 # HI6007 STATISTICS FOR BUSINESS DECISION

QUESTION 1:

1. Population Parameter and Sample Statistic

Population Parameter is a number or a value which describes the characteristics of the Populations. Parameter can be mean of the population, median of the population, mode, range etc.

Sample Statistic is similar to Population parameter. The number or value which is computed from the sample is called as Sample Statistic. It is believed that this statistic which has been computed from the sample holds the characteristics of the population and it is used to make decision on population.

1. Descriptive Statistics and Inferential Statistics

Descriptive Statistics is the study of the available data and interpreting the result for the set of the data. In other words, if we have taken data from 100 people and the result that we obtain is purely based on those 100 people’s input. We do not extrapolate this data to the extended group of this 100 people. Any study that we do for our sample stays with the sample. It is not applied to the population.

Inferential Statistics: The statistic that we compute from the sample is used to infer some result about the population. Then it is said to be Inferential Statistics. For example, we like to know the mean height of the people in UK. Instead of finding the heights of the people of the whole country, we take sample of 100 people or more and we find the average height of those people. The result thus obtained is used to understand the mean height of the population.

1. Nominal Scale and Ordinal Scale

Nominal Scale is the scale of measurement in which the numbers are just used a labels. For example, we might have come across survey’s in which they use a five point scale (1- 5) to describe from highly satisfied to highly dissatisfied. Here, 1 represents highly satisfied and 5 represents highly dissatisfied.

Ordinal Scale is the scale of measurement in which the variables are ordered based on the preferences. The distance between them may not be equal. It just ranks the data based on some preferences. The good example of the Ordinal scale is the ranking of people in Olympics or in a fashion show.

1. Primary Data Source and Secondary Data Source

If the person who is intending to do a study on a subject and if that person collects the data by use of survey or through some other means, then it is said to be primary data. In other words, the researcher he himself collects the data from the respondents or the source.

Secondary Data is the data which has been already collected by some government institution or an agency or any other entity or a person. The researcher just takes the data as it is for his work and if he uses the same, then it is said to be secondary data. For example, the price details that has been taken for a share from the stock market is the secondary data.

QUESTION 2 – WEEK 3:

1. Frequency Distribution and percent Frequency Distribution
 Class Interval (No in Million) Frequency Percent Frequency Distribution 0.0-2.5 15 30% 2.5-5.0 13 26% 5.0-7.5 10 20% 7.5-10.0 5 10% 10.0-12.5 1 2% 12.5-15.0 2 4% 15.0-17.5 0 0% 17.5-20.0 2 4% 20.0-22.5 0 0% 22.5-25.0 0 0% 25.0-27.5 1 2% 27.5-30.0 0 0% 30.0-32.5 0 0% 32.5-35.0 0 0% 35-37.5 1 2% Total 50 100%

Histogram:

1. Skewness in the distribution:

The Skewness for the above data is found to be 1.58. Since the skewness is > 0, it means that the data is positively skewed. All the data is distributed near 0 and the distribution is very low towards higher end.

From the histogram, we find that the population is very low in most of the states. 38 states have population less than 7.5 Million. There is only one state with the population in the range of 35 to 37.5 Million. From skewness, we understand that the more number of states have low population.

QUESTION 2 – WEEK 4:

1. Joint Probability Table:(GR Terrell – 2006, Mathematical statistics: A unified introduction)
 Particulars Uses Social Media and Other Websites to Voice Opinions About Television Programs Doesn’t Use Social Media and Other Websites to Voice Opinions About Television Programs Total Female 29.0% 21.3% 50.3% Male 23.7% 26.0% 49.7% Total 52.6% 47.4% 100.0%

1. Probability that a respondent is a female:

The Probability that a respondent is a female is 50.3%

1. Conditional Probability of a respondent uses social media and other websites to voice opinions about television programs given the respondent is female

P(A/B) = P(A∩B)/P(B) = 29%/50.3% = 57.65%

1. F = event that the respondent is a female

A = event that the respondent uses social media and other websites to voice opinions about television program.

Event F and A are said to be independent,

If P(F∩A) = P(F)*P(A)

P(F∩A) = 57.65%

P(F) = 50.3%

P(A) = 52.6%

P(F)*P(A) = 50.3%*52.6% = 26.46%

Since P(F∩A) ≠ P(F)*P(A). Events F and A are not independent.

QUESTION 3 – WEEK 5:

Mean (µ) = \$20,000

Standard Deviation (σ )= \$8,000

1. Probability that a randomly selected LU graduate will have a starting salary of atleast \$30,400

Z = (X-µ)/σ     = (30,400 – 20,000)/8000 = 10,400/8000 = 1.3

P(X>30,400) = P(Z>1.3) = 1- P(<1.3) = 1-0.9032 = 0.0968 (or) = 9.68%

(GR Terrell – 2006, Mathematical statistics: A unified introduction)

1. The probability that a randomly selected LU graduate will have a salary of \$30,400 is “Zero”.

The probability for a point in a normal distribution is zero.

1. Percentage of Graduates who will receive the tax break = P(X<15,600)

Z = (X-µ)/σ     = (15,600 – 20,000)/8000 =4,400/8000 = -0.55

= P(Z<-0.55)

= 0.5 – 0.2088

= 0.2912         = 29.12%

1. 189 recent graduates have salaries of atleast \$32,240.

Probability of the students getting salaries greater than \$32,240

= P(X>32,240)

Z = (X-µ)/σ     = (32,240 – 20,000)/8000 = 12,240/8000 = 1.53

P(Z>1.53) = 1-0.9370 = 0.063 = 6.3%

We understand that 6.3% is equal to 189 students.

The total number of students = 189/6.3% = 3000.

Number of students graduated this year from the university is 3000.

QUESTION 3 – WEEK 6:

Given: Mean score of Critical reading = 502

Mean score of Mathematics = 515

Mean score of Writing = 494

Population Standard Deviation = 100

1. Mean (µ) = 502

Standard Deviation (σ)= 100

n = 90

P(-10<x̅ -µ<10) = ?

Z =  (x̅ -µ)/(σ/ n) = (±10/(100/√90)) = ±10/10.542 = ±0.9487 =±0.95(HG Tucker – 2014, An introduction to probability and mathematical statistics)

Now, P(-10<x̅ -µ<10) = P(-0.95<Z<0.95)

= 0.3289+0.3289

= 0.6578 = 65.78%

1. b) Mean (µ) = 515

Standard Deviation (σ)= 100

n = 90

P(-10<x̅ -µ<10) = ?

Z =  (x̅ -µ)/(σ/ n) = (±10/(100/√90)) = ±10/10.542 = ±0.9487 =±0.95(HG Tucker – 2014, An introduction to probability and mathematical statistics)

Now, P(-10<x̅ -µ<10) = P(-0.95<Z<0.95)

= 0.3289+0.3289

= 0.6578 = 65.78%

The probability for both critical reading and maths subjects with the sample of 90 and the mean test score within 10 points of population mean is 65.78%