Question

Use the *t*-distribution to find a confidence interval
for a difference in means μ1-μ2 given the relevant sample results.
Give the best estimate for μ1-μ2, the margin of error, and the
confidence interval. Assume the results come from random samples
from populations that are approximately normally distributed.

A 90% confidence interval for μ1-μ2 using the sample results
x¯1=81.1, s1=10.3, n1=35 and x¯2=67.1, s2=7.9, n2=20

Enter the exact answer for the best estimate and round your answers
for the margin of error and the confidence interval to two decimal
places.

Answer #1

= 81.1, s1 = 10.3, n1 = 35

= 67.1, s2 = 7.9, n2 = 20

c = 90% , = 0.10

1)

Best point estimate =
= 81.1 - 67.1 = 14

**Best point estimate =
14**

2)

formula for margin of error is

where,

df = n1 + n2 - 2 = 35 + 20 2 = 53

1.674

Sp = 9.509528

= 4.462176451

**Margin of error =
4.46**

14 - 4.462176451 , 14 + 4.462176451

**Confidence interval = (
9.54 , 18.46 )**

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