HA1011 – Applied Quantitative Methods

Write about the Applied Quantitative Methods.

 

 

Answer:

Class	Frequency	Relative Frequency	Cumulative Relative Frequency	Class Midpoint
1 -780	32	0.533	0.533	390.5
781 - 1560	17	0.283	0.817	1170.5
1561 - 2340	7	0.117	0.933	1950.5
2341 - 3120	3	0.050	0.983	2730.5
3121 - 3900	0	0.000	0.983	3510.5
3901 - 4680	0	0.000	0.983	4290.5
4681 - 5460	0	0.000	0.983	5070.5
5461 - 6240	0	0.000	0.983	5850.5
6241 - 7020	0	0.000	0.983	6630.5
7021 -7800	1	0.017	1.000	3900.5

The data provided represent a population. This is because the whole set of students at Holmes is considered. A population includes all members of a defined group whose information is being collected for decisions that rely on data. Conversely, a sample represents a proportion of the population that is being studied (Anderson et al., 2012). In other words, a sample is made up of elements of a population and is scientifically drawn from the population such that it actually posses the same characteristics as the population.

1. The standard deviation of the weekly attendance is calculated using the formula:

 = “=STDEV.P(A2:A8)” in excel = 68.57

7216. The Inter Quartile Range (IQR) is given by the formula: IQR = Q₃– Q₁, where Q₁ gives the 25^thpercentile = 6114 and Q₃ gives the 75^th percentile = 7216. Therefore, IQR = 7216 – 6114 = 1102.

IRQ is more useful than the standard deviation when dealing with outliers and/or a skewed data set (Anderson et al., 2012).

1. The correlation coefficient was found to be, r = 0.97.

This shows that there is a strong positive linear relationship between weekly attendance at Holmes and the number of chocolates sold. That is, more chocolate is sold when there are more students attending class. Therefore, as the supermarket manager, I would keep fewer chocolate bars in stock when Holmes is closed over the upcoming holiday. 

1. The simple linear regression equation is given by: y = a + b x,

where, x = weekly attendance by students (independent variable)

            y = number of chocolates sold (dependent variable)

a = the y-intercept

b = the coefficient of x variable, which gives the slope of the line.

To find a and b we use the following equations: 

Based on the data we have:

Weekly attendance, X	No. of Chocolate bars sold, Y	XY	X^2	Y^2
472	6916	3264352	222784	47831056
413	5884	2430092	170569	34621456
503	7223	3633169	253009	52171729
612	8158	4992696	374544	66552964
399	6014	2399586	159201	36168196
538	7209	3878442	289444	51969681
455	6214	2827370	207025	38613796
X = 3392	Y =47618	XY = 23425707	X2 = 1676576	Y2 = 327928878

Therefore, a = 1628.69; b = 10.68

The regression equation is given by: y = 1628.69 + 10.68 x

Let’s take two scenarios:

When x = 0; y = 1628.69 + 10.68 * 0 = 1628.69

When x = 10; y = 1628.69 + 10.68 * 10 = 1735.49

This means that when Holmes is closed, the weekly attendance of students is zero, the number of chocolates that are sold remain a = 1628.69. the number of chocolates sold increases when the number of students increases.

1. The coefficient of determination, R², is given by the square of the r coefficient (DeGroot & Schervish, 2012). The formula for obtaining r is:

 = 0.9680

• R²= 0.9370 = 93.70%

The coefficient of determination is 93.70% which indicates that about 94% of the change in the number of chocolates sold is justified by the number of students within Holmes. This means that there is a very strong linear relationship between the weekly attendance of students at Holmes and the number of chocolates sold by the supermarket. 

From the given table we have;

	Scientific training	Grassroots training	Total
Recruited from Holmes students	35	92	127
				External recruitment	54	12	66
Total	89	104	193

Therefore, the probability that a randomly selected player is from Holmes, P(H) = 127/193 =0.66; is from External recruitment, P(E)= 66/193 = 0.34; is receiving Scientific training, P(S) = 89/193 = 0.46; and is receiving Grassroot receiving, P(G) = 104/193 = 0.54 

1. The probability that a randomly chosen player will be from Holmes OR receiving Grassroot training is calculated as:

P (H or G) = P(H) + P(G) – P (H and G),

But, P (H and G) = P(H) * P(G) = 0.66 * 0.54 = 0.3564

Therefore, P (H or G) = 0.66 + 0.54 – 0.3564 = 0.8436

1. The likelihood that player selected at random will be External and be in scientific training is calculated as:

P (E and S) = 54/193 = 0.1813

1. The probability that a player is in scientific training given that he is from Holmes is given by:

P(S|H) = 35/127 = 0.2756

1. To determine whether training is independent for recruitment, we check whether P(S|H) = P(S). That is, P(S|H) = 0.28 0.46 = P(S). Hence, the two events are dependent. Consequently, the type of training, whether scientific or grassroot, does affect whether the player is recruited from Holmes or externally. 

