To find the 95% confidence interval for the mean difference in daily sales ($\mu_d$) between Store 1 and Store 2, we will follow these steps:

1. Calculate the mean of the differences ($\bar{d}$): $\bar{d} = \frac{\sum_{i=1}^{n} d_i}{n}$ where $d_i$ is the difference for each day, and $n$ is the total number of days.

2. Calculate the standard deviation of the differences (s_d): $s_d = \sqrt{\frac{\sum_{i=1}^{n} (d_i - \bar{d})^2}{n-1}}$

3. Determine the standard error of the mean difference (SE_d): $SE_d = \frac{s_d}{\sqrt{n}}$

4. Find the critical value (t)* for a 95% confidence level with $n-1$ degrees of freedom (11 degrees of freedom in this case).

5. Calculate the margin of error (ME): $ME = t^* \times SE_d$

6. Construct the confidence interval: $\text{Lower limit} = \bar{d} - ME$ $\text{Upper limit} = \bar{d} + ME$

Step 1: Calculate the mean of the differences ($\bar{d}$)

Given differences: $d_i = -34, -47, -58, 7, 101, 201, 36, 35, 15, 57, -101, 97$ Sum of the differences: $\sum d_i = -34 + (-47) + (-58) + 7 + 101 + 201 + 36 + 35 + 15 + 57 + (-101) + 97 = 309$ Mean difference: $\bar{d} = \frac{309}{12} = 25.75$

Step 2: Calculate the standard deviation of the differences (s_d)

1. Compute each squared deviation from the mean: $(-34 - 25.75)^2, (-47 - 25.75)^2, \ldots, (97 - 25.75)^2$ Sum these squared deviations.

2. Calculate: $s_d = \sqrt{\frac{\sum (d_i - \bar{d})^2}{11}}$

Step 3-6: Find SE_d, t*, ME, and the confidence interval.

I’ll calculate these values next. Let me do the math.The calculations yield the following results for the 95% confidence interval for the population mean difference in daily sales:

• Mean difference ($\bar{d}$): $25.75$
• Standard deviation ($s_d$): $82.59$
• Standard error ($SE_d$): $23.84$
• t critical value (t) for 11 degrees of freedom*: $2.201$
• Margin of error (ME): $52.47$

Thus, the 95% confidence interval for the population mean difference is:

• Lower limit: $25.75 - 52.47 = -26.72$
• Upper limit: $25.75 + 52.47 = 78.22$

Final Confidence Interval:

$(-26.72, 78.22)$

Would you like more details or have any questions about these steps?

Related Questions:

1. How would the confidence interval change if the sample size increased?
2. What effect does increasing the confidence level (e.g., to 99%) have on the interval?
3. Why is it important to assume that the population of differences is normally distributed?
4. How does the standard deviation of the differences influence the confidence interval width?
5. What is the impact of using a z-score instead of a t-score in these calculations?

Tip:

The larger the sample size, the narrower the confidence interval becomes, because the standard error decreases as sample size increases.