Exercise 1
B:
The probability that B occurs is the sum of the probability that B occurs and A occurs, and the probability that B occurs
and A does not occur
a: It makes no sense to compute the mean of this probability distribution as shown, because it
doesn’t calculate the necessary data to represent the expected value.
For this consumption to make sense, the probabilities must be:
𝑃(0) = 𝑃(1) = ⋯ = 𝑃(5) = 0,2
b: µ = 0 ∗ 0,68 + 1 ∗ 0,19 + 2 ∗ 0,09 + 3 ∗ 0,03 + 4 ∗ 0,01 = 0,5
Mean: 164,5
Std d: 7,25
Requirement: 162,5
𝑋 − µ 162,5 − 164,5
=
= −0,276
𝑠𝑡𝑑 𝑑
7,25
Use appendix a table a “Standard Normal Cumulative Probabilities”
Or use JMP
By using JMP:
39,13% are not high enough
𝑜𝑏𝑠𝑒𝑟𝑣𝑎𝑡𝑖𝑜𝑛−𝑚𝑒𝑎𝑛
Z-score= 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛 =
162,5−164,5
7,25
= −0,276
By looking at the z table for Standard Normal Cumulative Probabilities, we see that our z-score is equal to 0.3897.
Therefore we can conclude that the proportion of men in India not tall enough to be in Nausena Bharti aviation unit is
38,97%
n: 3
x: number of ships who come home safely
5
P=6
Binomial distribution:
𝑃(𝑋 = 𝑥) =
𝑃(𝑋 = 0) =
𝑛!
−𝑥
∗ 𝑃 𝑥 (1 − 𝑝)𝑛 , 𝑥
𝑥! (𝑛 − 𝑥)!
3!
50
5 3−0
∗ ∗ (1 − )
= 0,0046
0! (𝑛 − 0)! 6
6
A: The sampling distribution of the sample mean for n = 3 has a mean of 85.
𝑠𝑡𝑑 𝑑
8
𝑠𝑒 =
=
= 4,62
√𝑛
√3
B: If the population distribution I approx. normal, then the sampling distribution is approx. normal
for all sizes
C: Standarlize the mean first:
95−85
4,62
= 2,1645
1 − 0,984787001702274 = 0,01521
The probability that arrival of ships in the harbor exceeds 95 is 1,521%
A:
1
µ: 𝑝 = 9 = 0, ,1111
1
1
√𝑝 ∗ (1 − 𝑝) √9 ∗ (1 − 9)
𝑠𝑑 =
=
= 0,0314
𝑛
100
B:
1 1
2 − 9 = 12,38
0,0314
Yes, it is very surprising. It is 12,38 standard deviations away from the mean.
C:
Population distribution: sets of 0’s and 1’s describing whether a container is green (1) or not (0).
11% 1’s and 89% 0’s.
Data distribution: Sets of 50 0’s and 50 1’s describing whether a container is green (1) or not (0).
Sample: The sample proportion: Mean=0,1111 and standard error= 0,0314.
A: 𝑝̂ = 208/602=0,345515
B:
Using the 97,5 and 2,5 normal distribution quantile we can be 95% confident that the proportions
of the worlds containers handled by the the BRIICK countries is between 28,35% and 30,75%
𝑠𝑒 = √0,3455(1 − 0,3455)/602 = 0,0193
𝑝̂ ± 1,96(0,0193) = 0,3455 ± 0,0378 𝑜𝑟 (0.3077, 0.3833)
C:
95% confidence refers to a probability that applies to the confidence interval method. If we use
this method over and over for numerous samples, in the long run we make correct inferences 95%
of the time.
This means that in 95% of the time, the chance of a random container being handled in a port
from the BRIICK countries is between 30,77 and 38,33%.
D:
We can conclude with 95% certainty that the BRIICK countries handled less than half of the worlds
containers because the ci is beneath 0,5
A:
We will find the values in the sample
X = pro merge (740)
N= 1400
𝑥
740
𝑝̂ = =
= 0,5286
𝑛 1400
𝑠𝑒 = √
𝑝̂ (1 − 𝑝̂ )
0,5286 (1 − 0,5286)
0,2491
=√
=√
= √0,00017793 = 0,01334
𝑛
1400
1400
Using this se, a 95% confidence interval for the population proportion is:
𝐶𝐼 = 𝑝̂ ± 1,96(𝑠𝑒) = 0,5286 ± 1,96(0,01334) = 0,5286 ± 0,0262 𝑜𝑟 (0.5024 , 0.5547)
We can predict that the merge would go through because 0,5 is beneathe the confidence
interval.
