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Hypothesis Testing
Establishing Hypotheses
- The null hypothesis is the hypothesis that cannot be viewed as false unless sufficient evidence to the contrary is obtained.
- The alternative hypothesis is the hypothesis against which the null hypothesis is tested and which is viewed as true when the null hypothesis is declared as false.
- Using the population mean as an example, there are three types of hypothesis tests o Two tailed (sided) alternatives
H 0 : μ =μ 0
Ha : μ ≠μ 0
o One tailed (sided) lower-tail alternatives
H 0 : μ ≥μ 0
Ha : μ <μ 0
o One tailed (sided) upper-tail alternatives
H 0 : μ ≤μ 0
Ha : μ >μ 0
Conclusion Drawn^ True Alternative from the Sample H^ 0 Ha H 0 (^) Correct conclusion Type II error H a (^) Type I error Correct conclusion
- A Type I error is the kind of error made when rejecting a null hypothesis when it is actually true.
- A Type II error is the kind of error made when not rejecting a null hypothesis when it is actually false.
- The probability of making a Type I error is called an α risk. α is also called the level of significance ,
α = P (Rejecting H 0 | H 0 is true) = P (Type I error).
- The probability of making a Type II error is called a β risk,
β = P (Not Rejecting H (^) 0 | H (^) 0 is false) = P (Type II error).
Decision Rule
- A statistical decision rule specifies for each possible outcome of the sample test statistic which alternative, H (^) 0 or Ha , should be selected.
- In a decision rule, values like L , R , t α (^2) ,ν , or z α 2 are specified. The values L and R are
called the lower and upper action limits , respectively. The value t α 2 , ν or z α 2 is called the critical value (for two tailed tests).
- In a statistical decision rule, the set of values of the sample test statistic for which the null hypothesis is not rejected is called the acceptance region. The set of values of the test statistic for which the alternative hypothesis is concluded is called the rejection region.
H 0
H a
Steps in Hypothesis Testing
- State the null and alternative hypotheses
- Establish the level of significance
- Determine the appropriate sampling distribution of the sample test statistic, e.g ., normal or t for the sample mean X.
- Establish the decision rule.
- Draw a sample and compute the sample test statistic.
- Draw conclusion based on the decision rule.
Hypothesis Test for the Population Mean
When the Population Standard Deviation is Known
- If the population is normally distributed, the standard normal variable Z is the appropriate test statistic because
Reject H 0 if| z |> z α 2.
- Example The mean monthly household income of all the households in a specific community was $2400 with a population standard deviation $480. An economist is interested in knowing if this mean monthly household income has changed. He selected a random sample of 144 households and the sample mean was found to be $2480. Conduct the hypothesis test at
- Soluiton H 0 : μ = 2400 Ha : μ ≠ 2400
Population variance is known, therefore, z = X σ−^ μ n^0 follows approximately a standard
normal distribution.
o Use the critical value to establish the decision rule: Because α = 0.05, z α 2 = 1.96and the
decision rule is Do not reject H 0 if | z | ≤ 1.96, Reject H 0 if | z | >1.96. (^0 2480 2400) 2. 480 144 z X n
μ σ
= −^ = − =.
The conclusion is to reject H 0 because | z | >1.96. o Use the action limits to establish the decision rule: Because (^02)
L z (^) α n 144 = μ − σ = − = and
(^02)
R z (^) α n 144 = μ + σ = + = ,
the decision rule is Do not reject H 0 if2321.6 ≤ X ≤2478.4, Reject H 0 if X < 2321.6or X > 2478.4. The conclusion is to reject H 0 because X > 2478.4.
- For a one tailed lower-tail test, the rejection region is in the lower tail side.
Reject Do not reject
L μ 0 X
Do not reject H 0 if X ≥ L , Reject H 0 if X < L ,
where, L = μ 0 − z (^) α^ σ n.
Or equivalently, Do not reject H 0 if z ≥ − z α, Reject H 0 if z < − z α,
where, z X^0 n
μ σ
- Example A machine filling soft drink cans, when appropriately adjusted, fills cans with a mean of at least 12 ounces and a standard deviation of 0.5 ounces. The contents of the cans are normally distributed. A random sample of 25 cans is selected and the sample mean is found to be 11.85 ounces. At a significance level of 0.05, test if there is sufficient evidence to believe that the population mean is less than 12 ounces.
- Solution The null and alternative hypotheses are H (^) 0 : μ ≥ 12 H (^) a : μ < 12
Or equivalently, Do not reject H 0 if z ≤ z α, Reject H 0 if z > z α,
where, z X^0 n
μ σ
- Example For the household monthly income example above, if the economist wants to know if there is sufficient evidence to prove that the population mean monthly household income has increased at a significance 0.01, the null and alternative hypotheses are H 0 : μ ≤ 2400 Ha : μ > 2400
- Solution
o Using the critical value to establish the decision rule: Because α = 0.01and z α = 2.33,
the decision rule is Do not reject H 0 if z ≤ 2.33, Reject H 0 if z >2.33. As computed above,. The conclusion is not to reject because. Therefore, we do not have sufficient evidence to prove that the population mean monthly household income has increased.
z = 2.00 H (^) 0 z <2.
o Using the action limit to establish the decision rule: Because
0
R z α (^) n 144 = μ + σ = + = ,
the decision rule is Do not reject H 0 if X ≤ 2493.2, Reject H 0 if X > 2493.2. The conclusion is not to reject H (^) 0 because X < 2493.2.
When the Population Standard Deviation is Unknown
- Whenever the population standard deviation is unknown, the statistic
t X s n = −^ μ
follows the student t distribution with n − 1 degrees of freedom if the population is normally distributed. Therefore the variable t is the appropriate test statistic.
