Solved Problems

Problem 1: Point Estimation (Mean)

A random sample of 10 students from a college class scored the following marks in an exam: 85, 78, 92, 80, 65, 90, 72, 88, 95, 83. Estimate the average score for the entire class.

Solution:

We can use the sample mean (average) as a point estimate for the population mean.

Sample Mean (x̄) = Σ(x_i) / n

where:

Σ (sigma) represents the sum

x_i represents the individual score of each student (i = 1 to 10)

n is the sample size (n = 10)

Calculating the sum of scores: Σ(xi)= 833

Therefore, Sample Mean (x̄) = 833 / 10 = 83.3

Interpretation: Based on this sample, we can estimate the average score for the entire class to be around 83.3.

Problem 2: Point Estimation (Proportion)

A survey of 200 customers at a grocery store revealed that 120 prefer brand A. Estimate the proportion of all customers who prefer brand A.

Solution:

We can use the sample proportion (p̂) as a point estimate for the population proportion (p).

Sample Proportion (p̂) = x / n

Therefore, Sample Proportion (p̂) = 120 / 200 = 0.6

Interpretation: Based on the survey, we can estimate that around 60% of all customers at the store prefer brand A.

Problem 3: Interval Estimation (Confidence Interval for Mean)

Continuing from Problem 1, suppose we want to estimate the average score for the entire class with a 95% confidence level.

Solution:

When a 95% confidence level with 9 degrees of freedom is used, the t-critical value (t*) must be calculated in order to express a mean value using a confidence interval.

Statistical software or internet tables can be used to find T*.

Assuming t* = 2.262 (for a 95% CI with 9 degrees of freedom), we can calculate the margin of error (ME):

ME = (t*) × (standard deviation / √n)

Since we don’t have the population standard deviation, we can estimate it using the sample standard deviation (s). You’ll need to calculate the sample standard deviation for the scores (around 5.8).

Example Calculation (assuming s = 5.8):

ME = (2.262) × (5.8 / √10) ≈ 4.3

Now, we can construct the confidence interval:

CI = Sample Mean (x̄) ± ME

CI = 83.3 ± 4.3

Therefore, the 95% confidence interval for the mean score is approximately (79, 87.6).

Interpretation: We can be 95% confident that the true average score for the entire class lies somewhere between 79 and 87.6.

Problem 4: Interval Estimation (Confidence Interval for Proportion)

Following the grocery store survey (Problem 2), construct a 90% confidence interval for the proportion of customers who prefer brand A.

Solution:

Similar to the previous problem, we can use the confidence interval for a proportion. We’ll need to find the z-critical value (z*) for a 90% confidence level (typically 1.645).

Calculation:

CI = p̂ ± (z*) × √(p̂(1-p̂) / n)

CI = 0.6 ± (1.645) × √(0.6 × (1-0.6) / 200) ≈ 0.6 ± 0.05

Therefore, the 90% confidence interval for the proportion who prefer brand A is approximately (0.55, 0.65).

Interpretation: We can be 90% confident that the true proportion of customers who prefer brand A in the entire population falls between 55% and 65%.

Estimation is a technique for calculating information about a bigger group from a smaller sample, and statistics are crucial to analyzing data. For instance, the average age of a city’s population may be obtained by taking the age of a sample of 1,000 residents. While estimates aren’t perfect, they are typically trustworthy enough to be of value.

In this article, we examine the significance of statistics, their function in the interpretation of data, and how efficient data analysis leads to the making of decisions based on accurate information.