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The owner of a popular coffee shop wants to determine if there is a difference between the proportion of customers who use their own cups when they purchase a coffee beverage and the proportion of customers who use their own cups when they purchase an espresso beverage. Customers using their own cups get a 5% discount, which is displayed on the receipt. The owner randomly selects 50 receipts from all coffee purchases and 50 receipts from all espresso purchases. For coffee purchases, 24 receipts showed that the customer used their own cup. For espresso purchases, 18 receipts showed that the customer used their own cup.

Assuming the conditions for inference have been met, what is the 99% confidence interval for the difference in proportion of customers who use their own cups?

A. [tex]\((0.48-0.36) \pm 2.58 \sqrt{\frac{0.48(1-0.48)}{50}+\frac{0.36(1-0.36)}{50}}\)[/tex]

B. [tex]\((0.48-0.36) \pm 2.33 \sqrt{\frac{0.48(1-0.48)}{50}+\frac{0.36(1-0.36)}{50}}\)[/tex]

C. [tex]\((0.52-0.64) \pm 2.58 \sqrt{\frac{0.52(1-0.52)}{100}+\frac{0.64(1-0.64)}{100}}\)[/tex]

D. [tex]\((0.52-0.64) \pm 2.33 \sqrt{\frac{0.52(1-0.52)}{100}+\frac{0.64(1-0.64)}{100}}\)[/tex]



Answer :

GinnyAnswer
To determine the 99% confidence interval for the difference in proportion of customers who use their own cups between coffee and espresso purchases, we will follow these steps:

1. Identify Proportions and Sample Sizes:
- Proportion of customers using their own cups for coffee [tex]\( p_1 \)[/tex]:
[tex]\[ p_1 = \frac{24}{50} = 0.48 \][/tex]
- Proportion of customers using their own cups for espresso [tex]\( p_2 \)[/tex]:
[tex]\[ p_2 = \frac{18}{50} = 0.36 \][/tex]
- Sample sizes for both coffee [tex]\( n_1 \)[/tex] and espresso [tex]\( n_2 \)[/tex]:
[tex]\[ n_1 = 50, \quad n_2 = 50 \][/tex]

2. Calculate the Difference in Proportions:
[tex]\[ \text{Difference} = p_1 - p_2 = 0.48 - 0.36 = 0.12 \][/tex]

3. Determine the Z-Score for 99% Confidence Level:
A 99% confidence level corresponds to a z-score [tex]\( z \)[/tex]:
[tex]\[ z = 2.58 \][/tex]

4. Calculate the Standard Error of the Difference in Proportions:
[tex]\[ \text{SE} = \sqrt{\left(\frac{p_1 (1 - p_1)}{n_1}\right) + \left(\frac{p_2 (1 - p_2)}{n_2}\right)} \][/tex]
Substituting the values:
[tex]\[ \text{SE} = \sqrt{\left(\frac{0.48 \cdot (1 - 0.48)}{50}\right) + \left(\frac{0.36 \cdot (1 - 0.36)}{50}\right)} \][/tex]
[tex]\[ \text{SE} = \sqrt{\left(\frac{0.48 \cdot 0.52}{50}\right) + \left(\frac{0.36 \cdot 0.64}{50}\right)} \][/tex]
[tex]\[ \text{SE} \approx 0.09797958971132713 \][/tex]

5. Calculate the Margin of Error:
[tex]\[ \text{Margin of Error} = z \cdot \text{SE} = 2.58 \cdot 0.09797958971132713 \approx 0.252787341455224 \][/tex]

6. Find the Confidence Interval:
The confidence interval is given by:
[tex]\[ (\text{Difference} - \text{Margin of Error}, \text{Difference} + \text{Margin of Error}) \][/tex]
Substituting the calculated values:
[tex]\[ (0.12 - 0.252787341455224, 0.12 + 0.252787341455224) \][/tex]
[tex]\[ (-0.13278734145522403, 0.372787341455224) \][/tex]

So, the 99% confidence interval for the difference in the proportion of customers who use their own cups between coffee and espresso purchases is [tex]\((-0.1328, 0.3728)\)[/tex].

From the given options, none match the values we have calculated precisely, but the correct closest matching form should reflect the detailed steps shown above.

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