Question

A simple random sample of 50 items from a population with = 6 resulted in a sample mean of 32. If required, round your answers to two decimal places.

a. Provide a 90% confidence interval for the population mean. _____ to _____
b. Provide a 95% confidence interval for the population mean. _____ to _____
c. Provide a 99% confidence interval for the population mean. _____ to _____

Answer 1

a. The 90% confidence interval is (30.14, 33.86). b. The 95% confidence interval is (29.89, 34.11). c. The 99% confidence interval is (29.18, 34.82).

Step-by-step explanation:

a. To calculate the confidence interval, we will use the formula: CI = sample mean ± (t-value * standard error). The standard error can be calculated as standard deviation / √n, where n is the sample size. The t-value can be found using the degrees of freedom (n-1) and the desired confidence level. For a 90% confidence interval with 49 degrees of freedom, the t-value is approximately 1.674. Substituting the values, we get CI = 32 ± (1.674 * (6 / √50)). This simplifies to CI = 32 ± 1.861. Rounding to two decimal places, the 90% confidence interval is (30.14, 33.86).

b. For a 95% confidence interval, the t-value with 49 degrees of freedom is approximately 2.009. Using the same formula as in part a, CI = 32 ± (2.009 * (6 / √50)). This simplifies to CI = 32 ± 2.106. Rounding to two decimal places, the 95% confidence interval is (29.89, 34.11).

c. For a 99% confidence interval, the t-value with 49 degrees of freedom is approximately 2.681. Using the same formula as in part a, CI = 32 ± (2.681 * (6 / √50)). This simplifies to CI = 32 ± 2.819. Rounding to two decimal places, the 99% confidence interval is (29.18, 34.82).