- For a two tailed test, there are two rejection regions, one on each side. The decision rule is
Do not reject H 0 if L ≤ X ≤ R , Reject H 0 if X < L or X > R ,
where, L (^) 0 t (^) 2 , n 1 s α n = μ − (^) − and R (^) 0 t (^) 2 , n 1 s α n = μ + −
Or equivalently, Do not reject H 0 if− t (^) α (^) 2 , n − 1 ≤ t ≤ t α 2 , n − 1 , Reject H 0 if t < − t (^) α (^) 2 , n − 1 or t > t α (^) 2 , n − 1 ,
where, t = Xs −^ μ n^0.
The latter can also be stated as Do not reject H 0 if| | t^^ ≤^ t α 2 , n − 1 , Reject H 0 if| t | > t α 2 , n − 1.
- For a one tailed lower-tail test, the rejection region is in the lower tail side. The decision rule is Do not reject H 0 if X ≥ L , Reject H 0 if X < L ,
where, L (^) 0 t (^) , n 1 s α n
Or equivalently, Do not reject H 0 if t ≥ − t (^) α , n − 1 , Reject H 0 if t < − t (^) α , n − 1 ,
where, t = Xs −^ μ n^0.
o Decision rule using action limit: Because R = μ 0 + t (^) α, n − 1 sn = 6 + 1.752 ^ 0.4 16 =6.
the decision rule is Do not reject H 0 if X ≤6. Reject H 0 if X > 6.1752. The conclusion is to reject H 0 because X > 6.1752. In this case, we have sufficient evidence to believe that the population mean has increased.
- When the population does not depart too markedly from normality and the sample size is not exceedingly small, the test statistic can be used approximately as an appropriate test statistic and the above test procedure can be used.
t
- When the random sample size is reasonably large, e.g ., n , the central limit theorem applies so that the standard normal random variable
Z may be used in the position of t.
Hypothesis Test for Population Proportion
- The random variable X representing the number of successes in the sample is a binomial or a hypergeometric random variable. Therefore, the population is not normal. The sample size n must be large.
- p = Xn is the sample proportion. We also used f to represent sample frequency. Therefore,
p f = (^) n because X = f.
- The null and alternative hypotheses are H (^) 0 : p = p 0 H (^) a : p ≠ p 0 for two tailed tests; H (^) 0 : p ≥ p 0 H (^) a : p < p 0 for one tailed lower tail tests; or
H (^) 0 : p ≤ p 0 H (^) a : p > p 0 for one tailed upper tail tests.
- Decision rule using the critical value: Do not reject H 0 if− z (^) α 2 ≤ z ≤ z α 2 Reject H 0 if z < − z α 2 or z > z α 2 for two tailed tests; Do not reject H 0 if z ≥ − z α Reject H 0 if z < − z α for one tailed lower tail test; or Do not reject H 0 if z ≤ z α Reject H 0 if z > z α for one tailed upper tail test, where 0 0 (1^0 )
z X^ np np p
or equivalently 0 0 (1^0 )
z p^ p p p n
- Decision rule using action limits: Do not reject H 0 if L ≤ p ≤ R Reject H 0 if p < L or p > R
for two tailed tests, where L = p 0 − z (^) α 2 p^0^^ (1 n^ − p^0 ) and R = p 0 + z (^) α 2 p^0^^ (1^ n − p^0 );
Do not reject H 0 if p ≥ L Reject H 0 if p < L
for one tailed lower tail test, where L = p 0 − z α p^0^ (1 n^ − p^0 ); or
Do not reject H 0 if p ≤ R Reject H 0 if p > R
for one tailed upper tail test, where R = p 0 + z α p^0^^ (1^ n − p^0 ).
P − value = P Z ( < z *).
- For a one tailed upper-tail test
P − value = P Z ( > z *).
P − value =.
2 ( ) if 0 2 ( ) if 0
P Z z z P Z z z
- Conduct hypotheses test with the P − value Whether the test is one tailed or two tailed, or lower-tail or upper-tail, the decision rule is always,
Do not reject H 0 if P − value≥ α
Reject H 0 if P −value< α
- Example In the last a few years, more than 80% of visitors to a theme park were travelers out of state. The marketing manager of the theme park wants to know if this population proportion has decreased or not. She selected a random sample of 100 visitors and found that 75 were from out of state. At a significance level 0.05, test if the population proportion has decreased or not. Use the P −value to state the decision rule and conduct the test.
- Solution The null and alternative hypotheses are H (^) 0 : p ≥ 0. 8 H (^) a : p <0. 8 The decision rule is Do not reject H 0 if P − value ≥ 0. Reject H 0 if P − value< 0.05.
In this problem, *^0 0 0
z X^ np np p
= −^ = −^ = − =
z^ *^ ) = ( Z < −1.25) = 0.5 − 0.3944 =0.
−. Therefore,
P − value = P Z ( <. The conclusion is not to reject H (^) 0 because P − value ≥ 0.05.
Relationship between Type I Error ( α ) and Type II Error ( β )
- A Type I error may be made only when is rejected. When is rejected, a Type II error
is not made.
H 0 H 0
- A Type II error may be made only when is not rejected. If is not rejected, a Type I
error is not made.
H 0 H 0
- The probability of making a Type II error ( β ) can be computed only when we assume a
specific true population mean μ a or a true population proportion pa.
Do not reject Reject
μ 0 R
μ a
X
X
- For a one tailed upper tail test, β = P X ( ≤ R )when μ a > μ 0.
- Example
For the monthly household income problem, if μ a = 2550 , then is false. A Type II error
is made if is not rejected. The probability of making a Type II error is
H 0
H 0
