Ha1011 Applied Quantitative Methods For Assessment Answers

Answer:

Measures of variability and association 

Based on the provided information, it could be said that the above could be considered as population and not sample. This is because all the students buying chocolate bars at Holmes Institute are considered in this case. In this context, Mendenhall & Sincich (2016) stated that population signifies the accumulation of all the components having similar characteristics. On the other hand, sample denotes a portion of the entire population selected to participate in the study. Since all the students of Holmes Institute are selected, the provided information denotes that this is a case of population. 

Based on the above table, it could be stated that standard deviation is obtained as 74.06. In this respect, Gravetter & Wallnau (2016) cited that standard deviation is used for the variation amount of data value set. 

For computing the inter-quartile range, the first quartile and the third quartile are computed initially. After these two values are computed, the first quartile is subtracted from the third quartile to arrive at the inter-quartile range. With the help of inter-quartile range, it is possible to make an initial projection of the outliers by assessing the values above 1 ½ times the distance of inter-quartile range above the third quartile or below the first quartile (Wan et al., 2014). This could not be estimated with the help of standard deviation. In the above case, based on the inter-quartile range, it could be stated that the bulk of the values lie at 1,102.  

Based on the above table, it could be stated that the correlation coefficient is obtained as 0.97. As commented by Rocco et al., (2017), correlation coefficient signifies the strength of the association between relative movements in two variables. In the above case, the weekly attendance and chocolate bars have been observed to have strong relationship with each other, since the value is close to 1.

Linear regression 

For finding out the linear regression equation, the following computations are made: 

In order to calculate the values of a and b, the following formulas are used: 

Thus, by using the above formula, the regression equation obtained is

y = 1,628.69 + 10.68x

This could be validated further with the help of Excel calculations in the form of data analysis tool used and these values are found to be the same as the coefficient values represented below:   

In this case, the weekly attendance is considered as the independent variable and the number of chocolate bars sold is taken into account as the dependent variable. According to the above tables, it could be said that p-value is lower than 0.05, which implies evidence of strong relationship between the above two variables. 

From the above table, it could be stated that R² is the coefficient of determination. The value is obtained as 0.9370, which implies that 93.70% of the dependent variable could be predicted by the independent variable.

Probability

The probability that a randomly selected player would be from Holmes or receiving grassroots training could be evaluated with the help of the following formula:

P (Holmes or grassroots training) = (35 + 92 + 12)/ (35 + 92 + 54 +12) = 139/193 = 0.7202

The probability that a randomly selected player would be external and scientific training could be evaluated with the help of the following formula:

P (External and scientific training) = 54/ (35 + 92 + 54 +12) = 54/193 = 0.2798

The probability that a player is in scientific training and from Holmes could be evaluated with the help of the following formula:

P (Scientific training and Holmes) = 35/ (35 + 92) = 35/127 = 0.2756

In case, independency could be found in two events A and B,

P (A) x P (B) = P (A & B)

P (External) = (54 + 12)/ (35 + 92 + 54 +12) = 66/193 = 0.3412

P (Scientific training) = (35 + 54)/ (35 + 92 + 54 +12) = 89/193 = 0.4611

P (External) x P (Scientific training) = 0.3412 x 0.4611 = 0.1573

P (External and scientific training) = 0.2756  0.1573

Therefore, the events are not independent. The training methodologies and player recruitment are not independent events, which imply that one variable has effect over occurrence profitability of the other event.

Bayes’ Rule 

Based on the provided information, the following could be illustrated as follows:

P (A) = 0.55, P (B) = 0.3, P (C) = 0.1 and P (D) = 0.05

Preference for X is provided as follows:

P (X/A) = 0.2, P (X/B) = 0.35, P (X/C) = 0.6 and P (X/D) = 0.9

According to the total probability law,

P (X) = P (A) P (X/A) + P (B) P (X/B) + P (C) P (X/C) + P (D) P (X/D)

P (X) = (0.55 x 0.2) + (0.3 x 0.35) + (0.1 x 0.6) + (0.05 x 0.9)

P (X) = 0.32

According to the Baynesian probability,

P (A/X) = [(P (A) x P (X/A)]/ P (X)

P (A/X) = (0.55 x 0.2)/0.32

P (A/X) = 0.3438 

The probability that the random customer’s first preference would be product X is P (X), which is obtained as 0.32. 

For this problem, X is considered as the random variable, which implies the number of individuals actually purchasing among those individuals entering the store.