B:
𝐶𝐼 = 𝑝̂ ± 1,96(𝑠𝑒) = 0,5286 ± 2,58(0,01334) = 0,5286 ± 0,0262 𝑜𝑟 (0.4942 , 0.5630)
Based on the 95% confidence interval, we estimate that the population proportion of
shareholders voting pro-merge was at least 50,25% but no more than 55,45%
The point estimate of 0,5285 has a margin of error of 0,0260.
D:
𝐷𝐹 = 𝑁 − 1 = 20 − 1 = 19
Df=20-1=19. A 95% confidence interval uses the t-score equal to 2,093 because that is the 97,5%
quantile for 19 degrees of freedom and a confidence interval of 95% (0,025 beyond the quantile
on either side)
E:
𝐶𝐼 = µ ± 𝑡0,975 (19) ∗ 𝑠𝑒 = 3,3 ± 2,093 ∗ 1,242 = [0.7; 5.9]
The true mean employee change is likely positive because no negative scores fall in the confidence
interval. The true mean employee change also could be very small, because the lower point is 0,7
is near 0
Shipping problems 4
Exercise 1
9.18
A, b, c, d:
E:
Exercise 2
A:
The relevant variable is how many hours they have worked in the previous week and the
parameter is the mean.
B:
ℎ0 = 𝑝 = 37
ℎ𝑎 = 𝑝 ≠ 37
C:
𝑡 = 𝑥̅ − µ/
D:
𝜎
√𝑛
= 41,6 − 37/
16,83
√67
= 4,6/
16,83
= 4,6/2,0561 = 2,2372
8,1853
Exercise 3
A:
For our study about the cadets reaction time when treated with TS and PP, the variable is the
cadets reaction time measured in ms, which is quantitative. The data is obtained using
randomization, here by a randomized experiment. The figure shows a box plot for the data
collected on the cadets reaction times. We see that the median almost equals the 1st quartile
(25%) There are also outliers on either side, but none that are dramatically removed from the rest
of the observations.
B:
1. Hypotheses
a. Null: 𝐻0 : 𝜇 = 𝜇0
b. Alternative: Ha: 𝜇 ≠ 𝜇0
2. Test statistic
a. 𝑡 =
b. 𝑡 =
𝑥̅ −𝜇0
, where 𝑠𝑒 =
𝑠𝑒
37,33−0
8,2067
𝑠
√
=
𝑛
31,7842
3,8729
= 8,2067
= 4,5492
3. P-value
a. Use t distribution with 𝑑𝑓 = 𝑛 − 1
4. Conclusion
a. Smaller P-values give stronger evidence against H0 and supporting Ha. If using a
significance level to make a decision, reject H0 if P-value is less than or equal to the
significance level (such as 0.05). Relate the conclusion to the context of the study
For the measurement study, we got a P-value of 0.0005 for testing H0: µ = 0 against 𝐻𝑎 : µ ≠ 0 for the mean reaction
times change with a scopolamine patch (TS) and with a placebo patch (PP) change.
With the 0.05 significance level, we would reject 𝐻0 . A 95% confidence interval for the population mean reaction times change µ is
𝑥̅ ± 𝑡0,025 (𝑠𝑒) = 37,33 ± 2,1448 𝑜𝑟 (19,7318 ; 54,9349) ms
The confidence interval shows just how different from 0 the population mean reaction times change is likely to be. It is
estimated to fall between 19,7318 and 54,9349 ms. We infer that the population mean reaction times change µ is
positive because all the numbers in this interval are greater than 0, but the effect of the treatment may be very small,
such as only 19,7318 ms gain in reaction time.
Exercise 4
A:
You should treat the samples on weekdays and weekends as dependent samples because the two
groups (weekdays and weekends) contains the same set of samples (450 oil fields).
B:
You should treat the samples in the two years as independent samples. Because the samples from
each year is randomly selected (450 random sampled oil fields), the samples from one year do not
provide information about the samples for the other year.
Exercise 5
A:
For the population from which this sample was taken, the proportion of people who died due to heart attack is
represented by p1 for taking placebo and p2 for taking aspirin. You divide x with n in each group to get the sample
proportion of people who died due to a heart attack.
