A test of the breaking strength of two different types of cables was conducted using samples of n, = n, = 100 pieces of each type of cable. Cable I Cable II X1 = 1,935 X2 = 1,900 s, = 39 s, = 32 Do the data provide sufficient evidence to indicate a difference in the mean breaking strength of the two cables? Use a = 0.05. State the null and alternative hypotheses. O Ho: (H1 - H2) = 0 versus H: (H1 - H2) + 0 Ho: (H2 - H2) * 0 versus H: (u, - H2) = 0 O Ho: (H1 - H2) = 0 versus H: (H1 - H2) < 0 O Ho: (H, - H2) = 0 versus H: (H, - H2) > 0 Ho: (H1 - H2) < 0 versus H,: (H1 - H2) >0 Find the test statistic and rejection region. (Round your answers to two decimal places. If the test is one-tailed, enter NONE for the unused region.) test statistic Z = rejection region z > z < ucion

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A test of the breaking strength of two different types of cables was conducted using samples of n, = n, = 100 pieces of
each type of cable.
Cable I
Cable II
X, = 1,935
X2 = 1,900
s, = 39
s, = 32
Do the data provide sufficient evidence to indicate a difference in the mean breaking strength of the two cables? Use
a = 0.05.
State the null and alternative hypotheses.
O Ho: (H1 – µ2) = 0 versus H: (H1 – H2) # 0
Ho: (Hz
- H2) # 0 versus H: (µ, - H2) = 0
О но: (и, - и2) %3D 0 versus H,: (и, — и2) < о
O Ho: (H1 – H2) = 0 versus H: (H, – H2) > 0
Но: (и, - и2) <O versus Ha: (и, — и2) > 0
Find the test statistic and rejection region. (Round your answers to two decimal places. If the test is one-tailed, enter NONE
for the unused region.)
test statistic
Z =
rejection region
z >
z <
ucion
Transcribed Image Text:A test of the breaking strength of two different types of cables was conducted using samples of n, = n, = 100 pieces of each type of cable. Cable I Cable II X, = 1,935 X2 = 1,900 s, = 39 s, = 32 Do the data provide sufficient evidence to indicate a difference in the mean breaking strength of the two cables? Use a = 0.05. State the null and alternative hypotheses. O Ho: (H1 – µ2) = 0 versus H: (H1 – H2) # 0 Ho: (Hz - H2) # 0 versus H: (µ, - H2) = 0 О но: (и, - и2) %3D 0 versus H,: (и, — и2) < о O Ho: (H1 – H2) = 0 versus H: (H, – H2) > 0 Но: (и, - и2) <O versus Ha: (и, — и2) > 0 Find the test statistic and rejection region. (Round your answers to two decimal places. If the test is one-tailed, enter NONE for the unused region.) test statistic Z = rejection region z > z < ucion
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