Statistical Analysis Techniques: From Mean and Standard Deviation to Hypothesis Testing in Business Scenarios

Questions:

1. Compute the mean and standard deviation for below:

a) Sample data: 1,2 and 3

b) Population data: 10,20 and 30

2. A researcher conducted an experiment with a sample size of 36. The sample standard deviation was found to be 12. What is the standard error in this case?

3. There are two stocks X and Y. You have invested 50% of your wealth in each of these stocks. The covariance of the two stocks is 0.1. The variance of stock X is 0.1 and that of stock Y is 0.2. What is the correlation?

4. Find a symmetrically distributed interval around µ that will include 95% of the sample means when µ = 368, σ = 15, and n = 25.

5. You are interested in the average response time of customer support calls at a telecommunications company. You randomly select a sample of 100 customer support calls and find that the mean response time is 8 minutes, with a standard deviation of 3 minutes. Calculate a 90% confidence interval for the average response time of customer support calls at the company.

6. Is the average number of sales per month for a new sales strategy significantly different from the average number of sales per month for the previous sales strategy? (Consider alpha = 0.05)

New Sales Strategy Previous Sales Strategy

50 45

55 48

60 50

65 52

70 55

7. A company wants to determine if the average salary of its employees has increased over the past 10 years, taking into account both full-time and part-time employees. They take a random sample of 50 employees, 25 full-time and 25 part-time, to see whether there is a significant change from the average salary of $50,000 ten years ago. Perform Hypothesis testing (one sample t-test) to come to a conclusion. (Consider alpha = 0.05)

Refer: 

Dataset 

Employees salary (in $)

80,000

70,000

77,000

75,000

72,000

76,000

79,000

73,000

81,000

78,000

82,000

71,000

74,000

85,000

83,000

69,000

72,000

75,000

68,000

72,000

82,000

79,000

73,000

81,000

78,000

29,000

32,000

33,000

30,000

27,000

25,000

31,000

28,000

26,000

29,000

30,000

24,000

28,000

26,000

31,000

32,000

29,000

25,000

27,000

33,000

33,000

24,000

28,000

26,000

31,000

Answer: 

Q1. Compute the mean and standard deviation for below:

a) Sample data: 1,2 and 3
b) Population data: 10,20 and 30 

Answer:

a. Sample Data 1, 2, 3

i. Mean (x̄) = μ = (Σx) / n = (1 + 2 + 3) / 3 = 6 / 3 = 2

Mean = 2

ii. Standard deviation (sample data): S = √s^2 = √(Σ(x - x̄)^2) / (n - 1)

(1 - 2)² = 1

(2 - 2)² = 0

(3 - 2)² = 1

Σ(x - x̄)^2 = 1 + 0 + 1 = 2

n -1 = 2

S = √(Σ(x - x̄)^2) / (n - 1) = √2/2 = 1

Sample. Standard Deviation = 1

b. Population Data: 10, 20 and 30:

i. Mean (μ) = (Σx) / n = (10 + 20 + 30) / 3 = 60 / 3 = 20

Mean (μ) = 20

ii. Standard deviation (Population) for 10, 20, 30

σ = √(σ^2) = √(Σ(x - μ)^2) / N
(10 - 20)² = 100
(20 - 20)² = 0
(30 - 20)² = 100
(Σ(x - μ)^2) = 200
N = 3
σ = √200/3 = 8.1649 = 8.16

Population. Standard Deviation = 8.16 

Q2. A researcher conducted an experiment with a sample size of 36. The sample standard deviation was found to be 12. What is the standard error in this case? 

Answer:

Standard error of mean = $\frac{1}{\sqrt{n}} \times \text{standard deviation}$

Q3: There are two stocks X and Y. You have invested 50% of your wealth in each of these stocks. The covariance of the two stocks is 0.1. The variance of stock X is 0.1 and that of stock Y is 0.2. What is the correlation?

Correlation: r = Cov(X, Y) / (σX * σY)

Correlation = Covariance (X, Y) / (Standard Deviation of Stock X * Standard Deviation of Stock Y)

As per provided details:

Covariance = 0.1
Variance of Stock X (σX) = 0.1
Variance of Stock Y (σY) = 0.2

Standard Deviation (Stock X) = √0.1 = 0.316
Standard Deviation (Stock Y) = √0.2 = 0.447

Correlation = 0.1 / (0.316 * 0.447)
≈ 0.1 / 0.141352 = 0.707

Thus, correlation between stock X and stock Y = 0.707.

Q4: Find a symmetrically distributed interval around μ that will include 95% of the sample means when μ = 368, σ = 15, and n = 25.

Answer:

Confidence Interval = μ ± (Z * (σ / √n))

μ (population mean) = 368  

95% confidence level corresponds to a Z-score of approximately 1.96  

σ (population standard deviation) = 15  

n (Sample size) = 25

Standard error = (σ / √n) = 15 / √25 = 15 / 5 = 3  

⇒ Standard error = 3

Confidence Interval = 368 ± (1.96 * (15 / √25))  

= 368 ± (1.96 * 3) = 368 ± 5.88

Thus, the symmetrically distributed interval around μ = 368 ± 5.88 = (362.12, 373.88)

Q5: You are interested in the average response time of customer support calls at a telecommunications company. You randomly select a sample of 100 customer support calls and find that the mean response time is 8 minutes, with a standard deviation of 3 minutes. Calculate a 90% confidence interval for the average response time of customer support calls at the company.

