Question

cnsider the following data rom wo populations are normally distributed. dependent samples th equal population var ances on struct a 9 % con der ce te val o es mate he dif rn ein po ulation means. Assume the popula on variances are equal and at the X1-36.42 S1-8.8 n1 #18 82 9.1 n2 19 ge 1 Click here to see the t distribution table page 2 The 90% confidence interval is( Round to two decimal places as needed.) DD

Answer 1

We will use t-statistic and two sample means (M₁ and M₂) to generate an interval estimate of the difference between two population means (μ₁ and μ₂).

The formula for estimation is:

μ₁ - μ₂ = (M₁ - M₂) ± ts_{(M1 - M2)}

where:

M₁ & M₂ = sample means
t = t statistic determined by confidence level
s_{(M1 - M2)} = standard error = √((s²_p/n₁) + (s²_p/n₂))

Detailed calculation is as shown below:

Pooled Variance
s²_p = ((df₁)(s²₁) + (df₂)(s²₂)) / (df₁ + df₂) = 2807.06 / 35 = 80.2 where df₁ = n₁ - 1 = 17 and df₂ = n₂ - 1 = 18

Standard Error
s_{(M1 - M2)} = √((s²_p/n₁) + (s²_p/n₂)) = √((80.2/18) + (80.2/19)) = 2.95

Confidence Interval
μ₁ - μ₂ = (M₁ - M₂) ± ts_{(M1 - M2)} = 4.2 ± (1.69 * 2.95) = 4.2 ± 4.977

Hence,

μ₁ - μ₂ = (M₁ - M₂) = 4.2, 90% CI [-0.777, 9.177].

You can be 90% confident that the difference between your two population means (μ₁ - μ₂) lies between -0.777 and 9.177.