contestada

For the following pairs of assertions, indicate which do not
comply with our rules for setting up hypotheses and why (the
subscripts 1 and 2 differentiate between quantities for two
different populations or samples): (a) H0: μ = 100, Ha: μ > 100
These hypotheses comply with our rules. H0 cannot include equality,
so these hypotheses are not in compliance. Each μ is a statistic,
so these hypotheses do not comply with our rules. The asserted
value in H0 should not appear in Ha, so these hypotheses are not in
compliance. (b) H0: σ = 20, Ha: σ ≤ 20 These hypotheses comply with
our rules. Ha cannot include equality, so these hypotheses are not
in compliance. Each σ is a statistic, so these hypotheses do not
comply with our rules. The asserted value in H0 should not appear
in Ha, so these hypotheses are not in compliance. (c) H0: p ≠ 0.25,
Ha: p = 0.25 These hypotheses comply with our rules. Ha cannot
include equality, so these hypotheses are not in compliance. Each p
is a statistic, so these hypotheses do not comply with our rules.
The asserted value in H0 should not appear in Ha, so these
hypotheses are not in compliance. (d) H0: μ1 − μ2 = 25, Ha: μ1 − μ2
> 100 These hypotheses comply with our rules. H0 cannot include
equality, so these hypotheses are not in compliance. Each μ is a
statistic, so these hypotheses do not comply with our rules. The
asserted value in H0 should also appear in Ha, so these hypotheses
are not in compliance. (e) H0: S12 = S22, Ha: S12 ≠ S22 These
hypotheses comply with our rules. H0 cannot include equality, so
these hypotheses are not in compliance. Each S is a statistic, so
these hypotheses do not comply with our rules. The asserted value
in H0 should not appear in Ha, so these hypotheses are not in
compliance. (f) H0: μ = 120, Ha: μ = 150 These hypotheses comply
with our rules. Ha cannot include equality, so these hypotheses are
not in compliance. Each μ is a statistic, so these hypotheses do
not comply with our rules. If μ appears in H0, then it should not
appear in Ha, so these hypotheses are not in compliance. (g) H0:
σ1/σ2 = 1, Ha: σ1/σ2 ≠ 1 These hypotheses comply with our rules. H0
cannot include equality, so these hypotheses are not in compliance.
Each σ is a statistic, so these hypotheses do not comply with our
rules. The asserted value in H0 should not appear in Ha, so these
hypotheses are not in compliance. (h) H0: p1 − p2 = −0.1, Ha: p1 −
p2 < −0.1 These hypotheses comply with our rules. H0 cannot
include equality, so these hypotheses are not in compliance. Each p
is a statistic, so these hypotheses do not comply with our rules.
The asserted value in H0 should not appear in Ha, so these
hypotheses are not in compliance.

Respuesta :

a)

Since alternative hypothesis can be of either right tailed or left tailed. Here alternative hypothesis is right tailed. And null hypothesis can include equality. Further mu is not a statistics,

So option a) i.e these data comply with our rules, is correct and all other are incorrect.

b)

Since alternative hypothesis can be of either right tailed or left tailed. Here alternative hypothesis is left  tailed. And null hypothesis can include equality. Further sigma is not a statistics,

So option d) is correct and all other are incorrect.

C)

Since alternative hypothesis can be of either right tailed or left tailed. Here alternative hypothesis is left  tailed. And null hypothesis can include equality. Further p is not a statistics,

So option (b) is correct and all other are incorrect.

d)

Since alternative hypothesis can be of either right tailed or left tailed. Here alternative hypothesis is left  tailed. And null hypothesis can include equality. Further mu1 and mu2  is not a statistics,

So option a) i.e these data comply with our rules, is correct and all other are incorrect.

e)

Since alternative hypothesis can be of either right tailed or left tailed. Here alternative hypothesis is left  tailed. And null hypothesis can include equality. Further sigma is not a statistics,

So option (c) correct and all other are incorrect.

f)

Since alternative hypothesis can be of either right tailed or left tailed or a simple hypothesis. Here alternative hypothesis is a simple  hypothesis  And null hypothesis can include equality. Further mu is not a statistics,

So option a) i.e these data comply with our rules, is correct and all other are incorrect.

g)

Since alternative hypothesis can be of either right tailed or left tailed or two tailed. Here alternative hypothesis is two  tailed. And null hypothesis can include equality. Further sigma1 and sigma 2 is not a statistics,

So option a) i.e these data comply with our rules, is correct and all other are incorrect.

h)

Since alternative hypothesis can be of either right tailed or left tailed. Here alternative hypothesis is left  tailed. And null hypothesis can include equality. Further p is not a statistics, and asseter value of H0 should not appear in asserted value in Ha.

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https://brainly.com/question/14724376

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