Toegepaste Statistiek 2 / Applied Statistics 2 (FEB12005)
1 Lecture 1.1
Four steps for hypothesis testing:
1. Formulate the hypotheses.
2. Calculate the test statistic.
3. Make a decision rule of when to reject the hypothesis.
− Fix significance level α
− Use rejection region or P-value.
4. Draw your conclusion.
1.1 One-sample z test
If we have observations from a normal distribution with unknown mean µ and known standard deviation σ, what can we say about µ? Does
µ = µ0 ? Look at the sample mean.
From a population with N (µ, σ) distribution, draw a random sample of n observations.
The sample mean X follows the N µ, √σn distribution.
X−µ
Then z = √
σ/ n
follows the standard normal N (0, 1) distribution.
Standardisation: Subtracting the mean and dividing by the standard deviation.
Central Limit Theorem: If the data are not normally distributed.
From any population (not necessarily normal) with mean µ and standard deviation σ. A random
sample of n observations,
√ where n is sufficiently large. The distribution of the sample mean X is
approximately N µ, σ/ n .
If we observe the sample mean of a random sample, which comes from a population with unknown mean µ and known standard deviation
σ, what can we say about µ? Does µ = µ0 ? If the distribution of the random sample is not normal, this test is only an approximation for large n.
Manufacturers want to know whether the average active ingredient in a type of pills is 50 mg. They take n = 100 samples from the pill
production and measures the sampled average active ingredient to be 50.22 mg.
Assume the active ingredient in each pill is normally distributed with unknown mean µ and known standard deviation σ = 1.
Suppose µ = 50.
1. H0 : µ = 50
Ha : µ 6= 50
√ √ √
2. Sample mean: X ∼ N µ, σ/ n with σ/ n = 1/ 100 = 0.1
X−µ
Standardized sample mean: Z = √
σ/ n
∼ N (0, 1)
What is the probability that
49.8 ≤ X ≤ 50.2?
P 49.8 ≤ X ≤ 50.2 = P 49.8−µ X−µ
√ ≤ 50.2−µ 49.8−50 50.2−50
√
σ/ n
≤ σ/ n
√
σ/ n
=P 0.1
≤Z≤ 0.1
= P (−2 ≤ Z ≤ 2) = 0.9544
What is the probability
that X > 50.2?
√ > 50.2−µ
X−µ 50.2−50
P X > 50.2 = P σ/ n
√
σ/ n
=P Z> 0.1
= P (Z > 2) = 0.0228
What is the probability
that X < 49.8?
X−µ
√ < 49.8−µ 49.8−50
P X < 49.8 = P σ/ n
√
σ/ n
=P Z< 0.1
= P (Z < −2) = 0.0228
X−µ
Possible decision rule: reject µ = 50 if X is below 49.8 or above 50.2, in other words, if Z = √
σ/ n
is below -2 or above 2.
Significance level: When H0 is true, Z ∼ N (0, 1), there is still a small probability that Z < −2 or Z > 2, hence H0 will
be rejected.
Critical values: −z ∗ = −1.96 and z ∗ = 1.96 at significance level α = 0.05.
1
,1.2 z test based on rejection region
Manufacturers want to know whether the average active ingredient in a type of pills is 50 mg. They take n = 100 samples from the pill
production and measures the sampled average active ingredient to be 50.22 mg.
Assume the active ingredient in each pill is normally distributed with unknown mean µ and known standard deviation σ = 1.
Suppose µ = 50.
1. H0 : µ = 50
Ha : µ 6= 50
Significance level α = 0.05
2. x = 50.22
σ=1
n = 100
x−µ
Test statistic: z = √
σ/ n
= 50.22−50
√ = 2.2
1/ 100
Critical value: zα/2
∗ ∗
= z0.025 = 1.96
3. Rejection region: if |z|> zα/2
∗ ∗
= z0.025
4. H0 is rejected at the 5% significance level. Conclusion: the amount of the active ingredient is likely to differ from 50 mg.
1.3 P-value
1.4 z test based on rejection region
Manufacturers want to know whether the average active ingredient in a type of pills is 50 mg. They take n = 100 samples from the pill
production and measures the sampled average active ingredient to be 50.22 mg.
