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Hypothesis Testing
Statistics 503 (Simonsen)
These steps are to be followed every time you do a hypothesis test.
Required Elements
- State the scientific question to be answered.
- Define the parameters of interest (Let μ 1 be …)
- State the null hypothesis H 0 both mathematically (in terms of parameters) and in words.
- Decide based on the problem statement whether the test is directional or not, and state the alternative hypothesis HA both mathematically and in words.
- Name the type of statistical test to be used and the distribution the test statistic follows under H 0.
- State the significance level α to be used and the corresponding critical value of the test statistic. Define the rejection region.
- Calculate the test statistic from the data.
- Compare the test statistic to the rejection region, or, give the P-value and compare it to α.
- Make a decision about the null hypothesis: a) If the test statistic is in the rejection region, state "reject H 0 ". b) If not, state "do not reject H 0 ".
- Form a scientific conclusion based on that decision. If 9a) then start with "This study provides evidence " if 9b) then start with "This study does not provide evidence" followed by "at the ___ significance level that" followed by the verbal statement of HA.
Remarks
Use a complete sentence with no symbols. This can be fairly general.
μ, or μ 1 and μ 2 , or π, or π 1 and π 2 , or …
Usually an equality.
Occasionally (e.g. F-test, WMW) it may be too cumbersome to express H (^) A mathematically.
Include the df (or other parameters) for that distribution if appropriate.
The critical value and rejection region depend on α, the df and distribution from 5., and whether HA is directional.
Show all calculations.
I.e. test statistic < or > critical value?
You MAY NOT use the word "accept" anywhere in this step. If using P-value, reject H 0 when P < α.
Use a complete sentence with no symbols. Use similar wording to step 4. You MUST NOT affirm H 0. See examples. Don't get too creative.
Null Alternative Hypothesis Hypothesis (^) non-directional directional T-tests (^) H 0 HA rejection
region
HA rejection region
df t (^) s (1-α) CI
One μ < μ 0 t (^) s < –tα Sample
μ = μ 0 μ ≠ μ 0 t (^) s < –tα/ t (^) s > tα/2 (^) μ > μ 0 t (^) s > tα
n-1 (^) − μ 0
y
y SE
for μ: ⎯y ± tα/2 SE⎯y
Two μ 1 < μ 2 t (^) s < –tα Independent Samples
μ 1 = μ 2 μ 1 ≠ μ 2 t (^) s < –tα/ t (^) s > tα/ μ 1 > μ 2 t (^) s > tα
1 2
1 2 −
y y
y y SE
for μ 1 - μ 2 : ⎯y1 – ⎯y 2 ± tα/2 SE⎯y1-⎯y
Paired μ 1 < μ 2 t (^) s < –tα Samples
μ 1 = μ 2 μ 1 ≠ μ 2 t (^) s < –tα/ t (^) s > tα/2 (^) μ 1 >^ μ 2 t^ s > tα
nd –1 (^1) − 2 diff
y y SE
for μ 1 - μ 2 : ⎯y ± tα/2 SEdiff
2 2 2 1 2 4 1 2 1 1 2 1
SE SE
df SE SE n n
4 < n^1 + n^2 – 2.
Action reject H 0 do not reject H 0 H 0 is true type I error “false positive”
correct non-rejection “true negative” Truth H 0 is false correct rejection “true positive”
type II error “false negative”
Example: "one-sample, non-directional t-test"
In a study of the effect of aluminum intake on the mental development of infants, a group of 92 infants who had been born prematurely were given a special aluminum-depleted intravenous-feeding solution. At age 18 months the neurologic development of the infants was measured using the Bayley Mental Development Index, which is designed so that 100 is the average score in the general population. How do premature infants fed this solution compare in neurologic development to the general population? BMDI score n mean s 92 97.95 21.
Solution [notes in square brackets are for your information and are not part of the solution]
Do premature infants fed a special aluminum-depleted intravenous-feeding solution have different neurological development than the general infant population?
Let μ be the mean BMDI score for premature infants fed the solution. Let μ 0 = 100; the mean BMDI score for the general infant population.
H 0 : μ = μ 0 ; premature infants fed the solution have the same mean BMDI score as the general population. HA: μ 1 ≠ μ 2 ; premature infants fed the solution have a different mean BMDI score than the general population.
