A client used two courier services on a random daily basis over a period of 3 months for similar destinations. Delivery times where noted by a clerk. During this time Courier Service A was used 60 times with a sample average delivery time calculated as 42 minutes. During this same time Courier Service B was used 48 times and a sample average delivery time is calculated as 38 minutes. Suppose that the population standard deviation for Courier Service A is 14 minutes and for Courier Service B is 10 minutes. Construct a 95% confidence interval for the difference between the mean delivery times of the two courier services. Give the lower limit.
To construct a 95% confidence interval for the difference between the mean delivery times of the two courier services, we can use the formula:
Confidence Interval = (sample mean difference) ± (critical value) * (standard error of the difference)
First, let's calculate the standard error of the difference:
SE = sqrt[(σ1^2 / n1) + (σ2^2 / n2)]
Where:
σ1 = population standard deviation for Courier Service A = 14 minutes
n1 = number of times Courier Service A was used = 60
σ2 = population standard deviation for Courier Service B = 10 minutes
n2 = number of times Courier Service B was used = 48
SE = sqrt[(14^2 / 60) + (10^2 / 48)]
SE = sqrt[(196/60) + (100/48)]
SE = sqrt[3.27 + 2.08]
SE = sqrt(5.35)
SE = 2.31 (rounded to two decimal places)
Next, let's calculate the critical value. Since we want a 95% confidence interval, we need the critical value for a 2-tailed test with alpha level 0.05/2 = 0.025. Looking up this value in the t-distribution table with (n1 + n2 - 2) degrees of freedom (60 + 48 - 2 = 106), we find the critical value to be approximately 1.984.
Now we can substitute the values into the confidence interval formula:
Confidence Interval = (sample mean difference) ± (critical value) * (standard error of the difference)
Confidence Interval = (42 - 38) ± 1.984 * 2.31
Confidence Interval = 4 ± 1.984 * 2.31
Confidence Interval = 4 ± 4.58
Confidence Interval = (-0.58, 8.58)
The lower limit of the 95% confidence interval for the difference between the mean delivery times of the two courier services is -0.58 minutes.
To construct a 95% confidence interval for the difference between the mean delivery times of the two courier services, we can use the formula:
CI = (x̄₁ - x̄₂) ± Z * √((σ₁² / n₁) + (σ₂² / n₂))
Where:
x̄₁ and x̄₂ are the sample average delivery times for Courier Service A and B, respectively.
σ₁ and σ₂ are the population standard deviations for Courier Service A and B, respectively.
n₁ and n₂ are the sample sizes for Courier Service A and B, respectively.
Z is the Z-score corresponding to the desired confidence level.
Given:
x̄₁ = 42 minutes, x̄₂ = 38 minutes,
σ₁ = 14 minutes, σ₂ = 10 minutes,
n₁ = 60, n₂ = 48,
and we want a 95% confidence interval.
First, let's calculate the Z-score for a 95% confidence level. We can use a standard normal distribution table or a calculator.
The Z-score for a 95% confidence level is approximately 1.96.
Now, substitute the values into the formula:
CI = (42 - 38) ± (1.96) * √((14² / 60) + (10² / 48))
Simplifying the equation:
CI = 4 ± (1.96) * √((196 / 60) + (100 / 48))
CI = 4 ± (1.96) * √(3.27 + 2.08)
CI = 4 ± (1.96) * √(5.35)
Calculating the square root:
CI = 4 ± (1.96) * 2.31
CI = 4 ± 4.52
Finally, calculate the lower limit of the confidence interval:
Lower Limit = 4 - 4.52
Lower Limit ≈ -0.52
Therefore, the lower limit of the 95% confidence interval for the difference between the mean delivery times of the two courier services is approximately -0.52 minutes.
To construct a 95% confidence interval for the difference between the mean delivery times of the two courier services, you can use the formula:
CI = (sample mean A - sample mean B) ± (critical value) * sqrt((sample variance A / sample size A) + (sample variance B / sample size B))
First, let's calculate the standard error of the mean for each courier service:
Standard error A = population standard deviation A / sqrt(sample size A)
Standard error A = 14 / sqrt(60)
Standard error B = population standard deviation B / sqrt(sample size B)
Standard error B = 10 / sqrt(48)
Next, we need to calculate the critical value. For a 95% confidence interval, with two samples, degrees of freedom equal to the smaller sample size minus 1 (df = min(sample size A, sample size B) - 1).
Looking up the critical value in a t-table or using a calculator, for a 95% confidence interval and 107 degrees of freedom, the critical value is approximately 1.983.
Now we can calculate the confidence interval:
CI = (42 - 38) ± (1.983) * sqrt((14^2 / 60) + (10^2 / 48))
Simplifying the equation:
CI = 4 ± 1.983 * sqrt(0.3267 + 0.4167)
Calculating the values inside the square root:
CI = 4 ± 1.983 * sqrt(0.7434)
Simplifying:
CI = 4 ± 1.983 * 0.8624
Calculating the upper and lower limits of the confidence interval:
Upper limit = 4 + 1.983 * 0.8624 = 5.712
Lower limit = 4 - 1.983 * 0.8624 = 2.288
Therefore, the 95% confidence interval for the difference between the mean delivery times of the two courier services is (2.288, 5.712). The lower limit is 2.288.