Course code 4485 solved Assignment

Bs Level Solve Assignment Course code 4485 spring 2023.Course name Introduction to Statistics Assignment no2

Level: BS

ASSIGNMENT No. 2

Q1:Discuss the different measures of dispersion. Also indicate their merits and demerits. Measures of Dispersion

Measures of dispersion are statistical measures that help us understand the spread or variability of data in a dataset. They complement measures of central tendency by providing insights into how the data points deviate from the average. Let’s discuss different measures of dispersion and their merits and demerits.

1. Range:
   1. Merits: Simple and easy to understand, provides a quick glimpse of data spread.
   1. Demerits: Sensitive to outliers, ignores the distribution of data between the minimum and maximum values.
2. Mean Absolute Deviation (MAD):
   1. Merits: Provides a robust measure of dispersion, less affected by extreme values compared to the range.
   1. Demerits: Requires additional computational steps, less commonly used in practice.
3. Variance:
   1. Merits: Takes into account the squared differences, giving more weight to extreme values, frequently used in statistical calculations.
   1. Demerits: Sensitive to outliers, the squared unit makes interpretation more challenging.
4. Standard Deviation:
   1. Merits: Clear and intuitive interpretation, used in various statistical techniques.
   1. Demerits: Sensitive to outliers, can be affected by the choice of measurement units.
5. Coefficient of Variation (CV):
   1. Merits: Provides a relative measure of variation, suitable for comparing datasets with different scales.
   1. Demerits: Inappropriate for datasets with a mean close to zero.

(b) Calculation of Variance and Coefficient of Variation

Given the following data:

Groups	35-39	40-44	45-49	50-54	55-59	60-64	65-69
f	13	15	17	28	12	10	5

To calculate the variance and coefficient of variation, we need to follow these steps:

Step 1: Calculate the mean (X̄). Step 2: Calculate the squared difference between each data point and the mean. Step 3: Sum up the squared differences. Step 4: Divide the sum by the total number of data points (n) to obtain the variance (σ^2). Step 5: Calculate the standard deviation (σ) by taking the square root of the variance. Step 6: Calculate the coefficient of variation (CV) by dividing the standard deviation by the mean and multiplying by 100.

Step 1: Calculate the mean (X̄)

X̄ = (35-39)*13 + (40-44)*15 + (45-49)*17 + (50-54)*28 + (55-59)*12 + (60-64)*10 + (65-69)*5 = (455 + 630 + 765 + 1400 + 660 + 640 + 325) / 100 = 5255 / 100 = 52.55

Step 2: Calculate the squared difference between each data point and the mean

Groups	Data Point (X)	(X – X̄)	(X – X̄)^2
35-39	13	-39.55	1564.9025
40-44	15	-37.55	1411.5025
45-49	17	-35.55	1264.9025
50-54	28	-24.55	603.7025
55-59	12	-40.55	1644.8025
60-64	10	-42.55	1808.5025
65-69	5	-47.55	2258.9025

Step 3: Sum up the squared differences

Σ (X – X̄)^2 = 1564.9025 + 1411.5025 + 1264.9025 + 603.7025 + 1644.8025 + 1808.5025 + 2258.9025 = 11557.3200

Step 4: Calculate the variance (σ^2)

σ^2 = Σ (X – X̄)^2 / n = 11557.3200 / 100 = 115.5732

Step 5: Calculate the standard deviation (σ)

σ = √(σ^2) = √(115.5732) ≈ 10.7517

Step 6: Calculate the coefficient of variation (CV)

CV = (σ / X̄) * 100 = (10.7517 / 52.55) * 100 ≈ 20.44%

2. Linear Regression Model and Assumptions

(a) What is a Linear Regression Model?

A linear regression model is a statistical technique used to model the relationship between a dependent variable (Y) and one or more independent variables (X). The model assumes that the relationship between the variables is linear, and it aims to find the best-fit line that minimizes the sum of squared residuals.

Assumptions Underlying the Linear Regression Model

For a linear regression model to be valid, several assumptions must be met:

1. Linearity: The relationship between the dependent variable and the independent variable(s) is linear.
2. Independence: The observations are independent of each other, meaning that one data point does not influence another.
3. Homoscedasticity: The variance of the residuals is constant across all levels of the independent variable(s). In simpler terms, the spread of the data points around the regression line should be consistent.
4. Normality: The residuals, which are the differences between the observed values and the predicted values, follow a normal distribution.

(b) Calculating the Correlation Coefficient between X and Y

Given: X = [78, 89, 97, 69, 59, 79, 68, 61] Y = [125, 137, 156, 112, 107, 136, 123, 108]

Step 1: Calculate the means of X and Y.

X̄ = (78 + 89 + 97 + 69 + 59 + 79 + 68 + 61) / 8 = 702 / 8 = 87.75

Ȳ = (125 + 137 + 156 + 112 + 107 + 136 + 123 + 108) / 8 = 1004 / 8 = 125.50

Step 2: Calculate the differences between each data point and the mean for both X and Y, as well as their squared differences.

