Question

A random sample of 9 fields of corn has a mean yield of 34.7 bushels per acre and standard deviation of 8.04 bushels per acre. Determine the 98% confidence interval for the true mean yield. Assume the population is approximately normal. Step 1 of 2 : Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places.

Answer 1

Answer:

98% Confidence interval:  (26.939,42.461)

Step-by-step explanation:

We are given the following in the question:

Sample mean, $\bar{x}$ = 34.7 bushels per acre

Sample size, n = 9

Alpha, α = 0.05

Sample standard deviation, s = 8.04 bushels per acre

Degree of freedom = n - 1 = 8

98% Confidence interval:

$\bar{x} \pm t_{critical}\displaystyle\frac{s}{\sqrt{n}}$  

Putting the values, we get,  

$34.7 \pm 2.896(\dfrac{8.04}{\sqrt{9}} )\\\\ = 34.7 \pm 7.761 = (26.939,42.461)$  

is the required confidence interval for the true mean yield.