It is evident that X ~ binomial (n, p), in which n is equal to the number of individuals entering the store and p denotes the probability that an individual would purchase any product. 

In this case, n is provided as 8 and p is provided as 0.1.

P (Less than 2 customers would purchase) = P (X ≤ 2) = P (X = 0) + P (X = 1) + P (X = 2)

In this case, P (X = x) =

At n = 8 and p = 0.1,

P (Less than 2 customers would purchase) =  +  +

P (Less than 2 customers would purchase) = 0.961 

It is provided as the number of individuals entering the store is denoted as X.  X ~ Poisson (λ) at the slot of 10 - 11 a.m. and λ is provided as 4 in a minute. As 4 individuals enter the store in 8 minutes, this implies that 8 individuals enter the store in 2 minutes. Therefore, Y could be denoted as the number of individuals entering the store in next two minutes, which is Poisson (2 λ).

P (9 individuals would enter in next two minutes) = P (Y = 9/ Y ~ Poisson (8)

P (9 individuals would enter in next two minutes) = e^-8 8⁹/ 9! = 0.124 

According to the provided situation, μ is equal to $1.1 million or $1,100,000 and σ is given as $385,000. 

Probability that the apartment would be sold for above $2 million, P (X > 2,000,000)

Z-score for (X = 2,000,000)

Z = (x - μ)/ σ

Z = (2,000,000 - 1,100,000)/ 385,000

Z = 2.34

Hence, it could be stated that P (X > 2,000,000) = P (Z > 2.34). By using the normal distribution table, P (Z > 2.34) is obtained as 0.0097 and this the probability of selling the apartment for above $2 million. 

Probability that the apartment would be sold for more than $1 million; however, below $1.1 million is P (1,000,000 < 1,100,000).

Z-score for (X = 1,000,000)

Z = (x - μ)/ σ

Z = (1,000,000 - 1,100,000)/ 385,000

Z = - 0.26

Z-score for (X = 1,100,000)

Z = (x - μ)/ σ

Z = (1,100,000 - 1,100,000)/ 385,000

Z = 0

Therefore, P (1,000,000 < 1,100,000) = P (-0.26 < Z < 0) = P (Z < 0) – P (Z < -0.26). According to the normal distribution table, P (Z < 0) – P (Z < -0.26) = 0.5 – 0.3975 = 0.1025. Therefore, the probability that the apartment would be sold for above $1 million; however, below $1.1 million is 0.1025. 

According to the theorem of central limit, the mean sampling distribution approaches normal distribution with the increase in sample size. Hence for a large sample greater than 30, normal distribution would be followed on the part of average price. Since the property sample size is 50, the average price would follow normal distribution and thus, Z-distribution test could be used for testing the research findings of the assistant against the individual (Anderson, Sweeney & Williams, 2014). 

Sample proportion of investors willing to invest = 11/45 = 0.2444

Standard error of the sample population = square root of [p (1 - p)/n]

Standard error of the sample population = square root of [0.2444 (1 – 0.2444)/45] = 0.064

Probability that 30% of the investors would be willing to commit is P [p > 0.30]

Probability that 30% of the investors would be willing to commit = P [z > (0.3 – 0.2444)/ 0.064)

Probability that 30% of the investors would be willing to commit = P [z > 0.87] = 0.192 

References and Bibliographies:

Anderson, D., Sweeney, D., & Williams, T. (2014). Modern business statistics with Microsoft Excel. Nelson Education.

Gravetter, F. J., & Wallnau, L. B. (2016). Statistics for the behavioral sciences. Cengage Learning.

Keller, G. (2016). Modern Business Statistics. Cengage Learning.

Mendenhall, W. M., & Sincich, T. L. (2016). Statistics for Engineering and the Sciences. Chapman and Hall/CRC.

Rocco, P., Cilurzo, F., Minghetti, P., Vistoli, G., & Pedretti, A. (2017). Simulation data for an estimation of the maximum theoretical value and confidence interval for the correlation coefficient. Data in brief, 14, 291.

Selvanathan, E. A., Selvanathan, S., & Keller, G. (2016). Business Statistics: Australia New Zealand with Student Resource Access for 12 Months. Cengage AU.

Siegel, A. (2016). Practical business statistics. Academic Press.

Wan, X., Wang, W., Liu, J., & Tong, T. (2014). Estimating the sample mean and standard deviation from the sample size, median, range and/or interquartile range. BMC medical research methodology, 14(1), 135.