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.