Assume the active ingredient in each pill is normally distributed with unknown mean µ and known standard deviation σ = 1.
Suppose µ = 50.
1. H0 : µ = 50
Ha : µ 6= 50
2. x = 50.22
σ=1
n = 100
x−µ
Test statistic: z = √
σ/ n
= 50.22−50
√ = 2.2
1/ 100
Idea: If H0 is true, z should not be an unlikely value.
The p-value is 2P (Z ≥ |z|) = 2P (Z ≥ 2.2) = 0.0278 < α = 0.05
3. If P-value < significance level α, then H0 is rejected.
4. Hence H0 is rejected at the 5% significance level. Conclusion: the amount of the active ingredient is likely to differ from 50 mg.
P-value: If H0 is true, the probability that the test statistic would be as extreme or more extreme than the
observed value. Extreme means deviating from H0 in the direction of Ha .
The smaller the P-value, the stronger the evidence against H0 .
Truth about population
H0 is true H0 is not true
Reject H0 Type I error Correct decision
based on
Decision
sample
Do not reject H0 Correct decision Type II error
Probability for a Type I error = significance level α.
Fix α: the fewer Type II errors, the better the test (higher power).
1.5 95% Confidence interval
The sample mean X can be used to construct a confidence interval for the unknown mean µ.
X−µ
√ ≤ 1.96 = P X − 1.96 √σ ≤ µ ≤ X + 1.96 √σ
0.95 = P (−1.96 ≤ Z ≤ 1.96) = P −1.96 ≤ σ/ n n n
95% confidence interval for µ: X ± 1.96 √σn .
2
,1.6 (1 − α) % Confidence interval
100 × (1 − α) % confidence interval for the mean µ: x ± zα/2
∗ √σ
n
1.7 One-sided vs two-sided z test
Manufacturers want to know whether the average active ingredient in a type of pills deviates from 50 mg.
1. H0 : µ = 50
Ha : µ 6= 50
Manufacturers want to know whether the average active ingredient in a type of pills might be bigger than 50 mg.
1. H0 : µ = 50
Ha : µ > 50
Significance level: α = 0.05
2. x = 50.22
σ=1
n = 100
x−µ
Test statistic: z = √0
σ/ n
= 50.22−50
√ = 2.2
1/ 100
Critical value: zα
∗ =z
0.05 = 1.645
Reject H0 only when z is large.
3. Rejection region: if z > zα
∗ = z∗
0.05
H 0 : µ = µ0 significance level α σ known σ unknown
x−µ x−µ
Test statistic z= √0
σ/ n
t= √0
s/ n
Rejection region ∗
z > zα t > t∗α (n − 1)
Ha : µ > µ 0
P value P (Z ≥ z) P (T ≥ t)
Rejection region ∗
z < −zα t < −t∗α (n − 1)
Ha : µ < µ 0
P value P (Z ≤ z) P (T ≥ T )
Rejection region ∗
|z|> zα/2 |t|> t∗α/2 (n − 1)
Ha : µ 6= µ0
P value 2P (Z ≥ |z|) 2P (T ≥ |t|)
Confidence interval ∗
x ± zα/2 √σ
n
x ± t∗α/2 (n − 1) √s
n
1.8 One-sample t test
The one-sample t test answers the question: If we have observations from a normal distribution with unknown mean µ and unknown standard
deviation σ, what can we say about µ? Does µ = µ0 ?
We can estimate σ with the sample standard deviation s.
From a population with the N (µ, σ) distribution, draw a random sample of n observations.
X−µ
Then T = √
s/ n
follows the t (n − 1) distribution.
As n grows large, the t (n − 1) distribution resembles the N (0, 1) distribution closely.
Manufacturers want to know whether the average active ingredient in a type of pills deviates from 50 mg.
1. H0 : µ = 50
Ha : µ 6= 50
Significance level: α = 0.05
2. x = 50.22
s = 1.08
n = 100
x−µ
Test statistic: t = √0
s/ n
= 50.22−50
√ = 2.037
1.08/ 100
Critical value: t∗α/2 (n − 1) = t0.025 (99) = 1.984
3. Rejection region: Reject H0 if |t|> t∗0.025 (99)
4. H0 is rejected at the 5% significance level. Conclusion: the amount of the active ingredient is likely to differ from 50 mg.
1.9 One-sample t test and non-normality
Test is based on normal distribution of the data.
Mean and standard deviation are sensitive to outliers, hence also the t statistic.
3
, But one-sample t test can still be used if there are no outliers and
− n sufficiently large (n > 100)
− n moderate (20 ≤ n ≤ 100) and little skewness
− n small (n < 20) and data approximately normally distributed.
1.10 Matched pairs t test
Matched pairs t test answers the question: There are two measurements for each individual in the population. Let µ1 and µ2 be their means.
If we observe these two measurements for a random sample, what can we say about µ1 and µ2 ? Does µ1 = µ2 ?
Prices of textbooks at UCLA bookstore and on Amazon.
µ1 : average price at UCLA bookstore
µ2 : average price on Amazon
1. H0 : µ1 = µ2
Ha : µ1 > µ2
Use differences d = x1 − x2 , where x1 is UCLA price and x2 is Amazon price
H0 : µ1 − µ2 = 0
Ha : µ1 − µ2 > 0 ⇒ One-sample t test
2. d = 13.512
sd = 14.972
n = 61
d−µ
Test statistic: t = √0
sd / n
= 13.512−0
√ = 7.049
14.972/ 61
Critical value: t∗α (n − 1) = ∗
t0.05 (60) = 1.671
3. Rejection region: Reject H0 if t > t∗0.05 (60)
4. H0 is rejected at the 5% significance level, the average prices appear to be higher at the UCLA bookstore.
SPSS Output
Tabel 1: Paired Samples Statistics
N Mean Standard deviation Standard Error Mean
UCLA 61 76.533 62.095 7.950
Amazon 61 63.021 51.076 6.540
Tabel 2: Paired Samples Test
Paired Differences
95% Confidence Interval
of the Difference
Mean Standard Standard Lower Upper t df Sig.
Deviation Error Mean (2-tailed)
UCLA-Amazon 13.512 14.972 1.917 9.677 17.346 7.049 60 0.000
4
1 Lecture 1.1
Four steps for hypothesis testing:
1. Formulate the hypotheses.
2. Calculate the test statistic.
3. Make a decision rule of when to reject the hypothesis.
− Fix significance level α
− Use rejection region or P-value.
4. Draw your conclusion.
1.1 One-sample z test
If we have observations from a normal distribution with unknown mean µ and known standard deviation σ, what can we say about µ? Does
µ = µ0 ? Look at the sample mean.
From a population with N (µ, σ) distribution, draw a random sample of n observations.
The sample mean X follows the N µ, √σn distribution.
X−µ
Then z = √
σ/ n
follows the standard normal N (0, 1) distribution.
Standardisation: Subtracting the mean and dividing by the standard deviation.
Central Limit Theorem: If the data are not normally distributed.
From any population (not necessarily normal) with mean µ and standard deviation σ. A random
sample of n observations,
√ where n is sufficiently large. The distribution of the sample mean X is
approximately N µ, σ/ n .
If we observe the sample mean of a random sample, which comes from a population with unknown mean µ and known standard deviation
σ, what can we say about µ? Does µ = µ0 ? If the distribution of the random sample is not normal, this test is only an approximation for large n.
Manufacturers want to know whether the average active ingredient in a type of pills is 50 mg. They take n = 100 samples from the pill
production and measures the sampled average active ingredient to be 50.22 mg.
Assume the active ingredient in each pill is normally distributed with unknown mean µ and known standard deviation σ = 1.
Suppose µ = 50.
1. H0 : µ = 50
Ha : µ 6= 50
√ √ √
2. Sample mean: X ∼ N µ, σ/ n with σ/ n = 1/ 100 = 0.1
X−µ
Standardized sample mean: Z = √
σ/ n
∼ N (0, 1)
What is the probability that
49.8 ≤ X ≤ 50.2?
P 49.8 ≤ X ≤ 50.2 = P 49.8−µ X−µ
√ ≤ 50.2−µ 49.8−50 50.2−50
√
σ/ n
≤ σ/ n
√
σ/ n
=P 0.1
≤Z≤ 0.1
= P (−2 ≤ Z ≤ 2) = 0.9544
What is the probability
that X > 50.2?
