92.6k views
0 votes
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.

User Ajevic
by
8.3k points

1 Answer

4 votes

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.

User Sergey Grigoriev
by
7.5k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories