Chapters 20-21 Summative Assessment Part 2

Question

2. An online grocery company is researching the best location to start their new delivery service. They are comparing two local regions, A and B. In a random sample of 505 residents in region A, 26.1% said that they would use an online grocery store. In a random sample of 388 residents in region B, 29.5% said they would use an online grocery store. 

1. Find a 95% confidence interval for the difference in the proportions of residents in these two regions who would use an online grocery store. 

1. The CEO believes that region B will have a higher proportion of potential customers than region A. Does your confidence interval support that conclusion? Explain. 

Answer 1

With the given information, wtih 2 regions, etc., it would be the following:

Region A:

1. n_A = 505
2. p_A = 0.261

Region B:

1. n_B = 388
2. p_B = 0.295

Confidence level: 95% → z = 1.96

1st part of the question:

Step 1: Difference in sample proportions

1. p_B - p_A = 0.295 - 0.261 = 0.034

Step 2: Standard Error

1. SE = √p_A(1 - p_A) / n_A + p_B(1 - p_B) / n_B
2. SE = √(0.261(0.739)) / 505 + (0.295(0.705)) / 388
3. SE = √0.000382 + 0.000536
4. SE = √0.000918 ≈ 0.0303

Step 3: Margin of error

1. ME = z x SE = 1.96(0.0303) ≈ 0.059

Step 4: Confidene interval

1. 0.034 ± 0.059

→ (-0.025, 0.093)

That means that 95% confidence interval would be: (-0.025, 0.093). That means we are 95% condident that the true difference in proportions of residents willing to use an online grocery service (Region B minue Region A) is between -2.5% and 9.3%

2nd part of the question: CEO's Claim

Since the CEO's claim was Region B having higher proportion of potenial customers than region A, the interval support would be no, because 0 is contained in the confidence interval, the data does not provide convincing evidence that Regions B's proportion is higher that Region A's at the 95% confidence level.