Question

In: Statistics and Probability

1. Your company makes an uber-bouncy-tennis ball for professional players that bounces an average of 57...

1. Your company makes an uber-bouncy-tennis ball for professional players that bounces an average of 57 inches when dropped from 100 inches (within international tennis regulations) over 900 drops when new. You’ve been tipped off that your material supplier may have changed the quality of their yellow fuzz product without any warning and you wonder how much of an effect, if any, this will have on the quality and bounciness of your tennis balls. You take a random sampling of n=20 samples. The data you collect (inches of bounce height) are below.

56.39

57.62

56.53

56.78

57.23

56.82

57.21

56.80

56.17

57.66

56.41

56.41

56.25

56.51

56.31

56.34

56.18

56.94

57.02

Double check your answer by finding the confidence bounds on the mean bounce height. Do your results agree? Comments?

Assuming that this was a screening experiment, what do you do next and/or what do you tell your boss about both your existing data and your ideas for the future?

Solutions

Expert Solution

56.39
57.62
56.53
56.78
57.23
56.82
57.21
56.8
56.17
57.66
56.41
56.41
56.25
56.51
56.31
56.34
56.18
56.94
57.02
mean 56.71474
sd 0.46311

Consider only Two tailed Values (given all models)

Hypothesis : α= 0.05
df 19 n-1 Given
Ho: μ1​ = μ0 X bar 56.7147
Ha: μ1​ not = μ0 (Two tailed) μ 57
S 0.4631
t Critical Value : n 20
tc -1.729132812 T.INV(D1,9) LEFT
1.729132812 T.INV(1-D1,9) RIGHT
2.093024054 T.INV.2T(D1,D2) TWO
ts <= tc LEFT To reject
ts >= tc RIGHT To reject
ts < for - tc TWO To reject
ts > for + tc TWO To reject
Test :
t -2.75512932 (X bar-μ )/(S/SQRT(n))
P value : CI
tc 2.093024 T.INV.2T(D1,D2)
p 0.006297046 T.DIST(ts,df,TRUE) LEFT
0.993702954 T.DIST.RT(ts,df) RIGHT Upper 56.93144 X bar + tc*(S/SQRT(n))
0.012594092 T.DIST.2T(-ts,df) TWO Lower 56.49796 X bar - tc*(S/SQRT(n))
Decision :
P value < α Reject H0
P value > α Do not reject

P value < 0.05, reject H0. There is enough evidence to conclude that population mean is different

CI = (56.49796, 56.93144) does not contain mean 57

Both results are same


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