Suppose you do a study of meditation to determine how effective it is in relieving pain. Sensory rates are measured for the sample of 20 subjects. The data yields a mean of 10.3 and a standard deviation of 2.0. Construct a 90% confidence interval for the mean sensory rate for the population from which you took the sample. Assume that population has a normal distribution of sensory rates.

Answer 1

To construct a confidence interval for the mean sensory rate for the population, we can use the formula:

Confidence Interval = sample mean ± (critical value * standard error)

First, calculate the standard error:
Standard Error = standard deviation / √n
Standard Error = 2.0 / √20
Standard Error ≈ 0.447

Next, determine the critical value for a 90% confidence level. Since the sample size is less than 30, we can use a t-distribution. The degrees of freedom for a sample size of 20 are 20 - 1 = 19. Using a t-table or calculator, we find the critical value for a 90% confidence level with 19 degrees of freedom is approximately 1.729.

Finally, plug in the values to calculate the confidence interval:
Confidence Interval = 10.3 ± (1.729 * 0.447)

Lower bound = 10.3 - (1.729 * 0.447) ≈ 9.502
Upper bound = 10.3 + (1.729 * 0.447) ≈ 11.098

Therefore, the 90% confidence interval for the mean sensory rate for the population is approximately (9.502, 11.098).