Use a one-sample, non-directional t-test.
t (^) s = 0
y − μ SE
has a t-distribution with df = n – 1 = 92 – 1 = 91 under H 0.
Test at significance level α = 0.05; critical value is t.025 = 1.986. [Use α = 0.05 unless a different level is specified in the problem.] If t (^) s < –1.986 or t (^) s > 1.986 then will reject H 0.
SE =
s n
t (^) s =
y SE
–1.986 < t (^) s < 1.986 so do not reject H 0 at level 0.05. [ If t (^) s were < –t.025 or > t (^) .025 then would reject H 0 .] This study does not provide evidence at the .05 significance level that premature infants fed the solution have a different mean BMDI score than the general population. [If H 0 was not rejected, conclude "This study provides evidence at the .05 significance level that premature infants fed the solution have a different mean BMDI score than the general population."]
[If you wish, you may also include the P-value in the concluding sentence. Place it in parentheses (P=___) immediately before the word "at", i.e. "This study does not provide evidence (P=0.37) at the .05 significance level that premature infants fed the solution have a different mean BMDI score than the general population." or "This study provides evidence (P = 0.037) at the .05 significance level that premature infants fed the solution have a different mean BMDI score than the general population "]
Example: "two-independent-sample, directional t-test"
A pain-killing drug was tested for efficacy in 50 women who were experiencing uterine cramping pain following childbirth. Twenty-five of the women were randomly allocated to receive the drug, and the remaining 25 received a placebo (inert substance). Capsules of drug or placebo were given before breakfast and again at noon. A pain relief score, based on hourly questioning throughout the day, was computed for each woman. The possible pain relief scores ranged from 0 (no relief) to 56 (complete relief for 8 hours). Summary results are shown in the table. Assuming that pain relief scores follow approximately a normal distribution, test for efficacy of the drug at reducing uterine cramping pain. PAIN RELIEF SCORE Treatment n Mean SD Placebo 25 31.96 12. Drug 25 25.32 13.
Solution [notes in square brackets are for your information and are not part of the solution]
Is the drug more effective than the placebo at reducing pain?
Let μ 1 be the mean pain relief score of women who take the placebo. Let μ 2 be the mean pain relief score of women who take the drug.
H 0 : μ 1 = μ 2 ; The drug and the placebo are equivalent at reducing uterine cramping pain. [or, " The mean pain relief score of women who take the drug is the same as that for women who take the placebo."] HA: μ 1 < μ 2 ; The drug is more effective than the placebo at reducing uterine cramping pain. [or "The mean pain relief score of women who take the placebo is less than that of women who take the drug."
Use a two-sample, directional t-test
t (^) s = (⎯y1 - ⎯y 2 ) / SE has a t-distribution with df =
2 2 2 1 2 4 1 2 1 1 2 1
4
− −
SE SE
SE SE
n n
= 47 under H 0.
Test at significance level α = 0.05; critical value is t.05 = 1. [Use α = 0.05 unless a different level is specified in the problem.] If t (^) s < –1.678 then will reject H 0. [Note: if H (^) A were ">" then this statement would be "If ts > 1.678 then will reject H 0 "]
(U)SE⎯y1-⎯y2 =
2 2 2 2 1 2 1 2
s s n n
t (^) s = 1 2
y y SE
t (^) s > –1.678 so do not reject H 0 at level 0. [If t (^) s < –1.678 then would reject H 0 .] This study does not provide evidence at the 0.05 significance level that the drug is more effective than the placebo at reducing uterine cramping pain. [If H 0 was rejected, conclusion would be "This study provides evidence at the 0.05 significance level that the drug is more effective than the placebo at reducing uterine cramping pain."] [The P-value for this example is P=0.962]
Example: “non-directional Wilcoxon–Mann–Whitney (WMW) test”
A plant physiologist conducted an experiment to determine whether mechanical stress affects the growth of soybean plants. Young plants were randomly allocated to two groups of 13 plants each. Plants in one group were mechanically agitated by shaking for 20 minutes daily, while plants in the other group were not agitated. After 16 days of growth, the total stem length (cm) of each plant was measured, with the results given in the accompanying table. Assuming that plant stem lengths do not follow a normal distribution, compare the treatments to determine whether mechanical stress changes the distribution of soybean stem length.
