Question

A random sample of 27 fields of rye has a mean yield of 43.1 bushels per acre and standard deviation of 5.31 bushels per acre. Determine the 80% confidence interval for the true mean yield. Assume the population is approximately normal. Step 2 of 2 : Construct the 80% confidence interval. Round your answer to one decimal place.

Answer 1

The 80% confidence interval = (41.8, 44.4)

Explanation:

From the above question, the random number of samples is 27, this sample size is below 30, hence it is small.

To solve for this, we have use t score instead of confidence interval.

Degrees of freedom = 27 - 1 = 26

t score for 80% confidence interval = 1.315

The 80% confidence interval = Mean ± t score × standard deviation/√n

Mean = 43.1 bushels

Standard deviation = 5.31

Hence,

43.1 ± 1.315 × 5.31/√27

43.1 ± 1.3438116191

Hence:

43.1 - 1.3438116191

= 41.756188381

≈ 41.8

43.1 + 1.3438116191

= 44.443811619

Approximately ≈ 44.4

Therefore, the 80% confidence interval = (41.8, 44.4)