Question

Every confidence interval we've talked about has the form "statistics plus or minus a quantile times a standard deviation". Suppose the population variance is unknown and the sample standard deviation is 1.88. Compute the following for a 80% confidence interval with a sample of size 6 (we can assume the sampling distribution of the sample average is normal): (Use 3 decimal places)

(a) The "quantile" part for the confidence interval.

(b) The "standard deviation" part for the confidence interval.

Answer 1

a) The "quantile" part for the confidence interval for an 80% confidence interval and a sample size of 6, with the degrees of freedom as 5, the t-value is approximately 1.249.

b) The standard deviation part for the confidence interval of 80% is 0.769.

The population variance = unknown

Sample standard deviation = 1.88

Confidence interval = 80%

Sample size = 6

Degree of freedom = 5 (6 - 1)

b) The "standard deviation" part for the confidence interval: The standard deviation part for the confidence interval is calculated as the sample standard deviation divided by the square root of the sample size.

Standard deviation part = 1.88 / √6 ≈ 0.769 (rounded to 3 decimal places)

Thus, we can conclude that for an 80% confidence interval with a sample of size 6, a) The "quantile" part is approximately 1.249, and b) The "standard deviation" part is approximately 0.769.