You are given the following information obtained from a random sample of 6

observations. Assume the population has a normal distribution.

13 14 17 14 17 15

a. What is the point estimate of µ?
b. Construct an 80% confidence interval for µ.
c. Construct a 98% confidence interval for µ.
d. Discuss why the 80% and 98% confidence intervals are different.

a.( 13 +14 +17 +14 +17 +15)/6 = 90/6 = 15

b. 15 -+ 1.28 * 1.67/ sqrt(6))
= [ 14.127, 15.873]

c. 15 -+ 2.33* 1.67/ sqrt(6))
= [13.411, 16.589]

a. To find the point estimate of µ (population mean), we will calculate the sample mean. Add up all the values in the sample and divide by the number of observations:

(13 + 14 + 17 + 14 + 17 + 15) / 6 = 90 / 6 = 15

Therefore, the point estimate of µ is 15.

b. To construct an 80% confidence interval for µ, we will use the t-distribution with n-1 degrees of freedom. Since we have 6 observations, the degrees of freedom will be 6 - 1 = 5.

First, we need to calculate the standard error of the mean (SE). The formula for the SE is:

SE = standard deviation / √n

To find the standard deviation, we calculate the sample standard deviation:

Step 1: Find the mean of the sample: (13 + 14 + 17 + 14 + 17 + 15) / 6 = 90 / 6 = 15
Step 2: Calculate the difference between each observation and the sample mean: (13-15), (14-15), (17-15), (14-15), (17-15), (15-15) = -2, -1, 2, -1, 2, 0
Step 3: Square each difference: (-2)^2, (-1)^2, (2)^2, (-1)^2, (2)^2, (0)^2 = 4, 1, 4, 1, 4, 0
Step 4: Find the sum of the squared differences: 4 + 1 + 4 + 1 + 4 + 0 = 14
Step 5: Divide the sum by n-1 (6-1=5): 14 / 5 = 2.8
Step 6: Take the square root: √2.8 ≈ 1.67

Now, we can calculate the SE:

SE = 1.67 / √6 ≈ 0.68

Next, we will find the t-value for an 80% confidence interval with 5 degrees of freedom. Using a t-table or calculator, the t-value is approximately 1.476.

Finally, we can construct the confidence interval:

Confidence Interval = sample mean ± (t-value * SE)
CI = 15 ± (1.476 * 0.68)
CI = 15 ± 1.005

The 80% confidence interval for µ is (13.995, 16.005).

c. To construct a 98% confidence interval for µ, we will follow the same steps as above, but this time we need to find the t-value for a 98% confidence interval with 5 degrees of freedom.

Using a t-table or calculator, the t-value is approximately 2.571.

Constructing the confidence interval:

CI = 15 ± (2.571 * 0.68)
CI = 15 ± 1.748

The 98% confidence interval for µ is (13.252, 16.748).

d. The 80% and 98% confidence intervals are different because the level of confidence affects the margin of error. In an 80% confidence interval, there is more room for error, so the interval will be wider. In a 98% confidence interval, there is less room for error, so the interval will be narrower. This relationship between confidence level and interval width is because increasing the level of confidence requires accounting for a larger range of possible sample means.

To answer these questions, we will need to calculate the point estimate of the mean and construct confidence intervals using the formulas.

a. The point estimate of µ (the population mean) is simply the average of the observed sample values. We add up all the values and divide by the total number of observations:

Point Estimate of µ = (13 + 14 + 17 + 14 + 17 + 15) / 6 = 90 / 6 = 15

b. To construct an 80% confidence interval for µ, we can use the formula:

Confidence Interval = Point Estimate ± Margin of Error

The margin of error depends on the desired confidence level and the standard deviation of the population (which we don't have). However, since the sample size is small (n ≤ 30), we can estimate the population standard deviation using the sample standard deviation.

In this case, the sample standard deviation is calculated by finding the square root of the variance.

Variance = [(13-15)² + (14-15)² + (17-15)² + (14-15)² + (17-15)² + (15-15)²] / (6-1)
= [4 + 1 + 4 + 1 + 4 + 0] / 5
= 14 / 5
= 2.8

Sample Standard Deviation = √(Variance) = √(2.8) ≈ 1.67

Next, we calculate the margin of error using the sample standard deviation and the t-value associated with the desired confidence level (80% confidence level corresponds to a t-value of 1.386 for a sample size of 6 - 1 = 5):

Margin of Error = t-value * (Sample Standard Deviation / √n)

Margin of Error = 1.386 * (1.67 / √6) ≈ 1.386 * (1.67 / 2.449) ≈ 0.945

Finally, we can construct the 80% confidence interval:

80% Confidence Interval = 15 ± 0.945
= (14.055, 15.945)

c. Similarly, to construct a 98% confidence interval, we will use the same process but with a different t-value. For a sample size of 5, the t-value corresponding to a 98% confidence level is approximately 4.604.

Margin of Error = 4.604 * (1.67 / √6) ≈ 4.604 * (1.67 / 2.449) ≈ 3.14

98% Confidence Interval = 15 ± 3.14
= (11.86, 18.14)

d. The 80% and 98% confidence intervals are different because they are calculated based on different t-values. The t-value is determined based on the desired confidence level, and higher confidence levels require larger t-values. As a result, the margin of error increases with higher confidence levels, resulting in wider confidence intervals. Therefore, the 98% confidence interval is wider than the 80% confidence interval, indicating a higher level of precision in estimating the population mean.