# Statistics for Business Decisions Assignment Samples | Assignment Help

Assignment: Tutorial Questions 1

Week 2

Question 1:

(a) The numeric properties of a population are known as population parameter.

The numeric properties of a sample are known as sample statistics.

Suppose that the mean electricity bill for all the families of a society is \$500 and the mean electricity bill for randomly selected families in society is \$477. Then \$500 is population parameter and \$477 is sample statistics.

(b) The summery measure for a data set calculated in the form of mean, median, mode or standard deviations to explain the characteristics of a sample is descriptive statistics.

When we do calculation with help of a sample and generalize the result for the population from which the sample is taken then it is inferential statistics.

(c) In Nominal scale, we cannot perform any arithmetic operations like addition, multiplications etc. These are categorical variables without any meaningful order. Example: Subject major of student, Favourite sports etc.

If we can arrange categorical data or numeric data in order then it is ordinal variable. For example: Our grade as A, B, C or D

(d) If a researcher directly collects the data to analyse it for the claim of research then the data source is primary.

If a researcher uses already published data by some valid and reputed sources then it is secondary data.

Week 3:

Question 2.

(a) Frequency distribution:

 Population Frequency (f) Percentage Frequency 0.6 - 3.0 21 42% 3.1 - 5.5 9 18% 5.6 - 8.0 9 18% 8.1 - 10.5 4 8% 10.6 - 13.0 3 6% 13.1 - 15.5 0 0% 15.6 - 18 0 0% 18.1 - 20.5 2 4% 20.6 - 23.0 0 0% 23.1 - 25.5 1 2% 25.6 - 28.0 0 0% 28.1 - 30.5 0 0% 30.6 - 33.0 0 0% 33.1 - 35.5 0 0% 35.6 - 38.0 1 2% Total 50 100%

Histogram for population:

(b) Shape of histogram is positively or right skewed.

(c) Most of the state populations are clustered around lower-class intervals except few states having very high population. These high population making the shape of distribution right skewed.

Week 4

Question 2:

(a) Joint Probability table:

Let A = Uses Social Media and Other Websites to Voice Opinions About Television Programs

B = Doesn’t Use Social Media and Other Websites to Voice Opinions About Television Programs

 A B Total Female 395/1364 = 0.2896 291/1364 = 0.2133 686/1364 = 0.5029 Male 323/1364 = 0.2368 355/1364 = 0.2603 678/1364 = 0.4971 Total 718/1364 = 0.5264 646/1364 = 0.4736 1

(b) P(Female) = 6861364=0.5029

(c) We need to find the conditional probability of respondent uses social media and other websites to voice opinions about television programs given the respondent is female.

P(A/Female) = 395686=0.5758

(d) P(A) = 0.5264

P(F) = 0.5029

P (A and F) = 395/1364 = 0.2896

Now P(A) * P(F) = 0.5264 * 0.5029 = 0.2647

So as P(A) * P(F) ? P (A and F), A and F are not independent events.

So, F and A are not independent.

Week 5:

Question (3) Given that population parameters as:

µ = 20,000 and SD = 8000

(a) z score for x = 30400,

Z = 30400-200008000=1.3

Hence,

P (at least 30400)

= P (z ? 1.3)  = 0.0968

(b) P (x = 30400) = 0

Getting a particular value under a normal curve is always 0 as it cannot represent any area.

(c) Income of 15600 is separating the low-income tax break. So,

P (low income tax break)

= P (x < 15600>

= P (z <15600-200008000)

= P (z < -0.55)

= 0.2912

(d) Proportion of graduates getting at least 32,240

P (x ? 32240)

= P (z ?32240-200008000)

= P (z ? 1.53)

= 0.063

If total number of students graduated is n then

N = 189/0.063 = 3000 students

Week 6:

Question 3.

(a) Here sample size n = 90.

Hence standard error of mean = 10090=10.541

Z score representing the boundaries within 10 of the population mean is

Z = ± 10/10.541 = ±0.95

P(-0.95 < z>

= 2 * 0.3289 = 0.6578

(b) Here sample size n = 90.

Hence standard error of mean = 10090=10.541

Z score representing the boundaries within 10 of the population means is

Z = ± 10/10.541 = ±0.95

P(-0.95 < z>

= 2 * 0.3289 = 0.6578

As the standard error of mean is same for both, and difference in x and population mean is also fixed as 10, no change in probability.