Control Stress 25.2 24. 29.5 25. 30.1 26. 30.1 27. 30.2 27. 30.2 27. 30.3 27. 30.6 27. 31.1 28. 31.2 28. 31.4 29. 33.5 30. 34.3 30.
Does mechanical stress change the growth of soybean plants?
H 0 : the distribution of soybean plant stem lengths is the same whether mechanically stressed or not. HA: the distribution of soybean plant stem lengths is different when mechanically stressed than when not stressed.
Use a non-directional WMW test. Us = max(K 1 , K 2 ) follows the WMW null distribution with n = 13, n’ = 13. Test at level α = 0.05; critical value is 124; if Us ≥ 124 then will reject H 0.
( )( )
1 (^1 ) 1 2 2 1 2 1 2
check: 148.5 20.5 169; 13 13 169 (okay)
K
K
K K n n
Us = max(K 1 , K 2 ) = 148. Us =148.5 ≥ 124 so reject H 0
This study provides evidence at the 0.05 significance level that the distribution of soybean plant stem lengths is different when mechanically stressed than when not stressed.
[P-value < 0.001]
[WMW tests can also be done with directional alternative hypotheses, in which case only K 1 or K (^2) (depending on direction) would be used. We did not cover directional WMW tests.]
Example: "paired-sample, non-directional t-test"
A study was conducted to compare the sodium content in the plasma of female southern fur seals with the level in the milk of these seals. The following observations of the sodium content (in mmol/L) were obtained in 10 randomly selected seals. Assuming that sodium content in milk and plasma follows a normal distribution, conduct an
appropriate hypothesis test with α = 0.04.
Seal Milk Plasma Difference 1 93 147 - 2 104 157 - 3 95 142 - 4 81.5 141 -59. 5 95 142 - 6 95 147 - 7 76.5 148 -71. 8 80.5 144 -63. 9 79.5 144 -64. 10 87 146 - mean 88.7 145.8 -57. s 8.99 4.61 7.
Solution [notes in square brackets are for your information and are not part of the solution]
Do milk and plasma differ in their sodium content in female southern fur seals?
Let μ 1 be the mean sodium content in milk. Let μ 2 be the mean sodium content in plasma.
H 0 : μ 1 = μ 2 ; milk and plasma have the same mean sodium content in female southern fur seals HA: μ 1 ≠ μ 2 ; milk and plasma have different mean sodium content in female southern fur seals.
Use a paired sample, non-directional t-test.
s^ =^ diff diff
y t SE
has a t-distribution with df = 10 – 1 = 9 under H 0.
Test at level α = 0.04; critical value is t.02 = 2. If t (^) s < -2.398 or if t (^) s > 2.398 then will reject H 0.
diff =^^ diff^ =^ =^ s =^ diff =^ =^ = − diff diff
s y (^) y y SE t n SE^ SE
t (^) s < –2.398 so reject H 0. This study provides evidence at the 0.04 significance level that milk and plasma have different mean sodium content in female southern fur seals
[The P-value is 2.96x10-9^ for this example.]
Example: “non-directional sign test”
A study was conducted to compare the sodium content in the plasma of female southern fur seals with the level in the milk of these seals. The following observations of the sodium content (in mmol/L) were obtained in 10 randomly selected seals. Assuming that sodium content in milk and plasma do not follow
a normal distribution, conduct an appropriate hypothesis test with α = 0.02.
Seal Milk Plasma Diff 1 93 147 – 2 104 157 – 3 95 142 – 4 81.5 141 – 5 95 142 – 6 95 147 – 7 76.5 148 – 8 80.5 144 – 9 79.5 144 – 10 87 146 –
[Note: the assumption of non-normality may be implicit in the problem, rather than stated explicitly as above. The “Diff” column is not calculated for you, but is part of your solution.]
Solution [notes in square brackets are for your information and are not part of the solution]
Do milk and plasma differ in their sodium content in female southern fur seals?
Let π be the probability, for any seal, that the milk has higher sodium content than does plasma.
