Final answer:
- a. The confidence interval for the difference between the two population means is (8.899, 11.301).
- b.The confidence interval for the difference between the two population means is (-5.599, 25.799).
- c.The 99% confidence interval for the difference between the two population means is approximately 6.82 to 13.38.
Step-by-step explanation:
a. To construct a 99% confidence interval for the difference between the two population means when σ₁ = 3, σ₂ = 2, n₁ = 60, and n₂ = 60, we can use the formula:
CI = (x₁ - x₂) ± Z * √((σ₁² / n1) + (σ₂² / n₂))
where x1 and x₂ are the sample means, σ1 and σ₂ are the population standard deviations, n1 and n₂ are the sample sizes, and Z is the z-value corresponding to the desired confidence level.
Substituting the given values into the formula:
CI = (47.9 - 37.8) ± 2.58 * √((3² / 60) + (2² / 60))
CI = 10.1 ± 2.58 * √(0.15 + 0.0667)
CI = 10.1 ± 2.58 * √(0.2167)
CI = 10.1 ± 2.58 * 0.4656
CI = 10.1 ± 1.201
The confidence interval for the difference between the two population means is (8.899, 11.301).
b. When σ₁ = σ₂, we can use the formula:
CI = (x₁ - x₂) ± t * √((s₁² / n₁) + (s₂² / n₂))
where x₁ and x₂ are the sample means, s₁ and s₂ are the sample standard deviations, n₁ and n₂ are the sample sizes, and t is the t-value corresponding to the desired confidence level.
Since we want a 99% confidence interval and the sample sizes are small (n₁ = 10, n₂ = 10), we need to use the t-distribution and the t-value will be approximately 3.250 (from the t-table).
Substituting the given values into the formula:
CI = (47.9 - 37.8) ± 3.250 * √((3² / 10) + (2²/ 10))
CI = 10.1 ± 3.250 *√(0.9 + 0.4)
CI = 10.1 ± 3.250 *√(1.3)
CI = 10.1 ± 3.250 * 1.1402
CI = 10.1 ± 3.699
The confidence interval for the difference between the two population means is (-5.599, 25.799).
c) To construct a 99% confidence interval for the difference between the two population means when σ1 ≠ σ₂, s₁ = 3, s₂ = 2, n₁= 10, and n₂ = 10, we can use the formula:
CI = (x₁ - x₂) ± t * √((s₁² / n₁) + (s₂²/ n₂))
where x₁ and x₂ are the sample means, s₁ and s₂ are the sample standard deviations, n₁ and n₂ are the sample sizes, and t is the t-value corresponding to the desired confidence level.
First, let's calculate the difference between the sample means:
x₁ - x₂ = 47.9 - 37.8 = 10.1
Next, let's determine the t-value for a 99% confidence level with (n₁+ n₂ - 2) degrees of freedom. Since both sample sizes are 10, the degrees of freedom will be 10 + 10 - 2 = 18. Using a t-table or a statistical software, the t-value for a 99% confidence level and 18 degrees of freedom is approximately 2.878.
Now, let's calculate the standard error of the difference:
√((s₁² / n1) + (s₂² / n₂)) =√((3² / 10) + (2²/ 10)) =√(0.9 + 0.4) = √(1.3) ≈ 1.14
Finally, we can construct the 99% confidence interval:
CI = (10.1) ± (2.878 * 1.14) = 10.1 ± 3.28
Therefore, the 99% confidence interval for the difference between the two population means is approximately 6.82 to 13.38...
Your question is incomplete, but most probably the full question was:
Two random samples were selected independently from populations having normal distributions. The statistics given below were extracted from the samples. Complete parts a through c.
x₁=47.9
x2=37.8.
a. If σ₁=3 and σ₂=2 and the sample sizes are n₁=60 and n₂=60, construct a 99% confidence interval for the difference between the two population means.
The confidence interval is
≤μ₁−μ₂≤
(Round to two decimal places as needed.)
b. If σ₁=σ₂, s₁=3, and s₂=2, and the sample sizes are n₁=10 and n₂=10, construct a 99% confidence interval for the difference between the two population means.
The confidence interval is?
nothing≤μ1−μ₂≤nothing.
(Round to two decimal places as needed.)
c. If σ₁≠σ₂, s1=3, and s₂=2, and the sample sizes are n₁=10 and n₂=10, construct a 99% confidence interval for the difference between the two population means.
The confidence interval is?
nothing≤μ₁−μ₂≤nothing.
(Round to two decimal places as needed.)