Assignment: Tutorial Questions 1

Week 2

Question 1:

(a) Population Parameter and Sample Statistic

Population Parameter: Quantitative property of a population is defined as population parameter: For example: population mean and population standard deviation.

Sample statistics: Quantitative property of sample is defined as sample statistics. For example: sample mean or sample standard deviation.

Suppose that in a statistics class total 100 students are enrolled. Mean GPA for all the 100 students is population parameter. If random sample of 20 students taken then mean of 20 students is sample statistics.

(b) Descriptive Statistics and Inferential Statistics

The process of summarizing the characteristics of a sample with help of mean, median, standard deviation, variance or range is descriptive statistics.

Making an inference or conclusion about population on the basis of calculations done for the sample is known as inferential statistics.

(c) Nominal Scale and Ordinal Scale

Nominal scale is defined as scale without numbers. These are just categories without an order. Particularly names, places or characters can be taken as nominal scale. Like colour, name of places etc.

Ordinal scale is defined as a scale in which the data values can be ordered. For example, ratings of a product from 1 to 10 can be ordinal.

(d) Primary Data Source and Secondary Data Source

In a primary data source, a researcher directly collects data from the target population for the analysis or testing the research hypothesis.

In Secondary data source, a researcher uses already collected or available data on some valid source or already published sources.

Week 3:

Question 2.

(a) Frequency distribution:

 Class interval Frequency % Frequency 0.6 - 3.0 21 42% 3.1 - 5.5 9 18% 5.6 - 8.0 9 18% 8.1 - 10.5 4 8% 10.6 - 13.0 3 6% 13.1 - 15.5 0 0% 15.6 - 18 0 0% 18.1 - 20.5 2 4% 20.6 - 23.0 0 0% 23.1 - 25.5 1 2% 25.6 - 28.0 0 0% 28.1 - 30.5 0 0% 30.6 - 33.0 0 0% 33.1 - 35.5 0 0% 35.6 - 38.0 1 2% Total 50 100%

Histogram:

(b) Yes, the shape of distribution looks like right skewed as there is longer tail in the right side clearly visible in histogram.

(c) Hence it looks like the population of states normally in lower end except some high outliers making the distribution skewed.

Week 4

Question 2:

(a) Joint Probability table:

 Uses Social Media and Other Websites to Voice Opinions About Television Programs (A) Doesn’t Use Social Media and Other Websites to Voice Opinions About Television Programs (B) Total Female 395/1364 = 0.2896 291/1364 = 0.2133 686/1364 = 0.5029 Male 323/1364 = 0.2368 355/1364 = 0.2603 678/1364 = 0.4971 Total 718/1364 = 0.5264 646/1364 = 0.4736 1

(b) P(Female) = 686/1364 = 0.5029

(c) Let a respondent uses social media and other websites to voice opinions about television programs given the respondent is female is event A.

Then P(A/Female) = 395/686 = 0.5758

(d) P(A) = 0.5264

P(A/F) ? P(A)

So, F and A are not independent.

Week 5:

Question (3)

µ = 20,000 and SD = 8000

(a) P (at least 30400) = P (x ? 30400)

= P (z ? (30400 – 20000)/8000) = P (z ? 1.3)

= 0.5 – 0.4032 = 0.0968

(b) P (x = 30400) = 0

As normal distribution represents an area, getting a value exactly under the normal curve is 0.

(c) P (low income tax break)

= P (x < 15600>

= P (z < (15600 – 20000)/8000)

= P (z < -0.55)

= 0.5 – 0.2088 = 0.2912

(d) P (x > 32240)

= P (z > (32240 – 20000)/8000)

= P (z > 1.53)

= 0.063

Now let number of students graduated is x

So 189 = 0.063*x

X = 189/0.063 = 3000

Week 6:

Question 3.

(a) Sample size n = 90.

Standard error = 100/?90 = 10.541

Now P (502 – 10 < x>

= P (492 < x>

= P (-10/10.541 < z>

= P (-0.95 < z>

= 2 * 0.3289 = 0.6578

(b) Standard error = 100/?90 = 10.541

Now P (515 – 10 < x>

= P (492 < x>

= P (-10/10.541 < z>

= P (-0.95 < z>

= 2 * 0.3289 = 0.6578

There is no any difference in both probabilities as standard error is same for both.

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