A recent broadcast of a television show had a 10 ​share, meaning that among 6000 monitored households with TV sets in​ use, 10​% of them were tuned to this program. Use a 0.01 significance level to test the claim of an advertiser that among the households with TV sets in​ use, less than 15​% were tuned into the program. Identify the null​ hypothesis, alternative​ hypothesis, test​ statistic, P-value, conclusion about the null​ hypothesis, and final conclusion that addresses the original claim. Use the​ P-value method. Use the normal distribution as an approximation of the binomial distribution. Identify the null and alternative hypotheses. Choose the correct answer below.   A. H0​:p=0.15 H1​: p>0.15   B. H0​:p=0.15 H1​:p≠0.150.15   C. H0​:p=0.85 H1​:p<0.850.85   D. H0​:p=0.15  H1​:p<0.150.15   E. H0​:p=0.85

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A recent broadcast of a television show had a 10
​share, meaning that among 6000 monitored households with TV sets in​ use, 10​% of them were tuned to this program. Use a 0.01 significance level to test the claim of an advertiser that among the households with TV sets in​ use, less than 15​%
were tuned into the program. Identify the null​ hypothesis, alternative​ hypothesis, test​ statistic, P-value, conclusion about the null​ hypothesis, and final conclusion that addresses the original claim. Use the​ P-value method. Use the normal distribution as an approximation of the binomial distribution.
Identify the null and alternative hypotheses. Choose the correct answer below.
 
A.
H0​:p=0.15
H1​: p>0.15
 
B.
H0​:p=0.15
H1​:p≠0.150.15
 
C.
H0​:p=0.85
H1​:p<0.850.85
 
D.
H0​:p=0.15 
H1​:p<0.150.15
 
E.
H0​:p=0.85
H1​:p>0.85
 
F.
H0​:p=0.85
H1​:p≠0.85

Should cover everything example does BELOW

A recent broadcast of a television show had a 20
​share, meaning that among 5500
monitored households with TV sets in​ use,
20​% of them were tuned to this program. Use a 0.01 significance level to test the claim of an advertiser that among the households with TV sets in​ use, less than
30​% were tuned into the program. Identify the null​ hypothesis, alternative​ hypothesis, test​ statistic, P-value, conclusion about the null​ hypothesis, and final conclusion that addresses the original claim. Use the​ P-value method. Use the normal distribution as an approximation of the binomial distribution.
Identify the null and alternative hypotheses.
 
The null​ hypothesis,
H0​, is a statement that the value of a population parameter is equal to some claimed value. The alternative​ hypothesis,
H1​, is a statement that the parameter has a value that somehow differs from the null hypothesis.
Identify the claim about the population proportion p. This is the hypothesis that will be tested.
 
p<0.30 he symbolic form that must be true when the claim is false is P≥0.30 Therefore, H0 is
P=0.30 and H1 is p<0.30
Identify the level of significance
alphaα.
alphaαequals=0.01
Identify the test statistic.
Solve for the test statistic using the formula​ below, where n is the sample​ size, p is the population proportion based on the​ claim, q is equal to
​(1minus−​p),
and
ModifyingAbove p with caretpequals=StartFraction number of successes x Over n EndFractionnumber of successes xn
is the sample proportion.
 
zequals=StartFraction ModifyingAbove p with caret minus p Over StartRoot StartFraction pq Over n EndFraction EndRoot EndFractionp−ppqn
Let
pequals=0.300.30​,
the population proportion given in the claim.
Solve for q. Substitute the value of
pequals=0.300.30
into the formula below.
 
qequals=1minus−p
qequals=1minus−0.300.30
qequals=0.700.70
The sample proportion
ModifyingAbove p with caretp
is equal to
0.200.20​,
because
2020​%
of
55005500
monitored households with TV sets in use were tuned to this program.
Calculate the test statistic. Substitute
pequals=0.300.30​,
qequals=0.700.70​,
ModifyingAbove p with caretpequals=0.200.20​,
and
nequals=55005500
into the formula below and​ simplify, rounding to two decimal places.
 
zz
equals=
StartFraction ModifyingAbove p with caret minus p Over StartRoot StartFraction pq Over n EndFraction EndRoot EndFractionp−ppqn
 
equals=
StartFraction 0.20 minus 0.30 Over StartRoot StartFraction left parenthesis 0.30 right parenthesis left parenthesis 0.70 right parenthesis Over 5500 EndFraction EndRoot EndFraction0.20−0.30(0.30)(0.70)5500
 
equals=
negative 16.18−16.18
Identify the​ P-value.
 
The​ P-value is the probability of getting a test statistic at least as extreme as the value representing the sample data.
Determine if the test is​ left-tailed, right-tailed, or​ two-tailed. If the alternative hypothesis has the
not equals≠
​symbol, then the test is​ two-tailed. If the alternative hypothesis has the
less than<
​symbol, then the test is​ left-tailed. If the alternative hypothesis has the
greater than>
​symbol, then the test is​ right-tailed.
 
The test is​ left-tailed.
The area of the​ P-value is determined by the type of test being conducted. In a​ left-tailed test, the​ P-value is equal to the area to the left of the test statistic z. In a​ right-tailed test, the​ P-value is equal to the area to the right of the test statistic z. In a​ two-tailed test, the​ P-value is equal to twice the extreme region bounded by the test statistic z.
While either technology or a standard normal distribution table could be used to find the​ P-value, for the purposes of this​ explanation, use technology. Recall that
zequals=negative 16.18−16.18.
Find the​ P-value for the​ left-tailed test, rounding to four decimal places.
 
​P-valueequals=0.0000
Identify the conclusion about the null hypothesis and the final conclusion that addresses the original claim.
 
If the​ P-value is less than or equal to the level of significance
alphaα​,
reject
H0​,
and conclude that there is sufficient evidence to warrant rejection of the claim.​ Otherwise, fail to reject
H0​,
and conclude that there is insufficient evidence to warrant the rejection of the claim.
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