Solutions of Exam: Management Research Methods 1 (MRM1)
Friday 27 October 2017, 9:00-12:00 (3 hours)
1a). Mean and standard deviation go down. 2p
Median and inter quartile range do not change. 2p
1b). Highest bars in (Q1, Q2). Lowest bars in (Q2, Q3). 3p [all or nothing!]
1c). You want to avoid falsely choosing H0, because then you would do an invalid t-test. So you
want a small probability of making a type II error. This is achieved by allowing for a relatively
high probability of making a type I error. 5p
2a). P(all 70 components work) = (1 – 0.001)70 = 0.99970 = 0.9324 5p
2b). P(A and B work) = 1 – P(A fails or B fails) = 1 – (0.06+0.03–0.01) = 1 – 0.08 = 0.92 5p
3a). P(X=0) = 0.9, P(X=1) = 0.1
𝜇𝜇 = 0 × 0.9 + 1 × 0.1 = 0.1 3p
𝜎𝜎 2 = (0 − 0.1)2 × 0.9 + (1 − 0.1)2 × 0.1 = 0.09 so 𝜎𝜎 = 0.3 4p
X − µ 0.0755 − 0.1
) P ( X < 0.0755=) P
3b). P ( X 1 + ... + X 1000 < 75.5= < =
σ / n 0.
P ( Z < −2.58) = 1 − 0.9951 = 0.0049 10p
[if a wrong distribution of the sample mean is used, for example the wrong standard deviation,
then maximum 5p]
Friday 27 October 2017, 9:00-12:00 (3 hours)
1a). Mean and standard deviation go down. 2p
Median and inter quartile range do not change. 2p
1b). Highest bars in (Q1, Q2). Lowest bars in (Q2, Q3). 3p [all or nothing!]
1c). You want to avoid falsely choosing H0, because then you would do an invalid t-test. So you
want a small probability of making a type II error. This is achieved by allowing for a relatively
high probability of making a type I error. 5p
2a). P(all 70 components work) = (1 – 0.001)70 = 0.99970 = 0.9324 5p
2b). P(A and B work) = 1 – P(A fails or B fails) = 1 – (0.06+0.03–0.01) = 1 – 0.08 = 0.92 5p
3a). P(X=0) = 0.9, P(X=1) = 0.1
𝜇𝜇 = 0 × 0.9 + 1 × 0.1 = 0.1 3p
𝜎𝜎 2 = (0 − 0.1)2 × 0.9 + (1 − 0.1)2 × 0.1 = 0.09 so 𝜎𝜎 = 0.3 4p
X − µ 0.0755 − 0.1
) P ( X < 0.0755=) P
3b). P ( X 1 + ... + X 1000 < 75.5= < =
σ / n 0.
P ( Z < −2.58) = 1 − 0.9951 = 0.0049 10p
[if a wrong distribution of the sample mean is used, for example the wrong standard deviation,
then maximum 5p]
