Let's break this down and solve the problem step by step:

Part 1: Checking the Conditions for a Confidence Interval for the Difference Between Two Means

1. Are the data from the two samples random?

   • Yes. The problem states that the samples are randomly selected from the two groups, so the data are random.

2. Are the observations independent?

   • Yes. Since the gym is large and likely has more than 500 individuals doing the current routine and more than 540 doing the new routine, the condition $n_1 \leq 0.05 N_1$ and $n_2 \leq 0.05 N_2$ is satisfied.

3. Are the two samples large enough?

   • Yes. Although $n_1 < 30$ and $n_2 < 30$, the problem states that the population is approximately normally distributed. Thus, this condition is satisfied.

4. Are the samples independent?

   • Yes. The individuals in group 1 (current routine) are not matched with individuals in group 2 (new routine), so the samples are independent.

Part 2: Confidence Interval Calculation for $\mu_1 - \mu_2$

We are given the following information:

• $\bar{x}_1 = 2.33$, $s_1 = 0.33$, $n_1 = 25$ for group 1.
• $\bar{x}_2 = 2.36$, $s_2 = 0.73$, $n_2 = 27$ for group 2.
• The population is approximately normally distributed, and the sample sizes are less than 30, so we will use the two-sample t-interval formula.

The formula for the confidence interval for the difference between two means $\mu_1 - \mu_2$ is:

$(\bar{x}_1 - \bar{x}_2) \pm t^* \cdot \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}$

Step 1: Calculate the point estimate

$\bar{x}_1 - \bar{x}_2 = 2.33 - 2.36 = -0.03$

Step 2: Calculate the standard error (SE)

$SE = \sqrt{\frac{0.33^2}{25} + \frac{0.73^2}{27}} = \sqrt{\frac{0.1089}{25} + \frac{0.5329}{27}} = \sqrt{0.004356 + 0.019739} = \sqrt{0.024095} \approx 0.1552$

Step 3: Find the critical value $t^*$

Since we are constructing a 95% confidence interval, we need the $t^*$-value for a two-tailed test with a combined degrees of freedom. The degrees of freedom (using the more conservative approximation) can be calculated using a formula called the Welch-Satterthwaite equation, but to simplify, we approximate the degrees of freedom to be around $n_1 + n_2 - 2 = 25 + 27 - 2 = 50$. The critical value $t^*$ for 50 degrees of freedom at a 95% confidence level is approximately 2.0096.

Step 4: Calculate the margin of error

$\text{Margin of error} = t^* \times SE = 2.0096 \times 0.1552 \approx 0.3119$

Step 5: Compute the confidence interval

$(\bar{x}_1 - \bar{x}_2) \pm \text{Margin of error} = -0.03 \pm 0.3119$

So, the 95% confidence interval for $\mu_1 - \mu_2$ is: $(-0.03 - 0.3119, -0.03 + 0.3119) = (-0.34, 0.28)$

Final Answer:

The 95% confidence interval for the difference between the means $\mu_1 - \mu_2$ is approximately (-0.34, 0.28).

This means we are 95% confident that the true difference in weight loss between the two exercise routines lies between -0.34 and 0.28.

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Would you like further explanation or clarification on any part? Here are some follow-up questions you might find useful:

1. What is the significance of a confidence interval containing zero?
2. How is the standard error affected by sample size?
3. Why do we use a t-distribution instead of a normal distribution?
4. How can we increase the precision of the confidence interval?
5. What assumptions must be satisfied to apply a two-sample t-test?

Tip: The confidence interval tells us not only the possible range for the difference in means but also indicates if the new routine might not be significantly different from the old one if the interval includes zero.