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Z0= 1.95 P (0 to Z) =0.4744 = 1- (0.4744 + 0.5) = 0.0256
Z0= -0.25 P (0 to Z) =0.0987 = (0.0987+ 0.5) = 0.5987

Section 2.6 Degree of Freedom= 12 + 12 - 2= 22 a. t0=2.30, 2.074, 2.508; .05,.02 b. t0= 3.41, 2.819, 3.792; 0.01, .001 c. t0= 1.95, 1.321; .200 d. t0= -2.45, -2.074, -2.508

Section 2.7

t0=2.31 t0=3.61 t0= 1.95 t0=2.19 2.18 a. H0: M=.255 H1: M>.255
Z0= .255- .2545/(0.0001)*sq rt(10)=1.58, z(0.05)
P=1-0.94295=0.05705
95%=1-0.94295=0.05705
2.24
H0: M1=16 H1: M1=M2

|Machine 1 |Machine 2 | |Mean |16.015 |16.005 | |Variance |0.00082 |0.000585 | |Sample Number |10 |10 | |

2.25 H0: M1>M2 H1: M1>10

Z0= (162.5-155.0)/sq rt {(1.062/10) + (1.0^2/12)} =1.75, = 0.95994 No based on the Z-value, we will not be rejecting the null hypothesis.

Hypothesis: The flight distance of the lighter paper airplane, M(g) is greater than the flight distance for the heavy paper airplane (with 2 paperclips)

H0: M1 =M2
H1: M1> M2

Flight # |Light Paper Airplane Distance (ft) |Heavy Paper Airplane Distance (ft) | |1 |16 |8 | |2 |17 |8 | |3 |17.25 |8.25 | |4 |17.25 |8 | |5 |17 |7.75 | |6 |16.75 |7.75 | |7 |16.50 |7.75 | |8 |16.25 |8 | |9 |17 |8.25 | |10 |17.5 |8.25 | |

Mean of Light Paper Airplane:

= (16 + 17 +17.25 + 17.25 + 17 + 16.75 +16.50 +16.25 + 17 +17.5) 10

= 16.85

Mean of Heavy Paper Airplane: = ( 8 + 8 + 8.25 + 8 + 7.75 + 7.75 + 7.75 +8 + 8.25 + 8.25) 10

= 8

Variance of Light Paper Airplane:

= [(16-16.85)^2 + (17--16.85 )^2 +(17.25-16.85)^2 + (17.25-16.85)^2 + (17 -16.85)^2 + (16.75 -16.85 )^2 +(16.50 -16.85 )^2 +(16.25 -16.85 )^2 + (17 -16.85 )^2 +(17.5 -16.85 )^2] 10

= 0.2025

Standard Deviation: = Sq Rt(0.2025) = 0.45

Variance of Heavy Paper Airplane: = [(8-8)^2 + (8 -8)^2+ (8.25 -8)^2 + (8 -8)^2 + (7.75 -8)^2+ (7.75 -8)^2 + (7.75 -8)^2 +(8 -8)^2 + (8.25 -8) ^2 + (8.25 -8) ^2] 10

= 0.0375 Standard Deviation: = Sq Rt(0.0375) = 0.193

Calculations |Light Paper Airplane |Heavy Paper Airplane | |Mean |16.85 |8 | |Variance |0.2025 |0.0375 | |Std Dev. |0.45 |0.193 | |N |10 |10 | |

Test Statistics:

=16.85-8 sq rt [(0.2025)/10}+(.0375/10)]

=8.85
.454
=19.49

Based on the test statistics, we will have to reject the null hypothesis, therefore the alternative process will be accepted.

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