Xi	Yi	(Xi – X̄)	(Yi – Ȳ)	(Xi – X̄)^2	(Yi – Ȳ)^2
78	125	-9.75	-0.50	95.0625	0.25
89	137	1.25	11.50	1.5625	132.25
97	156	9.25	30.50	85.5625	930.25
69	112	-18.75	-13.50	351.5625	182.25
59	107	-28.75	-18.50	828.125	342.25
79	136	-8.75	10.50	76.5625	110.25
68	123	-19.75	-2.50	390.0625	6.25
61	108	-26.75	-17.50	715.5625	306.25

Step 3: Calculate the sum of squared differences for both X and Y.

Σ (Xi – X̄)^2 = 95.0625 + 1.5625 + 85.5625 + 351.5625 + 828.125 + 76.5625 + 390.0625 + 715.5625 = 2544.0000

Σ (Yi – Ȳ)^2 = 0.25 + 132.25 + 930.25 + 182.25 + 342.25 + 110.25 + 6.25 + 306.25 = 2109.0000

Step 4: Calculate the correlation coefficient (r).

r = Σ [(Xi – X̄) * (Yi – Ȳ)] / √[Σ (Xi – X̄)^2 * Σ (Yi – Ȳ)^2] = (-9.75 * -0.50 + 1.25 * 11.50 + 9.25 * 30.50 + -18.75 * -13.50 + -28.75 * -18.50 + -8.75 * 10.50 + -19.75 * -2.50 + -26.75 * -17.50) / √(2544.0000 * 2109.0000) = (4.875 + 14.375 + 281.875 + 253.125 + 532.625 + 91.875 + 49.375 + 468.125) / √(5350656) ≈ 0.8612

Step 5: Interpret the correlation coefficient (r).

The correlation coefficient (r) between X and Y is approximately 0.8612. It indicates a strong positive linear relationship between the two variables, meaning that as the values of X increase, the values of Y tend to increase as well.

3. Definitions and Calculating Probability

(a) Define the Terms

1. Experiment: An experiment is an activity or process that produces a well-defined outcome. For example, rolling a die is an experiment with possible outcomes of 1, 2, 3, 4, 5, or 6.
2. Outcome: An outcome is a specific result that occurs as a result of an experiment. Each experiment can have one or more outcomes.
3. Event: An event is a set of one or more outcomes of an experiment. Events can be simple (a single outcome) or compound (more than one outcome).
4. Sample Space: The sample space is the set of all possible outcomes of an experiment. It includes all possible results, and every outcome in the sample space should be mutually exclusive.
5. Simple Event: A simple event is an event with only one outcome. For example, getting a 4 when rolling a die is a simple event.
6. Compound Event: A compound event is an event with two or more outcomes. For example, getting an even number (2, 4, or 6) when rolling a die is a compound event.
7. Impossible Event: An impossible event is an event that cannot occur under any circumstances. Its probability is zero.
8. Sure Event: A sure event is an event that is guaranteed to happen. Its probability is one.
9. Mutually Exclusive Events: Mutually exclusive events are events that cannot occur simultaneously. If one event happens, the other cannot happen at the same time.

(b) Calculating the Probability of Obtaining at Least One 6

To calculate the probability of obtaining at least one 6 when rolling a die, we can use the concept of complementary probability.

The probability of NOT getting a 6 in a single roll is 1 – (1/6) = 5/6.

To find the probability of not getting a 6 in n rolls, we raise the probability of not getting a 6 (5/6) to the power of n.

Let P(X) be the probability of getting at least one 6 in n rolls. Then, the probability of not getting a 6 in n rolls is 1 – P(X).

We want the probability of obtaining at least one 6 to be at least 0.99.

1 – P(X) ≥ 0.99

Solving for P(X):

P(X) ≤ 1 – 0.99

P(X) ≤ 0.01

Now we can find the value of n using the formula:

(5/6)^n ≤ 0.01

Taking the logarithm of both sides:

n * log(5/6) ≤ log(0.01)

n ≥ log(0.01) / log(5/6)

n ≥ -2.0 / -0.1823

n ≥ 10.98

Since we cannot have a fraction of a roll, we must round up to the nearest whole number.

Therefore, to ensure that the probability of obtaining at least one 6 is at least 0.99, we need to throw the dice at least 11 times.

(c) Probability of More Men Chosen Than Women

Let’s denote the probability of choosing a man as P(M) and the probability of choosing a woman as P(W).

P(M) = Number of men / Total number of people = 6 / 14 = 3 / 7 P(W) = Number of women / Total number of people = 8 / 14 = 4 / 7

Now, let’s calculate the probability of selecting more men than women when choosing 5 people at random.

The number of ways to choose 5 people out of 14 is given by the binomial coefficient C(14, 5).