Answer:

Confidence Interval = x̄ ± (Z * (s / √n))

x̄ (sample mean) = 8
Z is the Z-score corresponding to a 90% confidence is 1.645
s (sample standard deviation) = 3
n = 100

Confidence Interval = 8 ± (1.645 * (3 / √100))
Standard Error = 3 / √100 = 3 / 10 = 0.3
Margin of Error = 0.4935

⇒ Confidence Interval = 8 ± (1.645 * (3 / √100)) = 8 ± 0.4935

Confidence Interval = 8 ± 0.4935

Therefore, the average response time of customer support calls at the company should lie between 7.5065 and 8.4935 minutes. (@90% confidence)

Q6: Is the average number of sales per month for a new sales strategy significantly different from the average number of sales per month for the previous sales strategy? (Consider alpha = 0.05) 

Answer: Below is detail from excel for t-Test: Two-Sample Assuming Equal Variances

We can calculate the sample means and sample standard deviations for each group.

New Sales Strategy:

Mean 1 = (50 + 55 + 60 + 65 + 70) / 5 = 60

Standard Deviation (sd1) = 7.07

Previous Sales Strategy:

Mean 2 = (45 + 48 + 50 + 52 + 55) / 5 = 50

Standard Deviation (sd2) = 3.40

t ≈ (60 - 50) / 3.53 ≈ 2.83

Comparing t-value with the critical t-value at alpha 0.05 and (df) = (n1 + n2 - 2).

i.e.: df = 5 + 5 - 2 = 8.

Calculated critical t-value at alpha = 0.05 and df = 8 is 2.306.

Calculated t-value (2.83) is greater than the critical t-value (2.306), thus we reject the null hypothesis.

Therefore, we can observe to have that the average number of sales per month with new sales strategy

is significantly different from the average number of sales per month for the previous sales strategy at a

significance level of 0.05.

Q7. 7. A company wants to determine if the average salary of its employees has increased over the past

10 years, taking into account both full-time and part-time employees. They take a random sample of 50

employees, 25 full-time and 25 part-time, to see whether there is a significant change from the average

salary of $50,000 ten years ago. Perform Hypothesis testing (one sample t-test) to come to a

conclusion. (Consider alpha = 0.05)

Answer:

Hypothesis Setup

Null Hypothesis (H0): The average salary of the employees is equal to $50,000. (H0: μ = $50,000)

Alternative Hypothesis (Ha): The average salary of the employees is greater than $50,000. (Ha: μ > $50,000)

Test Statistics Calculation

Mean salary of the sample: $52,440

Sample variance: $591,108.57

Sample size (n): 50

Hypothesized mean (μ0): $50,000

Degrees of freedom (df): 49

t-Statistic: 0.7096

Step 3: P-Value Calculation

P(T <= t) one-tail = 0.2406

t Critical one-tail (at alpha 0.05) = 1.6766

==Calculation==

Data:

• Sample Mean ($\bar{x}$) = $52,440
• Hypothesized Mean ($\mu_0$) = $50,000
• Sample Variance ($s^2$) = 591,108,571.4
• Number of Observations (n) = 50
• Significance Level ($\alpha$) = 0.05

Formulating Hypotheses

• Null Hypothesis (H0): $\mu = 50,000$ (The average salary is $50,000)
• Alternative Hypothesis (Ha): $\mu > 50,000$ (The average salary is greater than $50,000)

Standard Error of the Mean (SE)

The standard error of the mean is calculated using the formula:

$$SE = \frac{s}{\sqrt{n}}$$

Where:

• $s$ is the standard deviation of the sample, which is the square root of the sample variance.
• $n$ is the number of observations.

First, calculate the sample standard deviation ($\mathrm{s):}$

$$s=\sqrt{591,108,571.4}\approx 24,307.59$$

Now, calculate the standard error:

$$SE = \frac{24,307.59}{\sqrt{50}} \approx \frac{24,307.59}{7.071} \approx 3,436.48$$

t-Statistic

The t-statistic is calculated using the formula:

$$t = \frac{\bar{x} - \mu_0}{SE}$$

Substitute the values:

$$t = \frac{52,440 - 50,000}{3,436.48} \approx \frac{2,440}{3,436.48} \approx 0.71$$

Step 4: Determine the Critical t-Value

Since this is a one-tailed test with $n - 1 = 50 - 1 = 49$ degrees of freedom and $\alpha = 0.05$, the critical t-value can be found using a t-distribution table or calculator. For $df = 49$ and $\alpha = 0.05$:

$$t_{\text{critical}} \approx 1.6766$$

Comparing t-Statistic to Critical t-Value

The calculated t-statistic (0.71) is less than the critical t-value (1.6766).

p-Value

Using the t-distribution, the p-value associated with the t-statistic of 0.71 is approximately 0.2406 (for one-tailed test).

====

Conclusion

Since the p-value (0.2406) is greater than the significance level (alpha = 0.05), we fail to reject the null hypothesis (H0).

There is no significant evidence to suggest that the average salary of the employees has increased from the $50,000 average 10 years ago. The data does not support the claim that the average salary is greater than $50,000.