B:
For p hat1 = 0,040936 and for p hat2= 0,026627.
𝑝̂ 1 − 𝑝̂ 2 = 0,040936 − 0,026627 = 0,0143085
The proportion of those who died due to heart attack was 0,014 higher for those who took
placebo. In percentage the difference was 4,1%-1,4%=2,7%
C:
If 0 is falling within the confidence interval, the population proportions might well be equal.
D:
Exercise 6
A:
1. Assumptions
a. A categorical response variable observed in each of two groups
b. Independent random samples, either from random sampling or a randomized
experiment
c. 𝑛1 and 𝑛2 are large enough that there are at least five successes and five failures in
each group if using a two-sided alternative
2. Hypotheses
a. Null 𝐻0 : 𝑝1 = 𝑝2 (that is, 𝑝1 − 𝑝2 = 0)
b. Alternative 𝐻𝑎 : 𝑝1 ≠ 𝑝2
3. Test Statistic
a. 𝑧 =
(𝑝̂1 −𝑝̂2 )−0
𝑠𝑒0
1
1
with 𝑠𝑒0 = √𝑝̂ (1 − 𝑝̂ ) (𝑛 + 𝑛 )
1
2
b. 𝑝̂ is the pooled proportion. This is the proportion if you put the samples together
i. 𝑝̂ =
𝑝̂1 ∗𝑛1 +𝑝̂2 ∗𝑛2
𝑛1 +𝑛2
4. P-value
a. P-value = Two-tail probability from standard normal distribution of values even
more extreme than observed z test statistic presuming the null hypothesis is true
5. Conclusion Smaller P-values give stronger evidence against H0 and supporting Ha. Interpret
the P-value in context. If a decision is needed, reject H0 if P-value … significance level (such
as 0.05).
Exercise 7
A:
We check wether or not 0 falls in the interval. When 0 falls in the interval, 0 Is a plausible value for
µ1-µ2, which means that possibly µ1=µ2. The mean rating for those who got caffeine may be 26
seconds lower or 21 seconds larger compared to those who got the placebo.
B:
The p value is above 0,05 (0,48) so we can not reject h0. We don’t have enough evidence to
conclude that in the setting of experiments like this, the population mean for those who got
caffeine differs for those who got the placebo.
C:
Because 0 falls within the confidence interval, 0 is a plausible value for µ1 - µ2, which means that
possibly µ1=µ2
Exercise 8
A:
1. Assumptions
a. A quantitative response variable observed in each of two groups
b. Independent random samples, either from random sampling or a randomized
experiment
c. Approximately normal population distribution for each group. (This is mainly
important for small sample sizes, and even then the two-sided test is robust to
violations of this assumption.)
d. The standard deviation could be said to be the same. 𝜎 = 𝜎1 = 𝜎2
i. This is true when the one of the standard deviation is no more than the
double of the other
2. Hypotheses
a. 𝐻0 : 𝜇1 = 𝜇2 they are the same
b. 𝐻𝑎 : 𝜇1 ≠ 𝜇2 They are not the same
3. Test Statistic
a. 𝑡 =
(𝑥̅ 1 −𝑥̅2 )−0
𝑠𝑒
𝑠2
𝑠2
1
2
where 𝑠𝑒 = √𝑛 + 𝑛
b. 𝑠 is equal to the pooled standard deviation
i. 𝑠 =
(𝑛1 −1)𝑠12 +(𝑛2 −1)𝑠22
𝑛1 +𝑛2 −2
4. P-value
a. P-value = Two-tail probability from t distribution of values even more extreme than
observed t test statistic, presuming the null hypothesis is true with df given by
software.
b. 𝑑𝑓 = 𝑛1 + 𝑛2 − 2
5. Conclusion
a. Smaller P-values give stronger evidence against H0 and supporting Ha. Interpret the
P-value in context and, if a decision is needed, reject H0 if P-value … significance
level (such as 0.05).
B:
The p value is above 0,05 (0,15) so we can not reject h0. We don’t have enough evidence to
conclude that in the setting of experiments like this, the population mean for CBS students study
hall time differs from KU students study hall time.
C:
Yes, in the test of 𝐻0 : 𝜇1 = 𝜇2 against 𝐻𝑎 : 𝜇1 ≠ 𝜇2 , the p-value is > 0,05. The test did not reject
𝐻0 at the 0,05 significance level and therefore conclude that the population means may be equal.
D:
Exercise 9
A:
B: