Load the data into CrunchIt! or Minitab and add an “Age” column of data to disti
ID: 3021201 • Letter: L
Question
Load the data into CrunchIt! or Minitab and add an “Age” column of data to distinguish old from new firms. (Use number 0 for old and number 1 for new.)
Do summary stats grouped by “Age” for Bill97, Bill98, Staff, Arch, and Eng. (You will compare the standard deviations of the two groups (old and new) to determine if it is OK to pool the standard deviations).
Shows steps taken to obtain information in Minitab or Crunch it. Then do two-sample t test factored by “age” for Bill97, etc.
Analyze the results. Determine for which data there is a difference between the old and the new firms. Please explain results.
See Data below to enter into Minitab or Crunch it:
Name TotalBill02 ArchBill02 ArchBill01 N_Arch N_Eng N_Staff Yr_Estab Age BSA 29.5 24.8 23.7 39 36 240 1975 1 CSO 12.1 9 11.1 17 1 66 1961 0 American 18.1 4.9 5.1 9 35 168 1966 0 Schmidt 10.5 10.5 9.4 17 5 80 1976 1 Browning 12.2 12.2 10.4 22 0 70 1967 0 OdleMcG 5.1 2.4 4.2 6 2 47 1916 0 Ratio 9.6 8.8 10.2 16 0 62 1983 1 Cripe 15.3 4.6 3.1 7 17 91 1937 0 InterDes 5.9 5.9 6.2 19 6 55 1975 1 RQAW 6.7 2 2 7 11 72 1954 0 Gibraltar 7.2 6.1 5.9 13 7 66 1996 1 Fanning 10.6 6.9 6.8 12 5 64 1983 0 Schenkel 4.4 2.6 2.3 5 0 17 1958 0 Sebree 2 2 1.5 2 0 12 1973 1 Woollen 2.4 2.4 2.6 8 0 15 1955 0 Rowland 3.3 2.9 2.9 6 0 27 1968 0 DLZ 7.5 2.5 1.5 6 15 58 1978 1 Gove 6 2 2.5 2 3 16 1985 0 MECA 1.7 1.7 1.3 1 0 13 1989 0 Axis 1.6 1.6 1.5 5 0 10 1996 1 Partenh 1.6 1.6 1.1 3 0 9 1987 1Explanation / Answer
Descriptive Statistics: NewTotalBill, C11, New ArchBill, C12, NewArchBill0, C13, ...
Variable Mean Variance
NewTotalBill02 8.38 74.12
C11 8.16 28.76
New ArchBill02 7.09 54.73
C12 4.467 10.981
NewArchBill01 6.78 52.73
C13 4.525 10.688
NewN_Arch 13.33 133.75
C14 8.50 35.91
N_Eng 7.67 137.75
C15 6.17 110.70
N_Staff 65.8 4996.7
C16 55.5 1995.9
Yr_Estab 1982.1 81.1
C17 1961.6 422.3
Age 1.0000 0.000000
C18 0.000000 0.000000
Two-Sample T-Test and CI: NewTotalBill02, C11 (can't assume equal variance)
Two-sample T for NewTotalBill02 vs C11
N Mean StDev SE Mean
NewTotalBill02 9 8.38 8.61 2.9
C11 12 8.16 5.36 1.5
Difference = (NewTotalBill02) - (C11)
Estimate for difference: 0.22
95% CI for difference: (-6.89, 7.32)
T-Test of difference = 0 (vs ): T-Value = 0.07 P-Value = 0.947 DF = 12
Two-Sample T-Test and CI: New ArchBill02, C12 (can't assume equal variance)
Two-sample T for New ArchBill02 vs C12
N Mean StDev SE Mean
New ArchBill02 9 7.09 7.40 2.5
C12 12 4.47 3.31 0.96
Difference = (New ArchBill02) - (C12)
Estimate for difference: 2.62
95% CI for difference: (-3.27, 8.52)
T-Test of difference = 0 (vs ): T-Value = 0.99 P-Value = 0.345 DF = 10
Two-Sample T-Test and CI: NewArchBill01, C13 (can't assume equal variance)
Two-sample T for NewArchBill01 vs C13
N Mean StDev SE Mean
NewArchBill01 9 6.78 7.26 2.4
C13 12 4.52 3.27 0.94
Difference = (NewArchBill01) - (C13)
Estimate for difference: 2.25
95% CI for difference: (-3.54, 8.04)
T-Test of difference = 0 (vs ): T-Value = 0.87 P-Value = 0.406 DF = 10
Two-Sample T-Test and CI: NewN_Arch, C14 (can't assume equal variance)
Two-sample T for NewN_Arch vs C14
N Mean StDev SE Mean
NewN_Arch 9 13.3 11.6 3.9
C14 12 8.50 5.99 1.7
Difference = (NewN_Arch) - (C14)
Estimate for difference: 4.83
95% CI for difference: (-4.47, 14.13)
T-Test of difference = 0 (vs ): T-Value = 1.14 P-Value = 0.277 DF = 11
Two-Sample T-Test and CI: N_Eng, C15 (can assume equal variance)
Two-sample T for N_Eng vs C15
N Mean StDev SE Mean
N_Eng 9 7.7 11.7 3.9
C15 12 6.2 10.5 3.0
Difference = (N_Eng) - (C15)
Estimate for difference: 1.50
95% CI for difference: (-8.70, 11.70)
T-Test of difference = 0 (vs ): T-Value = 0.31 P-Value = 0.762 DF = 19
Both use Pooled StDev = 11.0493
Two-Sample T-Test and CI: N_Staff, C16 (can't assume equal variance)
Two-sample T for N_Staff vs C16
SE
N Mean StDev Mean
N_Staff 9 65.8 70.7 24
C16 12 55.5 44.7 13
Difference = (N_Staff) - (C16)
Estimate for difference: 10.3
95% CI for difference: (-48.2, 68.8)
T-Test of difference = 0 (vs ): T-Value = 0.38 P-Value = 0.709 DF = 12
Two-Sample T-Test and CI: Yr_Estab, C17 (can't assume equal variance)
Two-sample T for Yr_Estab vs C17
N Mean StDev SE Mean
Yr_Estab 9 1982.11 9.01 3.0
C17 12 1961.6 20.5 5.9
Difference = (Yr_Estab) - (C17)
Estimate for difference: 20.53
95% CI for difference: (6.36, 34.70)
T-Test of difference = 0 (vs ): T-Value = 3.09 P-Value = 0.008 DF = 15
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.