Question

Construct a

99%

confidence interval for

mu 1 minus mu 2μ1−μ2

with the sample statistics for mean cholesterol content of a hamburger from two fast food chains and confidence interval construction formula below. Assume the populations are approximately normal with unequal variances.

Stats	x overbar 1 equals 134 mg comma s 1 equals 3.64 mg comma n 1 equals 20x1=134 mg, s1=3.64 mg, n1=20	x overbar 2 equals 88 mg comma s 2 equals 2.02 mg comma n 2 equals 16x2=88 mg, s2=2.02 mg, n2=16
ConfidenceConfidence interval wheninterval when variances arevariances are not equal	left parenthesis x overbar 1 minus x overbar 2 right parenthesis minus t Subscript c Baseline StartRoot StartFraction s Subscript 1 Superscript 2 Over n 1 EndFraction plus StartFraction s Subscript 2 Superscript 2 Over n 2 EndFraction EndRoot less than mu 1 minus mu 2 less than left parenthesis x overbar 1 minus x overbar 2 right parenthesis plus t Subscript c Baseline StartRoot StartFraction s Subscript 1 Superscript 2 Over n 1 EndFraction plus StartFraction s Subscript 2 Superscript 2 Over n 2 EndFraction EndRootx1−x2−tcs21n1+s22n2<μ1−μ2<x1−x2+tcs21n1+s22n2
	d.f. is the smaller of n 1n1minus−1 or n 2n2minus−1

Enter the endpoints of the interval.

nothing less than mu 1 minus mu 2 less than nothing<μ1−μ2<

​(Round to the nearest integer as​ needed.)

Answer 1

Using Two sample t test assuming unequal variance.

Given: = 0.01

Population 1:

n1 = 20, = 134, S1 = 3.64

Population 2:

n2 = 16, = 88, S2 = 2.02

Degrees of Freedom:

Critical value:

…………………..….using t table

99% Confidence Interval:

(43.37, 48.63)

Or

43 < 1-2 < 49               .........rounded to the nearest integer