remember that they provide a range of plausible values for the population parameter based on the sample data, not a definitive estimate of where the parameter lies.

Learn how to construct a 90% confidence interval for the difference in mean cholesterol content between two fast food chains. Includes step-by-step calculations using sample statistics and formulas for unequal variances. Suitable for advanced undergraduate students studying statistics.

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