Managerial Statistics Checkpoint 5 Essay

777 Words May 24th, 2012 4 Pages
Checkpoint 5

Complete Exercise 9.13 (The Video Game Satisfaction Case) on page 357 in your textbook

A. H0 ≤ 42 versus Ha >42

b: z = (42.954 - 42)/(2.64/√64) = 2.891

Z values at
0.10 = 1.28 - Reject
0.05 = 1.64 - Reject
0.01 = 2.33 - Reject
0.001 = 3.09 - Accept

Since we are dealing with the right side of the curve the z values are positive.

c:

Z = 2.891 therefore right side = 0.0019
Left side = 0.9981

Since p is = 0.0019 it less than 0.10, 0.05. 0.01 but greater than 0.001

D. Very strong evidence

Complete Exercise 9.19 on page 358 in your textbook. (

A. H0 = 16 versus Ha ≠ 16

b. α = 0.01
S = 0. 1
b.1 Xbar = 16.05 b.2 Xbar = 15.96
b.3 Xbar = 16.02
b.4 Xbar = 15.94

two tailed
…show more content…
a. H0: μ ≠ 42 versus Ha; μ > 42, α= 0.01

b. n = 65
Xbar = 42.95
S = 2.6424 α= 0.01 p-value of .0025

One-Sample T

Test of mu ≠ 42 vs > 42

95% Lower N Mean StDev SE Mean Bound T P
65 42.950 2.642 0.328 42.403 2.90 0.003

t0.01 = -2.33, since 2.90 is greater than 2.33 we can reject the H0 claim and accept that the mean rating is greater than 42.

For a p-value of 0.0025, is 0.1,0.05,0.01 are all greater than 0.0025 therefore reject at these α while 0.001 is less 0.0025 and we would accept α at this level There would be very strong evidence.

Complete Exercise 9.31 on page 362 in your textbook. (Points : 7)

A. H0: μ = 750 versus Ha; μ ≠ 750,
B.
Xbar = 811 s = 19.647
N=5
α =0.01 t = 6.942585 p- value = 0.002261

t = (xbar - µ)/(s/√n) t = (811-750)/(19.647/√5) t = 6.9425 t0.01/2 = 4.604
So reject H0 α = 0.01

A p- value = 0.002261 There would be very strong evidence against H0

5. Complete Exercise 9.42 on page 367 in your textbook.

A: H0: p = 0.95 versus Ha: p< 0.95 Phat = % of customer without repairs = 316/400 = 0.79%
Z = (Phat –P0)/√(P0(1- P0)/N)
Z = 0.79-0.95/√0.95(1-.0.95)/400
Z= -14.683
H0 would be rejected

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