Question

Two samples are taken with the following sample means, sizes, and standard deviations
¯x1 = 25 ¯x2 = 23
n1 = 54 n2 = 73
s1 = 5 s2 = 3

Estimate the difference in population means using a 88% confidence level. Use a calculator, and do NOT pool the sample variances. Round answers to the nearest hundredth.

Answer 1

Sample Size Sample mean Sample S.D na = 54 Sz = 5 Sample 1 Sample 2 X4 = 25 X2 =23 n2 = 73 S2 = 3 2 Si 2 S2 Standard Error = + = sqrt[(52/54)+(32/73)] = 0.7657 ni n2 2 si ni + sz n2 Degrees of freedom = 2 1 1 Si ni + sz n2 n1 n2 = [(52/54)+(3/73)]°/[(1/(54-1))*(52/54)2+(1/(73-1))*(32/73)?] = 80.77-80 Confidence Level = 88 Significance Level = a = (100-88)% = 0.12 Critical value = t* = 1.572 [using Excel =TINV(0.12,80)] Margin of Error = t*(standard error )= 1.572*0.7657 = 1.20363 Lower Limit = (X1-82 )- Margin of Error = (25 - 23) - 1.20363 = 0.7964 Upper Limit = (X1-X2 )+ Margin of Error = (25 - 23) + 1.20363 = 3.2036 Ans: 0.7964 <Mi - Hz <3.2036