Estimating the Mean Weight of Trucks on I-475

Explanation

Interval estimate can be found from x_bar +/-z*s/(n^0.5). Here we have n = 49, x_bar = 15.8 and z = 1.96 (for 95%) and s = 3.8. Therefore 15.8 +/- 1.96*3.8/7 and we get 14.76 and 16.84.

To calculate the 95 percent confidence interval for the population mean, we'll use the following formula:

Confidence Interval = Sample Mean ± (Critical Value × Standard Error)

1. Sample Mean: The sample mean is given as 15.8 tons.

2. Standard Deviation of the Sample: The standard deviation of the sample is given as 3.8 tons.

3. Sample Size: The sample size is 49 trucks.

4. Critical Value: The critical value is obtained from the t-distribution table based on the desired confidence level and degrees of freedom. For a 95 percent confidence level with a sample size of 49, the degrees of freedom (df) is (n - 1) = (49 - 1) = 48. Looking up the critical value in the t-distribution table, the value is approximately 2.01.

5. Standard Error: The standard error represents the standard deviation of the sampling distribution of the sample mean. It is calculated as:

Standard Error = Standard Deviation / √(Sample Size)

Standard Error = 3.8 / √(49)

Standard Error ≈ 0.542

Now, we can calculate the confidence interval:

Confidence Interval = 15.8 ± (2.01 × 0.542)

Confidence Interval = 15.8 ± 1.09

Lower Limit = 15.8 - 1.09 ≈ 14.71 Upper Limit = 15.8 + 1.09 ≈ 16.89

Therefore, the 95 percent confidence interval for the population mean weight of trucks traveling on the section of I-475 is approximately 14.71 to 16.89 tons.

The correct answer is A. 14.7 and 16.9.