Number of ways to choose 5 men and 0 women: C(6, 5) * C(8, 0) = 6 * 1 = 6 Number of ways to choose 4 men and 1 woman: C(6, 4) * C(8, 1) = 15 * 8 = 120 Number of ways to choose 3 men and 2 women: C(6, 3) * C(8, 2) = 20 * 28 = 560 Number of ways to choose 2 men and 3 women: C(6, 2) * C(8, 3) = 15 * 56 = 840 Number of ways to choose 1 man and 4 women: C(6, 1) * C(8, 4) = 6 * 70 = 420 Number of ways to choose 0 men and 5 women: C(6, 0) * C(8, 5) = 1 * 56 = 56

Total number of ways to choose 5 people from the group = C(14, 5) = 2002

Now, let’s calculate the probability of selecting more men than women:

P(More men than women) = (Number of favorable outcomes) / (Total number of outcomes) = (6 + 120 + 560 + 840 + 420) / 2002 = 1946 / 2002 ≈ 0.971

Thus, the probability of selecting more men than women when choosing 5 people at random from the group is approximately 0.971, or 97.1%.

4. Common Types of Sampling Techniques and Their Advantages and Disadvantages

(a) Common Types of Sampling Techniques

1. Simple Random Sampling:
   1. Advantages: Easy to implement, ensures every individual has an equal chance of being selected.
   1. Disadvantages: May not be representative of the entire population if the sample size is small, time-consuming for large populations.
2. Stratified Sampling:
   1. Advantages: Ensures representation from different subgroups, reduces sampling error, allows for better estimation within subgroups.
   1. Disadvantages: Requires prior knowledge of population characteristics, can be complex to implement.
3. Systematic Sampling:
   1. Advantages: Simple and easy to implement, ensures coverage of the entire population.
   1. Disadvantages: May introduce bias if there is a pattern in the population, less flexible in selecting specific samples.
4. Cluster Sampling:
   1. Advantages: Cost-effective for large populations, convenient for geographically dispersed populations.
   1. Disadvantages: Increased variability within clusters, potential for less precision compared to other methods.

(b) Sampling Distribution and Properties of the Sampling Distribution of the Means

Sampling Distribution: A sampling distribution is the distribution of a statistic (such as the mean) obtained from multiple samples of the same size drawn from the same population. It helps us understand the variability of the statistic and make inferences about the population parameter.

Properties of the Sampling Distribution of the Means:

1. Central Limit Theorem: Regardless of the shape of the population distribution, the sampling distribution of the mean approaches a normal distribution as the sample size increases.
2. Mean of Sampling Distribution: The mean of the sampling distribution of the means is equal to the population mean.
3. Standard Deviation of Sampling Distribution: The standard deviation of the sampling distribution of the means, also known as the standard error, is equal to the population standard deviation divided by the square root of the sample size.
4. Sample Size: Larger sample sizes lead to a narrower sampling distribution and more precise estimates.
5. Independence: Samples must be drawn independently and randomly from the population.

5. Testing Hypotheses on Proportions

(a) Hypothesis Testing on Proportions

Hypothesis testing on proportions is used to determine if there is a significant difference between the proportion observed in a sample and the proportion expected in the population. The steps involved in testing hypotheses on proportions are as follows:

1. Formulate Hypotheses: Define the null hypothesis (H0) and the alternative hypothesis (Ha) based on the research question.
2. Select the Significance Level: Determine the significance level (α) which represents the probability of making a Type I error (rejecting H0 when it is true).
3. Collect Data and Calculate Sample Proportion: Gather the data and calculate the sample proportion (p) from the sample.
4. Calculate Test Statistic: Compute the test statistic based on the sample proportion, population proportion, and sample size.
5. Determine Critical Region: Find the critical region based on the test statistic and significance level.
6. Make Decision: Compare the test statistic to the critical region to make a decision to either reject or fail to reject the null hypothesis.
7. Interpret Results: Interpret the results of the hypothesis test and draw conclusions about the population proportion.

(b) Testing Hypothesis on Association Between Total Income and Television Ownership

To test the hypothesis that there is no association between total income and television ownership in the three groups (A, B, and C), we can use a chi-square test for independence.

The null hypothesis (H0) assumes no association, and the alternative hypothesis (Ha) assumes an association between the two variables.

We compare the expected frequencies (based on independence) to the observed frequencies in the table to calculate the chi-square test statistic.

The degrees of freedom (df) for a 3×3 table are calculated as (number of rows – 1) * (number of columns – 1) = (3-1) * (3-1) = 2.

Next, we compare the calculated chi-square test statistic to the critical value of the chi-square distribution with 2 degrees of freedom at the chosen significance level.

If the calculated chi-square value is greater than the critical value, we reject the null hypothesis, indicating that there is an association between total income and television ownership. Otherwise, we fail to reject the null hypothesis, suggesting no significant association.