Final answer:
To construct 68.26% and 97% confidence intervals for the population mean, we use the sample mean of 128 ounces, the known population standard deviation of 1.7 ounces, the sample size, and the appropriate Z-scores for the specified confidence levels.
Step-by-step explanation:
To answer the student's question, we need to construct confidence intervals for the population mean weight of the cans of coffee using the sample mean, population standard deviation, and sample size provided.
For part (a), constructing a 68.26% confidence interval, we use the Z-score associated with the desired confidence level. Since the standard normal distribution is symmetric, a 68.26% confidence interval corresponds to ±1 standard deviation from the mean in a normal distribution. The formula for the confidence interval is given by:
Confidence Interval = μ ± (Z * (σ/√n))
Using the values provided: μ = 128 ounces, σ = 1.7 ounces, Z = 1 (for 68.26% confidence interval), and n = 196, the confidence interval is:
128 ± (1 * (1.7/√196))
The confidence interval can then be calculated and rounded to two decimal places.
For part (b), constructing a 97% confidence interval, we will look up the Z-score that corresponds to 97% confidence level, which is approximately 2.17. We then use the same formula and input the Z-score to calculate the interval:
128 ± (2.17 * (1.7/√196))
Again, the result is rounded to two decimal places to provide the student with the desired confidence interval.
It's important to note that these intervals are based on the assumption that the sampling distribution of the mean is normal or nearly normal, which is generally a reasonable assumption for large sample sizes due to the Central Limit Theorem.