√ > 50.2−µ
X−µ 50.2−50
P X > 50.2 = P σ/ n
√
σ/ n
=P Z> 0.1
= P (Z > 2) = 0.0228
What is the probability
that X < 49.8?
X−µ
√ < 49.8−µ 49.8−50
P X < 49.8 = P σ/ n
√
σ/ n
=P Z< 0.1
= P (Z < −2) = 0.0228
X−µ
Possible decision rule: reject µ = 50 if X is below 49.8 or above 50.2, in other words, if Z = √
σ/ n
is below -2 or above 2.
Significance level: When H0 is true, Z ∼ N (0, 1), there is still a small probability that Z < −2 or Z > 2, hence H0 will
be rejected.
Critical values: −z ∗ = −1.96 and z ∗ = 1.96 at significance level α = 0.05.
1
,1.2 z test based on rejection region
Manufacturers want to know whether the average active ingredient in a type of pills is 50 mg. They take n = 100 samples from the pill
production and measures the sampled average active ingredient to be 50.22 mg.
Assume the active ingredient in each pill is normally distributed with unknown mean µ and known standard deviation σ = 1.
Suppose µ = 50.
1. H0 : µ = 50
Ha : µ 6= 50
Significance level α = 0.05
2. x = 50.22
σ=1
n = 100
x−µ
Test statistic: z = √
σ/ n
= 50.22−50
√ = 2.2
1/ 100
Critical value: zα/2
∗ ∗
= z0.025 = 1.96
3. Rejection region: if |z|> zα/2
∗ ∗
= z0.025
4. H0 is rejected at the 5% significance level. Conclusion: the amount of the active ingredient is likely to differ from 50 mg.
1.3 P-value
1.4 z test based on rejection region
Manufacturers want to know whether the average active ingredient in a type of pills is 50 mg. They take n = 100 samples from the pill
production and measures the sampled average active ingredient to be 50.22 mg.
Assume the active ingredient in each pill is normally distributed with unknown mean µ and known standard deviation σ = 1.
Suppose µ = 50.
1. H0 : µ = 50
Ha : µ 6= 50
2. x = 50.22
σ=1
n = 100
x−µ
Test statistic: z = √
σ/ n
= 50.22−50
√ = 2.2
1/ 100
Idea: If H0 is true, z should not be an unlikely value.
The p-value is 2P (Z ≥ |z|) = 2P (Z ≥ 2.2) = 0.0278 < α = 0.05
3. If P-value < significance level α, then H0 is rejected.
4. Hence H0 is rejected at the 5% significance level. Conclusion: the amount of the active ingredient is likely to differ from 50 mg.
P-value: If H0 is true, the probability that the test statistic would be as extreme or more extreme than the
observed value. Extreme means deviating from H0 in the direction of Ha .
The smaller the P-value, the stronger the evidence against H0 .
Truth about population
H0 is true H0 is not true
Reject H0 Type I error Correct decision
based on
Decision
sample
Do not reject H0 Correct decision Type II error
Probability for a Type I error = significance level α.
Fix α: the fewer Type II errors, the better the test (higher power).
1.5 95% Confidence interval
The sample mean X can be used to construct a confidence interval for the unknown mean µ.
X−µ
√ ≤ 1.96 = P X − 1.96 √σ ≤ µ ≤ X + 1.96 √σ
0.95 = P (−1.96 ≤ Z ≤ 1.96) = P −1.96 ≤ σ/ n n n
95% confidence interval for µ: X ± 1.96 √σn .
2
,1.6 (1 − α) % Confidence interval
100 × (1 − α) % confidence interval for the mean µ: x ± zα/2
∗ √σ
n
1.7 One-sided vs two-sided z test
Manufacturers want to know whether the average active ingredient in a type of pills deviates from 50 mg.
1. H0 : µ = 50
Ha : µ 6= 50
Manufacturers want to know whether the average active ingredient in a type of pills might be bigger than 50 mg.