1. We know that a random customer comes from segment A is; P(A) = 0.55, while the likelihood that a person from Segment A prefers Product X is, P(X|A) = 0.20. Therefore, the probability of a random consumer comes from segment A prefers Product X is given by:

P (A  X) = P (A) * P (X|A) = 0.55 * 0.20 = 0.11

1. The probability that a random consumer’s first preference is Product X is determined as:

P (X|A  X|B   X|C  X|D) = 0.20 * 0.35 * 0.60 * 0.90 = 0.038 = 3.8% 

Probability that a customer will make a purchase, P(B) = 1/10 = 0.1

1. Given that there are only two possible outcomes when a customer enters the store, we as assume that the scenario represents a binomial process (DeGroot & Schervish, 2012). We let X represent the customers that actually make a purchase among the people who enter the store. Therefore X ~ B (n,p) where n = the number of people who enter the store and p = 0.1, the probability that a customer who enters the store will make a purchase

n = 8, p = 0.1

To find the probability that only 2 or fewer people will make a purchase, (P  ),we use the formula: P (X =x) =  

Hence (P  ) = P (X  ) + P (X  ) + P (X  )

Where, P (X  ) = = 0.4305

P (X  ) = = 0.3826

P (X  ) = = 0.1488

• (P ) = 0.4305 + 0.3826 + 0.1488 = 0.9619

Therefore, the probability that only 2 or less of those 8 people will buy anything is 0.9619 = 96.19%.

1.  Given that in this case X ~ Po(λ), where, λ= 4, the average number of people entering the store every minute (Anderson et al., 2012). In every 2 minutes, an average of 8 people enters the store.

We employ the poison distribution formula to ascertain the chance that 9 people will enter the store in the nest 2 minutes, P (X = x) =

• P (X = 9) = =0.1241 

Average selling price = $1.1 million = μ; Standard deviation of selling prices = $385,000 = σ

1. The probability that an apartment will sell for over $2 million is calculated by P (Z > z) = 1 – P (Z < z).

Where, z = (X – μ) / σ =  = 2.3377

• P (Z < 2.3377) = 0.9903
• P (Z > 2.3377) = 1 – 0.9903 = 0.0097

 Therefore, the likelihood that an apartment will sell over $2 million is 0.0097 = 0.97%.

1. The probability that an apartment will sell for over $1 million but less than $1.1 million is given by: P (z₁< Z < z₂) = P (Z < z₂) – (1 - P (Z < z₁))

Where, z₁ = = -0.2597

   z_{2 =} = 0

• P (Z < z₂) = P (Z < 0) = 0.5
• 1 - P (Z < z₁) = 1- 0.6025 = 0.3975
• P (-0.2597 < Z < 0) = 0.5 – 0.3975 = 0.1025

Therefore, the probability that an apartment will sell for over $1 million but less than $1.1 million is 0.1025 = 10.25%. 

Sample size, n = 50; average price, μ = $1.1 million; average price,  = $950,000; standard deviation, σ = s = $385,000.

30. Yes, the Z-distribution test can still be used to test the assistant’s research findings against mine. This is because the preferred sample size of 50 is greater than 30. The Central Limit Theorem indicates that as the sample size increases, the mean of the sample distribution approached a normal distribution (Mendenhall, Beaver, & Beaver, 2012).
31. Sample proportion of willing investors = 11/45 = 0.2444

Standard error of the sample proportion is given by:  

• Therefore, the probability that at least 30% of investors would be willing to commit, P (Z > z) is calculated as:

• z = (0.30 – 0.24444) / 0.06406 = 0.8679
• P (Z > 0.8679) = 1 – P (Z 8679) = 1 - 0.8073= 0.1927

Hence, the probability that 30% of the investors would be willing to commit $1 million or more to the fund is 0.1927 = 19.27%

References

Anderson, D. R., Sweeney, D. J., Williams, T. A., Camm, J. D., & Cochran, J. J. (2012). Quantitative methods for business. Cengage Learning.

DeGroot, M. H., & Schervish, M. J. (2012). Probability and statistics. Pearson Education.

Mendenhall, W., Beaver, R. J., & Beaver, B. M. (2012). Introduction to probability and statistics. Cengage Learning.