Ana Ligthart, I6299483
Assignments
WEEK3 - Seminar 3.1
Statistical concepts to be discussed: testing hypotheses, type I and type II error, statistical
power, p-value
Question 1
Pete claims that he can guess the symbol of a randomly selected playing card more than 1/4 of
the time. He randomly selected a card and guessed the symbol of that card. Out of 100 tires, he
guessed 44 times correct. The calculated p-value of observing at least 44 correct guessing if the
true probability was 1/4 is equal to p-value = 0.000027.
Explain why there is strong evidence that Pete can guess the symbol correctly more than 1/4 of
the time.
1
H 0 : p=
4
1
H 1 : p>
4
De kans dat Pete 44 keer juist gokt (of beter) is 0.000027. Dit is heel erg klein, dus is het erg
onwaarschijnlijk dat je net zo goed (44x op 100) random gokt of beter welke kaart het is ( H 0).
Dus Pete is vals aan het spelen of hij is echt paranormaal begaafd. De p-value is kleiner dan 0.05,
dus is er sterk bewijs tegen de nulhypothese en moet deze worden verworpen. \
P-value is de waarschijnlijkheid op een gebeurtenis als je aanneemt dat de nulhypothese klopt.
Het is NIET de kans dat de nulhypothese correct is. Het is de waarschijnlijkheid van jouw data
onder de voorwaarde dat de nulhypothese correct is. De p-value is hier de kans om 44x juist óf
nog meer te gokken wanneer je veronderstelt dat de nulhypothese (1/4) juist is.
X = aantal goed
P(X=44) = (100! /44! * 56!) (1/4)^44 * (3/4)^56
μ (x) = 25
σ = 100 * ¼ * ¾ = 300/16
N = 100
X́−μ (0)
Z=
σ ( x)
, Y = 0 of 1 (fout of goed)
Trek 100x en bereken het gemiddelde van die 0 of 1’tjes
y= proportie goed
μ0 =0.25
( )
1
∗3
2 4 3
σ =
y =
4 16
3
2
2 σy 16
σ́ y = =
n 100
ý−μ0
Z=
σ
Question 2
A physician measured the blood pressure of 100 randomly selected patients and found out that
the observed average blood pressure was X́ =117 mmHg. Suppose the population standard
deviation is known and equal to σ =10. We are interested in the following hypotheses:
H 0 : μ=120 mmHg versus the alternative
H a : μ=115 mmHg (the equal sign is not a typo)
X́−120
The distribution of the test statistic Z= is the standard normal (standard normal means:
σ /10
normally distributed with mean 0 and standard deviation 1).
a. Does the above distribution refer to the population, sample or sampling distribution
(why)?
De sampling distributie, je berekent de Z-score die je vergelijkt met de populatiescore.
b. What does the distribution of X́ look like approximately for large samples when the null
hypothesis is true and how does it look like when the alternative hypothesis is true (draw
these distributions in one figure)?
Gemiddelde nul hypothese = 120
Gemiddelde alternatieve hypothese = 120
σ =10/100 = 1
Assignments
WEEK3 - Seminar 3.1
Statistical concepts to be discussed: testing hypotheses, type I and type II error, statistical
power, p-value
Question 1
Pete claims that he can guess the symbol of a randomly selected playing card more than 1/4 of
the time. He randomly selected a card and guessed the symbol of that card. Out of 100 tires, he
guessed 44 times correct. The calculated p-value of observing at least 44 correct guessing if the
true probability was 1/4 is equal to p-value = 0.000027.
Explain why there is strong evidence that Pete can guess the symbol correctly more than 1/4 of
the time.
1
H 0 : p=
4
1
H 1 : p>
4
De kans dat Pete 44 keer juist gokt (of beter) is 0.000027. Dit is heel erg klein, dus is het erg
onwaarschijnlijk dat je net zo goed (44x op 100) random gokt of beter welke kaart het is ( H 0).
Dus Pete is vals aan het spelen of hij is echt paranormaal begaafd. De p-value is kleiner dan 0.05,
dus is er sterk bewijs tegen de nulhypothese en moet deze worden verworpen. \
P-value is de waarschijnlijkheid op een gebeurtenis als je aanneemt dat de nulhypothese klopt.
Het is NIET de kans dat de nulhypothese correct is. Het is de waarschijnlijkheid van jouw data
onder de voorwaarde dat de nulhypothese correct is. De p-value is hier de kans om 44x juist óf
nog meer te gokken wanneer je veronderstelt dat de nulhypothese (1/4) juist is.
X = aantal goed
P(X=44) = (100! /44! * 56!) (1/4)^44 * (3/4)^56
μ (x) = 25
σ = 100 * ¼ * ¾ = 300/16
N = 100
X́−μ (0)
Z=
σ ( x)
, Y = 0 of 1 (fout of goed)
Trek 100x en bereken het gemiddelde van die 0 of 1’tjes
y= proportie goed
μ0 =0.25
( )
1
∗3
2 4 3
σ =
y =
4 16
3
2
2 σy 16
σ́ y = =
n 100
ý−μ0
Z=
σ
Question 2
A physician measured the blood pressure of 100 randomly selected patients and found out that
the observed average blood pressure was X́ =117 mmHg. Suppose the population standard
deviation is known and equal to σ =10. We are interested in the following hypotheses:
H 0 : μ=120 mmHg versus the alternative
H a : μ=115 mmHg (the equal sign is not a typo)
X́−120
The distribution of the test statistic Z= is the standard normal (standard normal means:
σ /10
normally distributed with mean 0 and standard deviation 1).
a. Does the above distribution refer to the population, sample or sampling distribution
(why)?
De sampling distributie, je berekent de Z-score die je vergelijkt met de populatiescore.
b. What does the distribution of X́ look like approximately for large samples when the null
hypothesis is true and how does it look like when the alternative hypothesis is true (draw
these distributions in one figure)?
Gemiddelde nul hypothese = 120
Gemiddelde alternatieve hypothese = 120
σ =10/100 = 1