1. H0 : µ = 50
Ha : µ > 50
Significance level: α = 0.05
2. x = 50.22
σ=1
n = 100
x−µ
Test statistic: z = √0
σ/ n
= 50.22−50
√ = 2.2
1/ 100
Critical value: zα
∗ =z
0.05 = 1.645
Reject H0 only when z is large.
3. Rejection region: if z > zα
∗ = z∗
0.05
H 0 : µ = µ0 significance level α σ known σ unknown
x−µ x−µ
Test statistic z= √0
σ/ n
t= √0
s/ n
Rejection region ∗
z > zα t > t∗α (n − 1)
Ha : µ > µ 0
P value P (Z ≥ z) P (T ≥ t)
Rejection region ∗
z < −zα t < −t∗α (n − 1)
Ha : µ < µ 0
P value P (Z ≤ z) P (T ≥ T )
Rejection region ∗
|z|> zα/2 |t|> t∗α/2 (n − 1)
Ha : µ 6= µ0
P value 2P (Z ≥ |z|) 2P (T ≥ |t|)
Confidence interval ∗
x ± zα/2 √σ
n
x ± t∗α/2 (n − 1) √s
n
1.8 One-sample t test
The one-sample t test answers the question: If we have observations from a normal distribution with unknown mean µ and unknown standard
deviation σ, what can we say about µ? Does µ = µ0 ?
We can estimate σ with the sample standard deviation s.
From a population with the N (µ, σ) distribution, draw a random sample of n observations.
X−µ
Then T = √
s/ n
follows the t (n − 1) distribution.
As n grows large, the t (n − 1) distribution resembles the N (0, 1) distribution closely.
Manufacturers want to know whether the average active ingredient in a type of pills deviates from 50 mg.
1. H0 : µ = 50
Ha : µ 6= 50
Significance level: α = 0.05
2. x = 50.22
s = 1.08
n = 100
x−µ
Test statistic: t = √0
s/ n
= 50.22−50
√ = 2.037
1.08/ 100
Critical value: t∗α/2 (n − 1) = t0.025 (99) = 1.984
3. Rejection region: Reject H0 if |t|> t∗0.025 (99)
4. H0 is rejected at the 5% significance level. Conclusion: the amount of the active ingredient is likely to differ from 50 mg.
1.9 One-sample t test and non-normality
Test is based on normal distribution of the data.
Mean and standard deviation are sensitive to outliers, hence also the t statistic.
3
, But one-sample t test can still be used if there are no outliers and
− n sufficiently large (n > 100)
− n moderate (20 ≤ n ≤ 100) and little skewness
− n small (n < 20) and data approximately normally distributed.
1.10 Matched pairs t test
Matched pairs t test answers the question: There are two measurements for each individual in the population. Let µ1 and µ2 be their means.
If we observe these two measurements for a random sample, what can we say about µ1 and µ2 ? Does µ1 = µ2 ?
Prices of textbooks at UCLA bookstore and on Amazon.
µ1 : average price at UCLA bookstore
µ2 : average price on Amazon
1. H0 : µ1 = µ2
Ha : µ1 > µ2
Use differences d = x1 − x2 , where x1 is UCLA price and x2 is Amazon price
H0 : µ1 − µ2 = 0
Ha : µ1 − µ2 > 0 ⇒ One-sample t test
2. d = 13.512
sd = 14.972
n = 61
d−µ
Test statistic: t = √0
sd / n
= 13.512−0
√ = 7.049
14.972/ 61
Critical value: t∗α (n − 1) = ∗
t0.05 (60) = 1.671
3. Rejection region: Reject H0 if t > t∗0.05 (60)
4. H0 is rejected at the 5% significance level, the average prices appear to be higher at the UCLA bookstore.
SPSS Output
Tabel 1: Paired Samples Statistics
N Mean Standard deviation Standard Error Mean
UCLA 61 76.533 62.095 7.950
Amazon 61 63.021 51.076 6.540
Tabel 2: Paired Samples Test
Paired Differences
95% Confidence Interval
of the Difference
Mean Standard Standard Lower Upper t df Sig.
Deviation Error Mean (2-tailed)
UCLA-Amazon 13.512 14.972 1.917 9.677 17.346 7.049 60 0.000
4
