Final answer:
The 95% confidence interval for the population mean μ, given the sample mean of 26.1, standard deviation of 2.7, and sample size of 88, is calculated to be approximately (25.528, 26.672).
Step-by-step explanation:
To calculate the 95% confidence interval for the population mean μ, we will assume that the population is normally distributed or that the sample size is large enough for the Central Limit Theorem to apply. Given a sample mean (μ) of 26.1, a standard deviation (s) of 2.7, and a sample size (n) of 88, we first need to determine the standard error of the mean (SEM), which is s divided by the square root of n.
SEM = σ/√n = 2.7/√88 = 0.288
Next, because the sample size is above 30, we can use the Student's t-distribution for the confidence interval. To find the appropriate t-value for a 95% confidence level and 87 degrees of freedom (n-1), we can use a t-distribution table or a calculator. For this example, let's assume the t-value is approximately 1.989. The confidence interval is given by the formula:
μ ± (t * SEM)
Therefore, the 95% confidence interval for μ is:
26.1 ± (1.989 * 0.288) = (26.1 ± 0.572)
This results in a confidence interval of (25.528, 26.672).
The confidence interval provides a range where we expect the true population mean to fall in 95% of similar sample cases.