In a cooking competition called Heaven's Kitchen, two teams compete against each other to serve a restaurant full of customers. The competitors cook dishes and send them to an executive chef, where it is either approved and sent to the customer, or is deemed unacceptable and sent back to the kitchen. The competition has run for the past 19 years. During the 9th annual Heaven's Kitchen competition, the Ruby team cooked a total of 135 dishes, of which 84 were sent back to the kitchen by the executive chef. In the same period, the competing Sapphire team cooked a total of 189 dishes and had 157 return. Construct a confidence interval to estimate the difference in the proportion of dishes sent back to the Ruby team and the proportion of dishes sent back to the Sapphire team. Use a confidence level of $89 \%$. Round each answer to 4 decimal places. Do not round from one part to the next when performing the calculations, though. a. Find the confidence interval. b. Which of the following does your interval contain? only positive values both positive and negative values only negative values c. What does this mean in terms of the population proportions? We are $89 \%$ confident that, on average, the Ruby team has had more dishes return than the Sapphire team. We are $89 \%$ confident that, on average, the Ruby team has had less dishes return than the Sapphire team. We are $89 \%$ confident that, on average, the Ruby team has had just as many dishes return as the Sapphire team.
Solution
Step 1 ：Calculate the sample proportions for each team. For the Ruby team, the sample proportion (p1) is the number of dishes sent back divided by the total number of dishes cooked, which is \(\frac{84}{135} = 0.6222\). For the Sapphire team, the sample proportion (p2) is \(\frac{157}{189} = 0.8317\).
Step 2 ：Calculate the difference in sample proportions (p1 - p2), which is \(0.6222 - 0.8317 = -0.2095\).
Step 3 ：Calculate the standard error using the formula \(\sqrt{(p1(1-p1)/n1) + (p2(1-p2)/n2)}\), where n1 and n2 are the sample sizes for the Ruby and Sapphire teams, respectively. Substituting the values we have, the standard error is \(\sqrt{(0.6222(1-0.6222)/135) + (0.8317(1-0.8317)/189)} = 0.0561\).
Step 4 ：Find the z-score for an 89% confidence level, which is 1.645 (this value can be found in a standard z-table).
Step 5 ：Calculate the margin of error, which is the z-score times the standard error, or \(1.645 * 0.0561 = 0.0922\).
Step 6 ：Calculate the confidence interval, which is (p1 - p2) ± margin of error, or \(-0.2095 ± 0.0922\). This gives us a confidence interval of \((-0.3017, -0.1173)\).
Step 7 ：The interval contains only negative values.
Step 8 ：In terms of the population proportions, we are 89% confident that, on average, the Ruby team has had less dishes return than the Sapphire team. \(\boxed{-0.3017, -0.1173}\)
