Classify each one of the following variables as either measurable (continuous) or categorical.

Class of cabin on an airliner: 'first class', 'business class' and 'economy class: Ordinal categorical variable. The classes of cabin can be ranked based on preference first class to economy class. It is categorical since cabin class cannot be measured.

The carbon dioxide emissions of an airliner: Measurable variable. CO2 emissions can be measured in kilograms to several decimals.

The nationalities of airline passengers: Nominal categorical variable. The nationalities of individuals do not have a natural ranking order. Even if arranged in alphabetical order, it does not show any ranks.

State whether the following are true or false and give a brief explanation

Skewness of a distribution cannot be determined from a boxplot

FALSE: A box plot can describe the skewness of a distribution. The length of the upper and lower sections or whiskers indicate the skewness of the distribution. If they are not equal, then the distribution is skewed.

For any event A, it is possible for P(A) + P(Ac) < 1, where Ac denotes the complement set.

FALSE: If the two events are complements, then the sum of their probabilities must be equal to one (1).

A standardised normal random variable has a standard deviation equal to its variance.

TRUE: The standard deviation of a standardised normal random variable is 1. Since standard deviation is the square root of variance, I implies that variable must also be equal to one.

A lower level of confidence produces a wider confidence interval

FALSE: a lower level of confidence implies that a lower precision is required hence the confidence interval produced is narrow. The confidence interval is wider for a higher level of confidence.

A cross-sectional design collects data over time

FALSE: Cross-sectional design collects data at a given time. A longitudinal design is the one that collects data over time.

Determine the quotas to be selected from each age group

18 - 29 = 20% 500 = 100

30 - 49 = 35% 500 = 175

50 - 69 = 30% 500 = 150

70+ = 15% 500 = 75

Probability distribution of random variable Y

Expected value of Y

E(Y) = i=1ny.p(y)= (2 0.1) + (4 0.3) + (6 0.4) + (8 0.2)

= 0.2 + 1.2 + 2.4 + 1.6

= 5.4

Standard Deviation of Y

Variance of Y = E(Y2) - E(Y)2

Standard deviation = VarianceE(Y2) = i=1ny2.p(y) = (22 0.1) + (42 0.3) + (62 0.4) + (82 0.2)

= 0.4 + 4.8 + 14.4 + 12.8

= 32.4

Variance of Y = 32.4 - 5.42 = 3.24

Standard deviation = 3.24 = 1.8

Probability that Y is within 1 standard deviation of the expected value

Z = Y- sIf y is within one standard deviation of the mean, then Z = 1

P(Z <1) = 0.8413

Explain why Y does not have a discrete uniform distribution

Y does not have a discrete uniform distribution since the probability varies for each value of Y as provided in the table. If Y had a discrete uniform distribution, then every value of Y would have an equal probability.

A population is normally distributed with = 138 and = 21. Given a simple random sample of size n = 25, determine the probability that the sample mean, X, will be less than 128

Z = X- sn = 128- 1382125 = -2.38

P(Z < -2.38) = 0.0087

Two cards are chosen at random from a standard 52-card deck.

Are the events A and B independent?

Events A and B are independent. Since the cards are chosen with replacement, getting A does not affect the probability of event B happening.

Are the events A and B Mutually Exclusive?

Events A and B are not mutually exclusive. Since there is replacement, event A occurring does not exclude event B from occurring.

Without replacement

Are the events A and B Mutually Exclusive

Events A and B are mutually exclusive since there is no replacement. There is only one ace of spade. It implies that if the first card chosen is an ace of spade, there will be no more ace of spade. Thus, the probability of B is zero. Either A or B can occur, but not both A and B.

Are the events A and B Independent?

Events A and B are not independent since there is no replacement. If A occurs, B cannot occur since there will be no ace of spade left. Therefore, event B occurring is dependent on A not occurring.

SECTION B

Question 2

Tutoring and Performance in Physics

Based on the data in the table, and without doing a significance test, how would you describe the relationship between receiving private tutoring and performance in Physics?

Based on the numbers and percentages, there seems to be a positive relationship between receiving private tutoring and performance in physics. The percentages suggest that the performance in physics if higher for those who have received private tutoring than for those who have not. For instance, for students who have not received tutoring, 11% recorded low performance in physics. For the group that receives some tutoring, the percentage of low performers decline to 5%. For the students who receive frequent tutoring, the percentage of low performances drops further to 2%. Similarly, only 48% of the students that do not receive tutoring record high performance. The percentage of high performers increases to 63% for students who receive some tutoring, and 79% for the group that receives frequent tutoring.

Calculate the kh2 statistic and use it to test for independence, using a 10% significance level. What do you conclude?

