Question

A random sample of 25 fields of rye has a mean yield of 31.2 bushels per acre and standard deviation of 6.99 bushels per acre. Determine the 80% 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:

Critical value is 1.318 and the confidence interval is 29.357 to  33.042

Step-by-step explanation:

n = 25

x = 31.2

s = 6.99

Degree of freedom = n-1= 25-1 =24

Confidence level = 0.8

So, α = 1- 0.8= 0.2

t critical = $t_{\frac{\alpha}{2},df}=t_{\frac{0.2}{2},24}=t_{0.10,24}=1.318$

Formula of confidence interval : $\bar{x}-t_{\frac{\alpha}{2},df} \times \frac{s}{\sqrt{n}}$ to  $\bar{x}+t_{\frac{\alpha}{2},df} \times \frac{s}{\sqrt{n}}$

Confidence interval : $31.2-1.318 \times \frac{6.99}{\sqrt{25}}$ to  $31.2+1.318 \times \frac{6.99}{\sqrt{25}}$

Confidence interval : $29.357$ to  $33.042$

Hence critical value is 1.318 and the confidence interval is 29.357 to  33.042