Construct a 95% 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 123 mg comma s 1 equals 3.79 mg comma n 1 equals 20x1=123 mg, s1=3.79 mg, n1=20 x overbar 2 equals 87 mg comma s 2 equals 2.05 mg comma n 2 equals 15x2=87 mg, s2=2.05 mg, n2=15 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 Question content area bottom Part 1 Enter the endpoints of the interval. nothing less than mu 1 minus mu 2 less than nothingenter your response here<μ1−μ2<enter your response here Assume the populations are approximately normal with unequal variances.Assume the populations are approximately normal with unequal variances.

Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018

18th Edition

ISBN:9780079039897

Author:Carter

Publisher:Carter

Chapter10: Statistics

Section10.3: Measures Of Spread

Problem 1GP

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Related questions

Question

Construct a

95%

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 123 mg comma s 1 equals 3.79 mg comma n 1 equals 20x1=123 mg, s1=3.79 mg, n1=20	x overbar 2 equals 87 mg comma s 2 equals 2.05 mg comma n 2 equals 15x2=87 mg, s2=2.05 mg, n2=15
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