Null hypothesis, H0: There is no association between performance in physics and of private tutoring

Alternative hypothesis, HA: There is an association between performance in physics and of private tutoring

Performance in Physics

Tutoring Low Medium High Total

No Tutoring 46 168 196 410

Some Tutoring 100 572 1148 1820

Frequent Tutoring 32 248 1076 1356

Total 178 988 2420 3586

Calculating kh2

Cell OiEi(Oi - Ei)2 (Oi - Ei)2/EiR1C1 46 20.35 657.85 32.32

R1C2 168 112.96 3029.23 26.82

R1C3 196 276.69 6510.41 23.53

R2C1 100 90.34 93.31 1.03

R2C2 572 501.44 4978.86 9.93

R2C3 1148 1228.22 6435.39 5.24

R3C1 32 67.31 1246.68 18.52

R3C2 248 373.60 15775.25 42.23

R3C3 1076 915.09 25891.38 28.29

Expected values:

R1C1 = (410 178)/3586 = 20.35

R1C2 = (410 988)/3586 = 112.96

R1C3 = (410 2420)/3586 = 276.69

R2C1 = (1820 178)/3586 = 90.34

R2C2 = (1820 988)/3586 = 501.44

R2C3 = (1820 2420)/3586 = 1228.22

R3C1 = (1356 178)/3586 = 67.32

R3C2 = (1356 988)/3586 = 373.60

R3C3 = (1356 2420)/3586 = 915.09

kh2 = (Oi-Ei)2Ei= 32.32 + 26.82 + 23.53 + 1.03 + 9.93 + 5.24 + 18.52 + 42.23 + 28.29

= 187.91

Degrees of freedom = (r - 1)(c - 1) = (3 - 1)(3 -1) = 4

The critical value of Chi-square at 10% significance level and 4 degrees of freedom = 7.779.

187.91 > 7.779.

The test statistic is greater than the critical value. Therefore, the null hypothesis is rejected. It can be concluded that there is a significant association between receiving private tutoring and performance in physics.

Would you conclude that private tutoring improves the performance in Physics? Briefly justify your answer.

The hypothesis test is (ii) above shows that there is a significant association between receiving private tutoring and performance in Physics. It implies that there is a significance difference between the performance of those who receive tutoring and those who do not receive tutoring. Based on the analysis, it can be concluded that private tutoring improves the performance in Physics.

Design a nationwide survey

Sampling scheme and sampling frame

I would use a simple random sampling approach. This approach is suitable since it is possible to obtain the list of all twitter users nationwide using tools such as Google Tags, among other tools. Random sampling is appropriate since it gives each of the university students using twitter an equal chance of selection for the survey. The sampling frame for this survey includes all university students who are twitter users in the country.

Although random sampling reduces selection bias, there would still be a potential for selection bias. The simple random sample may be skewed towards some universities. This implies that there can be a disproportionate representation of the universities and other groups in the total sample. To mitigate this bias, an initial simple random sample can be selected and details of the selected students such as the name of the university, region, gender, age, among other details analysed to determine if the sample is a good representative of the population.

Design for longitudinal survey

The survey can be redesigned to a longitudinal survey by collecting data on the same students collected over time. For instance, the sampling frame can be restricted to students who have been active users on twitter for at least three years. Data can be collected on the twitter usage for each of the last three years or more.

Question 3

Relationship between the obtained yield of apple trees and the amount of weeds found in their roots

Scatter diagram

Sample Correlation Coefficient

r = i=1nxiyi-nxy(i=1nxi2-nx2)(i=1nyi2-ny2)i=110xi = 285

i=110yi = 355

i=110xi2 = 8419

i=110yi2 = 13349

i=110xiyi = 9718

n = 10

x = 285/10 = 28.5

x = 355/10 = 35.5

r = 9,718-(10 28.5 35.5)(8419-10 28.52)(13349-10 35.52)= 9,718-10,1177.5 296.5 746.5= -399.5221,337.25= -0.8492

The correlation coefficient is negative and close to 1. It implies that there is a strong negative correlation between yield and weed in roots. Thus, a large amount of weed in roots is associated with lower yields.

Calculate the least squares line of y on x and draw the line on the scatter diagram

Y = a + bx, where Y is yield and x is weed in roots

b = i=1nxiyi-nxyi=1nxi2-nx2= 9,718-(10 28.5 35.5)8419-10 28.52= -399.5 296.5= -1.3474

a = y-bx = 35.5 - (-1.3474 28.5) = 73.901

Equation:

Yield = 73.901 - 1.3474Weed

Y = 73.901 - 1.3474x

Plotting the Equation

If x = 40, y = 73.901 - 1.3474 40 = 20.005

If y = 50, x = (50 - 73.901)/1.3474 = 18

y x

20 40

50 18

Based on the regression equation in part (iii.), what will be the predicted yield of apple for a tree with 37 grams of weeds in its roots? Will you trust this value? Justify your answer

Yield = 73.901 - (1.3474 37)

= 73.901 - 49.8538

= 24.05 kilograms

Suitability of the simple linear regression model for the data of this question

The simple linear regression model is suitable for this data. As shown by the scatter plot and correlation coefficient, there is a strong correlation between the two variables. Besides, the association between the two variable appear to be linear. Therefore, the data meets the essential conditions for linear regression. The correlation coefficient also suggests that the model's R2 is about 0.72. It indicates that changes in weed in roots explain 72% of the changes in yield. Therefore, the explanatory power of the model is high.

Quality of Customer Service

Difference between two means

Null hypothesis, H0: x2013 = x2012Alternative hypothesis, HA: x2013 x2012Calculating the test-statistic:

The population standard deviation is not known, hence the t-statistic is used.

t = (x1- x2)s12n1+ s22n2 = (6.1-5.8)0.6234+ 0.5241 = 2.193

Degrees of freedom = the smaller of (41 -1) and (34 - 1) = 33

Test at 99% confidence Interval

Critical value of t0.005, 33 = 2.733

2.193 < 2.733

The test statistic is less than the critical value, indicating that the null hypothesis cannot be rejected. Therefore, it can be concluded that the mean waiting times were not different between these two years

Test at 95% confidence Interval

Critical value of t0.025,...