Question

Consider a normal population distribution with the value of σ known. (a) What is the confidence level for the interval x ± 2.88σ/ n ? (Round your answer to one decimal place.) Incorrect: Your answer is incorrect. % (b) What is the confidence level for the interval x ± 1.43σ/ n ? (Round your answer to one decimal place.) % (c) What value of zα/2 in the CI formula below results in a confidence level of 99.7%? (Round your answer to two decimal places.) x − zα/2 · σ n , x + zα/2 · σ n zα/2 = (d) Answer the question posed in part (c) for a confidence level of 75%. (Round your answer to two decimal places.) zα/2 =

Answer 1

The confidence levels for the z-scores 2.88 and 1.43 are approximately 99.6% and 84.8%, respectively. For a 99.7% confidence interval, a z-score of about ±3.00 is used, and for a 75% confidence interval, the z-score is approximately ±1.15.

Step-by-step explanation:

To calculate the confidence level for a given interval in a normal distribution where the standard deviation (σ) is known, you use a z-score, which corresponds to the specified confidence level. The z-score represents the number of standard deviations you are from the mean. To find these, we refer to the standard normal distribution (z-distribution) tables or by using statistical software.

(a) If x ± 2.88σ/√n gives us the interval, then we're dealing with a z-score of 2.88. By consulting a z-table, we can find that this z-score corresponds to a confidence level that is approximately 99.6% (rounded to one decimal place).

(b) For x ± 1.43σ/√n, a z-score of 1.43 points to a confidence level close to 84.8% (rounded to one decimal place).

(c) To get a 99.7% confidence level, we would be looking for a z-score such that the area between -z and +z under the standard normal curve is 0.997. This value is typically around ± 3.00 (as per the empirical rule).

(d) For a 75% confidence level, we need a smaller z-score, which would be approximately ± 1.15.