H 0 : π = ½; For any seal, it is equally likely that milk or plasma will have higher sodium content. [OR, milk and plasma have the same sodium content] HA: π ≠ ½;. For any seal, it is not equally likely that milk or plasma will have higher sodium. [OR, milk and plasma have different sodium content]
Use a non-directional sign test. Bs =max (^) ( N (^) + , N − (^) )has a folded binomial(10, 0.5) distribution under H 0.
Test at level α = 0.02; critical value is B.02 = 10 [from Table 7]; will reject H 0 if Bs ≥ 10. N + = 0, N − = 10, so Bs = N −= 10. Bs ≥ 10 so reject H 0.
This study provides evidence at the 0.02 significance level that milk and plasma have different sodium content.
P-value calculation [do if requested]: Where Y is Binomial(n=10, π = 0.5),
P-value = ( ) ( ) ( ) ( ) = 0.
2 Pr 10 2 Pr 10 2 0.5 0. 10
Y Y
P-value justification: [this is why it works; don’t actually do this in your problem!] notice that N (^) + + N (^) −= n , and we can think of these as the number of successes and failures
( ) ( ) ( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( )
P-value Pr 0 Pr 0 Pr 0 Pr 0 Pr 0
Pr 0.5 Pr 0.5 2 Pr 0.5 (by symmetry since 0.5)
− = =
= ≥ π = + ≤ − π = = ≥ π = π =
⎛ ⎞ ⎛ ⎞ = (^) ⎜ ⎟ = ⎜ ⎟ ⎝ ⎠ ⎝ ⎠
∑ ∑ s s
H s H s H s H s H s
s s s n (^) y n y n n
y B y B
B B N B N B N B n N B
N B N n B N B n n y y
Example: "directional sign test"
RHR Subject Before After Diff 1 75 80 – 2 92 79 + 3 89 89 0 4 88 83 + 5 88 79 + 6 90 85 + 7 87 84 + 8 73 71 + 9 82 73 +
A random sample of 9 middle-aged men who did not exercise regularly was selected to participate in a study to test the effect of regular exercise on resting heart rate (RHR). First, the RHR of each subject was measured. Then, each man was required to exercise on a treadmill for 30 minutes, 4 times a week, for 6 weeks. Afterwards, the RHR of each man was measured again. The results are shown in the table. Assuming that RHR does not have a normal distribution, use these data to test, at the 0.01 significance level, whether regular exercise decreases mean RHR.
[Note: the assumption of non-normality may be implicit in the problem, rather than stated explicitly as above. The “Diff” column is not calculated for you, but is part of your solution.]
Solution [notes in square brackets are for your information and are not part of the solution]
Does regular exercise decrease RHR? (^) (1) Let π be the probability, for any man, that he has higher RHR before exercising regularly. (1)
H 0 : π = 0.5 ; regular exercise has no effect on RHR, [OR, men are equally likely to have higher RHR before (2) or after exercising regularly] HA: π > 0.5 ; regular exercise decreases RHR. [OR, men are more likely to have higher RHR before (2) exercising regularly than after exercising regularly]
Use a directional sign test. (^) (1) Bs = N (^) +has a binomial(8, 0.5) distribution under H 0. (1)
[We completely discard subject 3 because of a tie, so n is reduced by 1] [If H (^) A were “π < 0.5”, then we would use Bs = N − instead]
Test at level α = 0.01; Critical value is Bcritical = 8 [Table 7]. (1) If Bs ≥ 8 then will reject H 0. (2) Bs = N + = 7 .Since Bs < 8 we do not reject H 0 (2)
This study does not provide evidence at the 0.01 significance level that regular exercise decreases mean RHR.
P-value calculation [do if requested]: Where Y is Binomial(n = 8, π = 0.5),
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( )
P-value Pr 7 Pr 7 Pr 8 0.5 0.5 0.5 0.5 8 1 0.5 0. 7 8
Y Y Y
P-value justification: [this is why it works; don’t actually do this in your problem! If HA were “π < 0.5”, then would do this with N −]
( ) ( ) ( )
( ) ( ) ( )
P-value Pr 0 Pr 0 Pr 0.
− = =
= ≥ = ≥ = ≥ π =
⎛ ⎞ ⎛ ⎞ = (^) ⎜ ⎟ = ⎜ ⎟ ⎝ ⎠ ⎝ ⎠
∑ ∑ s s
H s H s s n (^) y n y n n
y B y B
B B N B N B
n n y y
