A popular Italian eatery has had a dispute among the wait staff regarding the amount of tips waiters receive on the lunch shift and on the dinner shift. Before making any adjustments in schedules or other​ policies, the restaurant manager wants to estimate the difference in the average tips earned by waiters on the lunch schedule and on the dinner schedule. To do​ this, she randomly selected 6060 tip reports for lunch servers over the past year and 5050 tip reports from dinner servers. The results given below were observed. Assuming equal population​ variances, develop and interpret aa 9595​% confidence interval estimate for the difference between the population mean tips earned by lunch servers and by dinner servers.
Lunch
Dinner
Mean Tips
​$64.964.90
​$101.3101.30
St. Dev. Tips
​$11.811.80
​$13.5913.59
Question content area bottom
Part 1
Let sample 1 be the sample from the lunch servers and let sample 2 be the sample from the dinner servers.
The 9595​% confidence interval is enter your response hereless than or equals≤left parenthesis mu 1 minus mu 2 right parenthesisμ1−μ2less than or equals≤enter your response here.
​(Round to two decimal places as​ needed.)
Part 2
The mean tips earned
▼
are not
are
the same during the lunch and dinner​ shifts, because the confidence interval
▼
does not contain
contains
the value
▼
negative 1.−1.
0.0.
1.1.

To construct a 95% confidence interval for the difference between the mean tips earned by lunch servers ($ \mu_1 $) and dinner servers ($ \mu_2 $), we can follow these steps:

### Step 1: Identify the sample statistics

- **Lunch (Sample 1)**:
  - Mean ($ \bar{x}_1 $) = $64.90
  - Standard deviation ($ s_1 $) = $11.81
  - Sample size ($ n_1 $) = 6060

- **Dinner (Sample 2)**:
  - Mean ($ \bar{x}_2 $) = $101.30
  - Standard deviation ($ s_2 $) = $13.59
  - Sample size ($ n_2 $) = 5050

### Step 2: Calculate the pooled standard deviation

The pooled standard deviation ($ s_p $) can be calculated using the formula:

\[
s_p = \sqrt{ \frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2} }
\]

Substituting the values:

\[
s_p = \sqrt{ \frac{(6060 - 1)(11.81^2) + (5050 - 1)(13.59^2)}{6060 + 5050 - 2} }
\]

### Step 3: Calculate the confidence interval

The confidence interval for the difference in means is given by:

\[
(\bar{x}_1 - \bar{x}_2) \pm t^* \cdot s_{d}
\]

Where:
- $ t^* $ is the critical value from the t-distribution for a 95% confidence level.
- $ s_d = s_p \sqrt{ \frac{1}{n_1} + \frac{1}{n_2} } $

### Step 4: Determine the degrees of freedom

The degrees of freedom ($ df $) can be calculated as:

\[
df = n_1 + n_2 - 2
\]

### Step 5: Perform the calculations

1. **Calculate the pooled standard deviation ($ s_p $)**.
2. **Find $ t^* $** for $ df $ at 95% confidence.
3. **Calculate the standard error ($ s_d $)**.
4. **Construct the confidence interval**.

Let's calculate the values step-by-step.

### Calculation

1. **Pooled standard deviation ($ s_p $)**:

   \[
   s_p = \sqrt{ \frac{(6060 - 1)(11.81^2) + (5050 - 1)(13.59^2)}{6060 + 5050 - 2} }
   \]

   Calculate $ s_p $:
   - $ s_p \approx \sqrt{ \frac{(6059)(139.6561) + (5049)(184.7081)}{11108} } $
   - $ s_p \approx \sqrt{ \frac{846206.882 + 932110.1529}{11108} } $
   - $ s_p \approx \sqrt{ \frac{1778317.035}{11108} } \approx 11.91 $

2. **Degrees of freedom**:

   \[
   df = 6060 + 5050 - 2 = 11108
   \]

   Using a t-table or calculator, find $ t^* $ for 95% confidence level and 11108 degrees of freedom (approximately 1.96).

3. **Calculate the standard error ($ s_d $)**:

   \[
   s_d = s_p \sqrt{ \frac{1}{n_1} + \frac{1}{n_2} } = 11.91 \sqrt{ \frac{1}{6060} + \frac{1}{5050} }
   \]

   \[
   s_d \approx 11.91 \sqrt{0.000165 + 0.000198} \approx 11.91 \sqrt{0.000363} \approx 11.91 \times 0.01904 \approx 0.227
   \]

4. **Calculate the confidence interval**:

   \[
   (\bar{x}_1 - \bar{x}_2) \pm t^* \cdot s_d
   \]

   \[
   (64.90 - 101.30) \pm 1.96 \cdot 0.227
   \]

   \[
   -36.40 \pm 0.445
   \]

   Thus, the 95% confidence interval is:

   \[
   (-36.84, -35.96)
   \]

### Final Answer

**Part 1**: The 95% confidence interval is:

\[
\mu_1 - \mu_2: -36.84 \leq (\mu_1 - \mu_2) \leq -35.96
\]

**Part 2**: The mean tips earned **are not** the same during the lunch and dinner shifts, because the confidence interval **does not contain** the value **0.0**.

### Questions for Further Understanding

1. What factors might influence the difference in tips between lunch and dinner shifts?
2. How would the confidence interval change if the sample sizes were different?
3. What assumptions are we making by using the t-distribution in this analysis?
4. How would you interpret the confidence interval in a practical context for the restaurant?
5. What additional data might help in understanding the tipping patterns more comprehensively?

### Tip
When estimating differences in means, always ensure that the samples are representative and that the assumptions of normality and equal variances are checked.

A popular Italian eatery has had a dispute among the wait staff regarding the amount of tips waiters receive on the lunch shift and on the dinner shift. The restaurant manager wants to estimate the difference in the average tips earned by waiters on the lunch schedule and on the dinner schedule. A random selection of 6060 tip reports for lunch servers and 5050 tip reports from dinner servers shows lunch tips mean $64.90 and dinner tips mean $101.30. Construct a 95% confidence interval for the difference in population means.

Explore the construction of a 95% confidence interval for the difference in average tips earned by waiters during lunch and dinner shifts at a popular Italian eatery. With lunch mean tips at $64.90 and dinner at $101.30, learn the statistical methods involved.

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