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,="">
Identify the level of significance
alphaα.
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.
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.
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.